🧠:Okay, so I need to prove these two identities involving divisor functions. Let me start with part a). The problem states that the sum of the number of divisors from 1 to n, which is τ₁ + τ₂ + τ₃ + … + τₙ, is equal to the sum [n/1] + [n/2] + [n/3] + … + [n/n], where [x] denotes the floor function. Hmm. First, I recall that τₖ, the number of divisors of k, counts how many positive integers divide k. For example, τₖ = 2 if k is prime, since primes have exactly two divisors. But how does summing τ₁ to τₙ relate to these floor functions?Let me think about a small example. Let’s take n=3. Then τ₁=1 (only 1 divides 1), τ₂=2 (1 and 2), τ₃=2 (1 and 3). So the left side is 1 + 2 + 2 = 5. The right side is [3/1] + [3/2] + [3/3] = 3 + 1 + 1 = 5. Okay, it works for n=3. Let me check another one. Maybe n=4. Left side: τ₁=1, τ₂=2, τ₃=2, τ₄=3. Sum is 1+2+2+3=8. Right side: [4/1] + [4/2] + [4/3] + [4/4] = 4 + 2 + 1 + 1 = 8. Okay, that works too. So the identity seems to hold for these examples. Now, how to generalize this? I need to find a connection between counting divisors and the floor functions. Let me think about what each term [n/k] represents. For a fixed k, [n/k] is the number of multiples of k that are less than or equal to n. For example, [n/1] counts all numbers from 1 to n, [n/2] counts multiples of 2 up to n, etc. But how does that relate to the number of divisors? Let me consider the divisor function τₖ. For each number m from 1 to n, τₘ counts how many divisors m has. So, if we sum τ₁ + τ₂ + … + τₙ, we are essentially counting the total number of divisors of all numbers from 1 to n. Wait, but each divisor d of a number m is such that d ≤ m. So, if we consider all pairs (d, m) where d divides m and 1 ≤ m ≤ n, then the sum of τ₁ to τₙ is the total number of such pairs. Alternatively, we can think of this as counting for each divisor d, how many multiples of d are there up to n. But that might be the key. Let me see. If I fix a divisor d, how many numbers m ≤ n are divisible by d? That's exactly [n/d]. So, for each d from 1 to n, [n/d] counts how many multiples of d are there in 1 to n. But each multiple of d corresponds to a divisor d of that multiple. So, if I sum over d from 1 to n the number of multiples of d (which is [n/d]), this should count the total number of divisors for all numbers from 1 to n. Wait, that seems right. Because for each divisor d, every multiple of d (i.e., numbers m = d, 2d, 3d, ..., [n/d]d) will have d as one of their divisors. So, summing [n/d] over d from 1 to n gives the total number of divisors across all numbers from 1 to n. Therefore, the sum τ₁ + τ₂ + ... + τₙ is equal to the sum of [n/1] + [n/2] + ... + [n/n]. That makes sense. So part a) is proven by switching the order of summation: instead of summing over numbers and their divisors, sum over divisors and their multiples. Let me formalize that. Let S = τ₁ + τ₂ + ... + τₙ. Each τₖ is the number of divisors of k, so S is the total number of divisors of all numbers from 1 to n. Alternatively, S can be expressed as the sum over all divisors d of the number of multiples of d up to n. For each d, the number of multiples is [n/d], so summing over d from 1 to n gives the total. Hence, S = Σₐ=1ⁿ [n/a], which is the right-hand side. Therefore, part a) is proven.Now moving to part b). The sum σ₁ + σ₂ + ... + σₙ is supposed to equal [n/1] + 2[n/2] + 3[n/3] + ... + n[n/n], where σₖ is the sum of divisors of k. Let me check with the same examples. Take n=3. Left side: σ₁=1, σ₂=1+2=3, σ₃=1+3=4. Sum is 1+3+4=8. Right side: [3/1] + 2[3/2] + 3[3/3] = 3 + 2*1 + 3*1 = 3 + 2 + 3 = 8. Good. Another example: n=4. Left side: σ₁=1, σ₂=3, σ₃=4, σ₄=1+2+4=7. Sum is 1+3+4+7=15. Right side: [4/1] + 2[4/2] + 3[4/3] +4[4/4] =4 + 2*2 + 3*1 +4*1=4+4+3+4=15. Correct. So, similar idea. Now, instead of counting the number of divisors, we are summing the divisors themselves. So, σₖ is the sum of divisors of k. Then σ₁ + ... + σₙ is the total sum of all divisors of all numbers from 1 to n. Again, we need to relate this to the sum on the right-hand side, which is Σₐ=1ⁿ a[n/a]. Let me think. For part a), each term [n/a] counts the number of multiples of a, which corresponded to counting each divisor a for each multiple. Here, instead of just counting the divisor a, we need to sum the divisors. So, if for each divisor d, instead of just counting how many times d appears (which was [n/d]), we sum the divisor d each time it appears. Wait, but σₖ is the sum of all divisors of k. So the total sum over σ₁ to σₙ is the sum over all divisors of all numbers from 1 to n. Alternatively, for each divisor d, how many times does d appear in the sum? It's equal to the number of multiples of d up to n, which is [n/d]. But if we want to sum d each time it appears, then the total contribution of d to the sum is d * [n/d]. Therefore, summing over all d from 1 to n gives Σₐ=1ⁿ a[n/a]. Wait, that seems to be the case. Let me verify with the example. For n=3, right side is [3/1] + 2[3/2] +3[3/3] =3 +2*1 +3*1=3+2+3=8. Left side: σ₁=1, σ₂=3, σ₃=4, sum=8. Correct. For each divisor d=1, it appears in all numbers from 1 to 3, so [3/1]=3 times, contributing 1*3=3. For d=2, it appears in 2 and 4 (but 4>3), so only [3/2]=1 time, contributing 2*1=2. For d=3, appears only in 3, so [3/3]=1, contributing 3*1=3. So total sum 3+2+3=8. But wait, the sum of σ's is 1+3+4=8. Wait, but 1=1, 3=1+2, 4=1+3. So total sum is (1) + (1+2) + (1+3) = 1+1+2+1+3=8. So each divisor 1 is added three times (from numbers 1,2,3), divisor 2 once (from number 2), divisor 3 once (from number 3). So yes, 1*3 + 2*1 +3*1=8. So the total sum is the sum over d of d multiplied by the number of multiples of d up to n, which is [n/d]. Therefore, the identity holds. Thus, the reasoning is similar to part a), but instead of counting the number of divisors, we are summing the divisors themselves. Therefore, the total sum is the sum over each divisor d, multiplied by the number of times it appears as a divisor (which is [n/d]), and then summed over all d. Hence, Σₖ=1ⁿ σₖ = Σₐ=1ⁿ a[n/a]. Therefore, part b) is proven by switching the order of summation: instead of summing over numbers and their divisors, sum over divisors and their contributions. Each divisor d contributes d to each multiple of d, and there are [n/d] such multiples. So the total contribution is d*[n/d], and summing over all d gives the total sum of divisors. So, both identities are proven by interchanging the order of summation, recognizing that summing over numbers and their divisors is equivalent to summing over divisors and their multiples. For part a) we count the number of divisors, which is equivalent to counting the number of multiples for each divisor. For part b), we sum the divisors, which is equivalent to summing each divisor multiplied by the number of multiples it has up to n. I need to make sure there are no edge cases or mistakes here. Let me check n=1. For part a), τ₁=1, and [1/1]=1, so equality holds. For part b), σ₁=1, and 1*[1/1]=1*1=1, which matches. Another check with n=2. Part a): τ₁ + τ₂ =1+2=3. Right side: [2/1] + [2/2] =2+1=3. Part b): σ₁ + σ₂=1 + (1+2)=4. Right side:1*2 +2*1=2+2=4. Correct. Seems consistent. Therefore, the proofs are valid.

✅:Proof of Part (a):We aim to show that the sum of the number of divisors of all natural numbers from 1 to ( n ), denoted ( tau_1 + tau_2 + cdots + tau_n ), equals the sum ( leftlfloor frac{n}{1} rightrfloor + leftlfloor frac{n}{2} rightrfloor + cdots + leftlfloor frac{n}{n} rightrfloor ).Key Insight: Instead of summing the number of divisors for each number ( m in {1, 2, ldots, n} ), we count how many times each divisor ( d in {1, 2, ldots, n} ) appears across all numbers ( leq n ). For a fixed divisor ( d ), it appears in every multiple of ( d ). The number of multiples of ( d ) up to ( n ) is ( leftlfloor frac{n}{d} rightrfloor ).Formal Proof: The total number of divisors for all numbers ( 1 ) to ( n ) is:[sum_{m=1}^n tau_m = sum_{m=1}^n sum_{d mid m} 1.]By interchanging the order of summation:[sum_{m=1}^n tau_m = sum_{d=1}^n sum_{substack{m=1 d mid m}}^n 1 = sum_{d=1}^n leftlfloor frac{n}{d} rightrfloor.]Thus, ( tau_1 + tau_2 + cdots + tau_n = leftlfloor frac{n}{1} rightrfloor + leftlfloor frac{n}{2} rightrfloor + cdots + leftlfloor frac{n}{n} rightrfloor ).Proof of Part (b):We aim to show that the sum of the sum of divisors of all natural numbers from 1 to ( n ), denoted ( sigma_1 + sigma_2 + cdots + sigma_n ), equals the weighted sum ( 1leftlfloor frac{n}{1} rightrfloor + 2leftlfloor frac{n}{2} rightrfloor + cdots + nleftlfloor frac{n}{n} rightrfloor ).Key Insight: Instead of summing the divisors for each number ( m in {1, 2, ldots, n} ), we sum the contribution of each divisor ( d in {1, 2, ldots, n} ) across all its multiples ( leq n ). For a fixed divisor ( d ), it contributes ( d ) to every multiple of ( d ). The total contribution of ( d ) is ( d cdot leftlfloor frac{n}{d} rightrfloor ).Formal Proof: The total sum of divisors for all numbers ( 1 ) to ( n ) is:[sum_{m=1}^n sigma_m = sum_{m=1}^n sum_{d mid m} d.]Interchanging the order of summation:[sum_{m=1}^n sigma_m = sum_{d=1}^n sum_{substack{m=1 d mid m}}^n d = sum_{d=1}^n d cdot leftlfloor frac{n}{d} rightrfloor.]Thus, ( sigma_1 + sigma_2 + cdots + sigma_n = 1leftlfloor frac{n}{1} rightrfloor + 2leftlfloor frac{n}{2} rightrfloor + cdots + nleftlfloor frac{n}{n} rightrfloor ).Final Answera) (boxed{sum_{k=1}^{n} tau_k = sum_{d=1}^{n} leftlfloor frac{n}{d} rightrfloor}) b) (boxed{sum_{k=1}^{n} sigma_k = sum_{d=1}^{n} d leftlfloor frac{n}{d} rightrfloor})

🔑:Let's denote by ( tau_n ) the number of divisors of the natural number ( n ), and by ( sigma_n ) the sum of the divisors of ( n ).# Part (a): Proving the relationship for the number of divisors1. Base Case: ( n = 1 ) [ tau_1 = 1 quad text{and} quad left[ frac{1}{1} right] = 1 ] Thus, for ( n = 1 ), the equality holds: [ tau_1 = left[ frac{1}{1} right] ]2. Induction Hypothesis: Suppose that the relationship holds for some natural number ( n ): [ tau_1 + tau_2 + tau_3 + ldots + tau_n = left[ frac{n}{1} right] + left[ frac{n}{2} right] + left[ frac{n}{3} right] + ldots + left[ frac{n}{n} right] ]3. Inductive Step: We need to show that if the hypothesis holds for ( n ), it also holds for ( n + 1 ): [ tau_1 + tau_2 + tau_3 + ldots + tau_{n+1} = left[ frac{n+1}{1} right] + left[ frac{n+1}{2} right] + left[ frac{n+1}{3} right] + ldots + left[ frac{n+1}{n+1} right] ] We consider the function ( left[ frac{n+1}{k} right] ): - If ( n+1 ) is not divisible by ( k ): [ n + 1 = qk + r quad text{where} quad 1 leq r leq k-1 ] Then, [ left[ frac{n+1}{k} right] = q quad text{and} quad left[ frac{n}{k} right] = q ] - If ( n+1 ) is divisible by ( k ): [ n + 1 = qk quad ( r = 0 ) ] Then, [ left[ frac{n+1}{k} right] = q quad text{and} quad left[ frac{n}{k} right] = q - 1 ] Thus, [ left[ frac{n+1}{k} right] = left[ frac{n}{k} right] + 1 ]4. Summing over ( k ) from 1 to ( n+1 ): [ left[ frac{n+1}{1} right] + left[ frac{n+1}{2} right] + ldots + left[ frac{n+1}{n+1} right] = left[ frac{n}{1} right] + left[ frac{n}{2} right] + ldots + left[ frac{n}{n+1} right] + tau_{n+1} ] Hence, [ left[ frac{n}{1} right] + left[ frac{n}{2} right] + ldots + left[ frac{n+1}{n+1} right] = tau_1 + tau_2 + ldots + tau_{n+1} ] By the principle of mathematical induction, we conclude that [ tau_1 + tau_2 + tau_3 + ldots + tau_n = left[ frac{n}{1} right] + left[ frac{n}{2} right] + left[ frac{n}{3} right] + ldots + left[ frac{n}{n} right] ]# Part (b): Proving the relationship for the sum of divisors1. Observation: Each number ( k ) that a number ( n ) may be divided by contributes a sum equal to ( k cdot left[ frac{n}{k} right] ).2. Summing the expression: The sum of divisor function ( sigma_n ) adds up all the divisors of ( n ): [ sigma_1 + sigma_2 + sigma_3 + ldots + sigma_n = sum_{k=1}^n k left[ frac{n}{k} right] ]3. Base Case: ( n=1 ): [ sigma_1 = 1 quad text{and} quad left[ frac{1}{1} right] = 1 ]4. Inductive Step: Suppose the relationship holds for some ( n ): [ sigma_1 + sigma_2 + sigma_3 + ldots + sigma_n = left[ frac{n}{1} right] + 2 left[ frac{n}{2} right] + 3 left[ frac{n}{3} right] + ldots + n left[ frac{n}{n} right] ] For ( n+1 ), we need to show: [ sigma_1 + sigma_2 + ldots + sigma_{n+1} = left[ frac{n+1}{1} right] + 2 left[ frac{n+1}{2} right] + 3 left[ frac{n+1}{3} right] + ldots + (n+1) left[ frac{n+1}{n+1} right] ] Using the previous steps where we discussed ( left[ frac{n+1}{k} right] ): [ left[ frac{n+1}{1} right] + 2left[ frac{n+1}{2} right] + ldots + (n+1) left[ frac{n+1}{n+1} right] = left[ frac{n}{1} right] + 2 left[ frac{n}{2} right] + ldots + (n+1) left[ frac{n}{n+1} right] + sigma_{n+1} ] By the principle of mathematical induction, the relationship holds: [ sigma_1 + sigma_2 + ldots + sigma_n = left[ frac{n}{1} right] + 2 left[ frac{n}{2} right] + ldots + n left[ frac{n}{n} right] ]Conclusion:[boxed{tau_1 + tau_2 + tau_3 + ldots + tau_n = left[ frac{n}{1} right] + left[ frac{n}{2} right] + left[ frac{n}{3} right] + ldots + left[ frac{n}{n} right]}][boxed{sigma_1 + sigma_2 + sigma_3 + ldots + sigma_n = left[ frac{n}{1} right] + 2left[ frac{n}{2} right] + 3left[ frac{n}{3} right] + ldots + nleft[ frac{n}{n} right]}]

🧠:Okay, let's tackle this probability problem. So, the set A is {1,2,3,4}, and Ander is picking numbers from this set every second with replacement. The process stops when the sum of the last two numbers selected is a prime number. We need to find the probability that the last number selected is "1". First, let me understand the problem step by step. The key points are:1. Numbers are selected with replacement, so each time, the probability of picking any number (1,2,3,4) is 1/4, independent of previous picks.2. The stopping condition is when the sum of the last two numbers is prime.3. We need the probability that when the process stops, the last number is 1.So, essentially, we need to model this as a Markov process where each state depends on the previous number picked, since the sum of the last two is what determines whether we stop or not. Wait, because the stopping condition depends on the last two numbers, the state should probably be the last number selected, because the next number picked will combine with it to potentially form a prime sum. Let me think. If I model the states as the last number selected, then each time we pick a new number, the transition depends on the previous state. If the sum of the previous state (number) and the new number is prime, the process stops. Otherwise, the state transitions to the new number. Therefore, the process can be represented by states corresponding to the last number picked. The states are 1, 2, 3, 4. Each time, we transition from the current state to a new state based on the next number picked. However, if the sum of the current state and the next number is prime, the process terminates. Our goal is to find the probability that when termination occurs, the last number (the one that caused the termination) is 1. So, perhaps we can model this using probabilities for each state leading to termination with 1. Let me denote by P(i) the probability that, starting from state i, the process eventually stops with the last number being 1. Wait, but actually, the process starts from scratch, meaning there's no prior number. So, the first pick is just a number, and then the second pick will combine with the first to check if their sum is prime. If not, then we continue, keeping the second number as the new state. Wait, hold on. The problem says "the sum of the last two selected numbers is a prime number". So, actually, the process can't stop until at least two numbers have been selected. So, the first number is picked, then the second, check the sum. If it's prime, stop; else, pick the third, check sum of the second and third, etc. Therefore, the initial step is to pick the first number, then the second, check sum. If not prime, pick the third, check sum of second and third. So, the state after each step is the last number picked, because the next number will be combined with it to check for the prime sum. Therefore, the process can be modeled as follows:- Start: No numbers picked yet.- Pick first number: state is now that number.- Pick second number: check sum with first. If prime, stop. Else, state is the second number.- Pick third number: check sum with second. If prime, stop. Else, state is the third number.- And so on.Therefore, except for the very first step, each subsequent step depends only on the previous number (the state). So, the process is a Markov chain with states 1,2,3,4, and transitions based on the next number picked. Therefore, to model this, let's consider that once we are in state i, the next number j is picked with probability 1/4. If i + j is prime, the process stops, and we record j as the terminating number. If not, we transition to state j. Therefore, the probability we are looking for is the probability that, starting from the first number (which is picked uniformly at random), the process eventually stops with the terminating number being 1. But actually, the process starts by picking the first number, then the second. So maybe we need to adjust the model. Let's think again:1. First pick: number X1. Since there's no prior number, we can't check any sum yet. So we pick X1, then pick X2. Check X1 + X2. If prime, stop; else, continue by picking X3, check X2 + X3, and so on. Therefore, the initial step is a bit different because the first transition requires two picks. However, after the first two picks, each subsequent pick only depends on the previous one. Hmm, this complicates things a bit. So maybe the first step is to pick two numbers, then from there on, each step only depends on the previous number. Alternatively, since after the first two numbers, each subsequent number pairs with the previous one to check for the sum. Therefore, once you have two numbers, the next number is paired with the second, and so on. Therefore, the process can be modeled as a Markov chain where each state is the last number picked. However, the initial step requires two numbers. But perhaps we can model this by considering that the first number is picked, leading to a state where we need to pick the next number. But once the next number is picked, we check the sum. If it's prime, stop; else, we transition to the state corresponding to the second number, and repeat. Therefore, maybe we can model this as follows:- The initial state is "Start", which transitions to states 1,2,3,4 with probability 1/4 each, representing the first number picked.- From each state i (representing the first number), we pick another number j with probability 1/4. If i + j is prime, the process stops with j as the last number. If not, we transition to state j, and repeat.Therefore, the overall probability can be computed by considering the probability starting from each state i, leading to termination with last number 1. But since the first number is picked uniformly at random, we need to average over all possible initial states. Therefore, let me formalize this:Let P(i) be the probability that, starting from state i (i.e., the previous number is i), the process eventually stops with the last number being 1. Our goal is to compute the average over the initial state, which is (P(1) + P(2) + P(3) + P(4))/4, since the first number is picked uniformly.Alternatively, since after the first number is picked, the second number is picked, and from there on, it's a Markov process. Wait, but actually, if we start from state i, we pick j next. If i + j is prime, stop with j; else, transition to j. So P(i) is the probability that, starting from i, the process stops with 1. Therefore, we can set up equations for P(1), P(2), P(3), P(4). Similarly, for each state i, P(i) = sum over all j in A of [probability of picking j] * [if i + j is prime, then indicator if j is 1; else, P(j)]So mathematically,For each i in {1,2,3,4},P(i) = Σ_{j=1}^4 [ (1/4) * [ if i + j is prime then δ_{j,1} else P(j) ] ]Where δ_{j,1} is 1 if j = 1, else 0.Therefore, we can write four equations with four variables (P(1), P(2), P(3), P(4)).Let me note down the primes that can be formed by the sum of two numbers from A. The possible sums are 1+1=2, 1+2=3, 1+3=4, 1+4=5, 2+2=4, 2+3=5, 2+4=6, 3+3=6, 3+4=7, 4+4=8. The primes in these sums are 2, 3, 5, 7. So the prime sums are 2,3,5,7.Therefore, for any i and j, if i + j is in {2,3,5,7}, then the process stops when j is picked. Otherwise, it continues with state j.Therefore, let's list for each i, the j's such that i + j is prime:For i = 1:j = 1: 1+1=2 (prime)j = 2: 1+2=3 (prime)j = 3: 1+3=4 (not prime)j = 4: 1+4=5 (prime)So, for i=1, j=1,2,4 lead to termination.For i = 2:j=1: 2+1=3 (prime)j=2: 2+2=4 (not prime)j=3: 2+3=5 (prime)j=4: 2+4=6 (not prime)Thus, j=1,3 lead to termination.For i = 3:j=1: 3+1=4 (not prime)j=2: 3+2=5 (prime)j=3: 3+3=6 (not prime)j=4: 3+4=7 (prime)Thus, j=2,4 lead to termination.For i =4:j=1: 4+1=5 (prime)j=2: 4+2=6 (not prime)j=3: 4+3=7 (prime)j=4: 4+4=8 (not prime)Thus, j=1,3 lead to termination.So, for each i, the possible j that cause termination are:i=1: j=1,2,4i=2: j=1,3i=3: j=2,4i=4: j=1,3Therefore, for each state i, we can write P(i) as:P(i) = sum_{j: i+j is prime} (1/4)*δ_{j,1} + sum_{j: i+j not prime} (1/4)*P(j)So let's compute each P(i):Starting with i=1:P(1) = [for j=1: (1/4)*1] + [j=2: (1/4)*0] + [j=4: (1/4)*0] + [j=3: (1/4)*P(3)]Wait, hold on. Wait, for i=1, the j's that terminate are 1,2,4. For each of these j, we check if j=1 (to contribute to the probability), otherwise, it terminates but doesn't contribute. Then for j=3, which doesn't terminate, we have to transition to state j=3 and multiply by P(3).Wait, let's clarify:For i=1:Terminating j's: 1,2,4. For each of these, if j=1, add 1/4 *1; otherwise, add 1/4 *0. For the non-terminating j (only j=3), add 1/4 * P(3).Therefore,P(1) = (1/4)*1 + (1/4)*0 + (1/4)*0 + (1/4)*P(3)Similarly, breaking it down:- j=1: sum is prime, and j=1, so contributes 1/4 *1- j=2: sum is prime, j=2, contributes 1/4 *0- j=4: sum is prime, j=4, contributes 1/4 *0- j=3: sum is not prime, so contributes 1/4 *P(3)Therefore:P(1) = 1/4 + 0 + 0 + (1/4)P(3) = 1/4 + (1/4)P(3)Similarly for P(2):Terminating j's are 1 and 3. So j=1: 1/4 *1 if j=1, else 0. Non-terminating j's are 2 and 4, but wait, for i=2:Wait, for i=2, the possible j's:j=1: 2+1=3 (prime), so termination. If j=1, contributes 1/4 *1j=2: 2+2=4 (not prime), transition to P(2)j=3: 2+3=5 (prime), j=3, contributes 1/4 *0j=4: 2+4=6 (not prime), transition to P(4)Wait, no: Wait, for i=2, j=1 and j=3 lead to termination. So:Terminating j=1 and j=3. For j=1: add 1/4 *1, j=3: add 1/4 *0. Then for non-terminating j=2 and j=4, transition to P(j). Therefore:P(2) = (1/4)*1 + (1/4)*0 + (1/4)*P(2) + (1/4)*P(4)So:P(2) = 1/4 + (1/4)P(2) + (1/4)P(4)Similarly for P(3):Terminating j's are 2 and 4. So j=2: contributes 1/4 *0, j=4: contributes 1/4 *0. Non-terminating j's are 1 and 3, transition to P(1) and P(3).Wait, for i=3:j=1: 3+1=4 (not prime) → P(1)j=2: 3+2=5 (prime) → contributes 0j=3: 3+3=6 (not prime) → P(3)j=4: 3+4=7 (prime) → contributes 0Therefore:P(3) = (1/4)*P(1) + (1/4)*0 + (1/4)*P(3) + (1/4)*0 = (1/4)P(1) + (1/4)P(3)And for P(4):Terminating j's are 1 and 3. So j=1: contributes 1/4 *1, j=3: contributes 1/4 *0. Non-terminating j's are 2 and 4, transition to P(2) and P(4).For i=4:j=1: 4+1=5 (prime) → contributes 1/4 *1j=2: 4+2=6 (not prime) → P(2)j=3: 4+3=7 (prime) → contributes 1/4 *0j=4: 4+4=8 (not prime) → P(4)Thus:P(4) = (1/4)*1 + (1/4)*P(2) + (1/4)*0 + (1/4)*P(4) = 1/4 + (1/4)P(2) + (1/4)P(4)Now, let's write all four equations:1. P(1) = 1/4 + (1/4)P(3)2. P(2) = 1/4 + (1/4)P(2) + (1/4)P(4)3. P(3) = (1/4)P(1) + (1/4)P(3)4. P(4) = 1/4 + (1/4)P(2) + (1/4)P(4)Now, we have four equations with four variables. Let's solve them step by step.Starting with equation 3:P(3) = (1/4)P(1) + (1/4)P(3)Subtract (1/4)P(3) from both sides:P(3) - (1/4)P(3) = (1/4)P(1)(3/4)P(3) = (1/4)P(1)Multiply both sides by 4:3P(3) = P(1)Thus,P(1) = 3P(3) --- Equation ASimilarly, equation 1:P(1) = 1/4 + (1/4)P(3)But from Equation A, substitute P(1) with 3P(3):3P(3) = 1/4 + (1/4)P(3)Subtract (1/4)P(3) from both sides:3P(3) - (1/4)P(3) = 1/4(12/4 - 1/4)P(3) = 1/4(11/4)P(3) = 1/4Multiply both sides by 4:11P(3) = 1Thus,P(3) = 1/11Then from Equation A:P(1) = 3*(1/11) = 3/11Now, moving to equation 4:P(4) = 1/4 + (1/4)P(2) + (1/4)P(4)Subtract (1/4)P(4) from both sides:P(4) - (1/4)P(4) = 1/4 + (1/4)P(2)(3/4)P(4) = 1/4 + (1/4)P(2)Multiply both sides by 4:3P(4) = 1 + P(2)Thus,P(2) = 3P(4) - 1 --- Equation BNow, equation 2:P(2) = 1/4 + (1/4)P(2) + (1/4)P(4)Multiply both sides by 4 to eliminate denominators:4P(2) = 1 + P(2) + P(4)Subtract P(2) and P(4) from both sides:4P(2) - P(2) - P(4) = 13P(2) - P(4) = 1 --- Equation CFrom Equation B: P(2) = 3P(4) - 1Substitute into Equation C:3*(3P(4) - 1) - P(4) = 19P(4) - 3 - P(4) = 18P(4) - 3 = 18P(4) = 4P(4) = 4/8 = 1/2Then from Equation B:P(2) = 3*(1/2) -1 = 3/2 -1 = 1/2So now we have all P(i):P(1) = 3/11 ≈ 0.2727P(2) = 1/2 = 0.5P(3) = 1/11 ≈ 0.0909P(4) = 1/2 = 0.5But wait, this seems a bit odd. Let me check the equations again to ensure no mistakes.First, solving for P(3) and P(1):We had equation 3 leading to P(1) = 3P(3), then substituting into equation 1 gave P(3) = 1/11 and P(1) = 3/11. That seems correct.Equation 4 led to P(2) = 3P(4) -1, and equation 2 substitution into equation C led to P(4) = 1/2, hence P(2) = 1/2. That seems okay.But let's check equation 4 with P(4) = 1/2 and P(2) = 1/2:Original equation 4: P(4) = 1/4 + (1/4)P(2) + (1/4)P(4)Plug in P(4) = 1/2 and P(2)=1/2:Left side: 1/2Right side: 1/4 + (1/4)(1/2) + (1/4)(1/2) = 1/4 + 1/8 + 1/8 = 1/4 + 2/8 = 1/4 + 1/4 = 1/2. Correct.Equation 2: P(2) = 1/4 + (1/4)P(2) + (1/4)P(4)Plug in P(2)=1/2, P(4)=1/2:Right side: 1/4 + (1/4)(1/2) + (1/4)(1/2) = same as above, 1/2. Correct.So equations are consistent.Now, recall that the initial state is the first number picked, which is uniform over 1,2,3,4. Therefore, the total probability we want is the average of P(1), P(2), P(3), P(4):Total probability = (P(1) + P(2) + P(3) + P(4))/4Plugging in the values:(3/11 + 1/2 + 1/11 + 1/2)/4Compute numerator:3/11 + 1/11 = 4/111/2 + 1/2 = 1So total numerator = 4/11 + 1 = 15/11Thus, total probability = (15/11)/4 = 15/44 ≈ 0.3409But let me verify this, because I want to ensure that this is indeed the correct approach.Wait, but hold on. The process starts by picking the first number, then the second. The termination occurs at the second pick if their sum is prime. If not, then the third pick is made, checking with the second, and so on. However, in our model, we considered P(i) as the probability that starting from state i (i.e., the previous number is i), the process stops with last number 1. Then, since the first number is picked uniformly, we average the P(i) over i=1,2,3,4. However, is this accurate?Wait, actually, the first number is picked, then the second. The termination can happen at the second pick. So the total probability should account for the possibility that the process stops at the second pick with last number 1, or continues, etc. However, in our model, when we average P(i), we are considering starting from each first number, then the subsequent process. But actually, the first two picks are both part of the initial process.Wait, perhaps the initial step is different. Let's think again.The process is:1. Pick X1.2. Pick X2. If X1 + X2 is prime, stop. Last number is X2.3. If not, pick X3. If X2 + X3 is prime, stop. Last number is X3.4. Etc.Therefore, the initial step isn't just starting from a state X1, but the first transition is X1 -> X2, which may terminate. So actually, the overall probability can be considered as:Total probability = Probability that the last number is 1.This can happen at step 2 (X2=1 and X1 +1 is prime), or step 3 (X3=1 and X2 +1 is prime, and the process didn't stop at step 2), or step 4 (X4=1 and X3 +1 is prime, and didn't stop at steps 2,3), etc.Therefore, the total probability is the sum over n >=2 of the probability that the process stops at step n with the nth number being 1.This seems complex, but maybe our original approach is still valid because the Markov chain model inherently accounts for these possibilities by setting up the recursive equations. However, perhaps the error is in the initial averaging.Wait, in the Markov chain model, when you start from a state i (the previous number), and compute P(i), which is the probability of eventually stopping with 1 starting from i. However, in reality, the first number picked is X1, then X2 is picked. The probability that the process stops at X2 with X2=1 is (1/4) * [sum over X1 such that X1 +1 is prime]. Then, if it doesn't stop at X2, it transitions to state X2, and the probability from there is P(X2). So the total probability is:Total probability = Probability(stop at X2 with X2=1) + Probability(not stopping at X2) * E[P(X2)]Where E[P(X2)] is the average of P(j) over j=1,2,3,4, since X2 is picked uniformly regardless of X1. Wait, but actually, if X1 is given, then X2 is picked uniformly, but if X1 + X2 is not prime, then X2 becomes the new state. Therefore, the total probability can be written as:Total probability = (1/4) * [sum_{i=1}^4 sum_{j=1}^4 (1/4) * [if i + j is prime, then δ_{j,1} else 0 ] ] + (1/4) * [sum_{i=1}^4 sum_{j=1}^4 (1/4) * [if i + j is not prime, then P(j) else 0 ] ]Wait, this seems messy. Alternatively, since the first number is picked uniformly (1/4), then the second number is picked uniformly (1/4). The probability that the process stops at the second pick with X2=1 is:For each X1, probability that X1 +1 is prime, and X2=1.So:Sum_{X1=1}^4 [ (1/4) * (1/4) * δ_{X1 +1 is prime} * δ_{X2=1} ]But X2=1 is fixed, so δ_{X2=1} is 1 only if j=1. Therefore, for each X1, the probability is (1/4)*(1/4) if X1 +1 is prime.But wait, no. If X2=1, then the probability is (1/4) for X2=1, and for each X1, the probability that X1 +1 is prime. So overall:Probability(stop at step 2 with X2=1) = (1/4) * [sum_{X1=1}^4 (1/4) * I_{X1 +1 is prime} ]Where I_{...} is 1 if true, else 0.Similarly, the total probability is:Sum_{n=2}^infty Probability(stop at step n with X_n=1)Which can be complex, but the Markov model approach might still work because the process after the first two steps is memoryless.But actually, the initial step is different because the first number is arbitrary, then the second is arbitrary. However, once you pass the first step, it becomes a Markov chain. Therefore, the total probability can be considered as:Total probability = Probability(stop at step 2 with X2=1) + Probability(not stopping at step 2) * Average P(j)Where Average P(j) is the average of P(j) over j=1,2,3,4, since after not stopping at step 2, the state becomes X2, which is uniform over 1,2,3,4 if we didn't stop. Wait, no. If we didn't stop at step 2, X2 is not uniform. Because X2 is picked uniformly, but conditioned on X1 + X2 not being prime. Therefore, X2 is not uniform anymore.Ah, here's a mistake in the previous approach. If we model P(i) as the probability starting from state i, then the initial step after the first two numbers is not uniform. Because if we didn't stop at step 2, the distribution of X2 is not uniform—it depends on X1 and the fact that X1 + X2 is not prime. Therefore, the initial approach of averaging P(i) over i=1,2,3,4 is incorrect because after the first step, the distribution of states is not uniform.Therefore, our previous calculation is flawed because we assumed that after the first number, we simply average over all possible starting states, but in reality, the second number is dependent on the first. Therefore, we need to adjust the model to account for the correct initial conditions.This complicates things. Let's re-express the problem correctly.Let’s denote Q as the total probability we want. The process is:- First, pick X1 uniformly from {1,2,3,4}.- Then, pick X2 uniformly from {1,2,3,4}.- If X1 + X2 is prime, stop. The probability that X2=1 in this case is equal to the probability that X2=1 and X1 +1 is prime.- If X1 + X2 is not prime, then we transition to state X2, and from there, the probability of eventually stopping with 1 is P(X2), as previously defined.Therefore, the total probability Q can be written as:Q = Probability(stop at step 2 with X2=1) + Probability(not stopping at step 2) * E[P(X2) | not stopped at step2]Now, let's compute each part.First, Probability(stop at step 2 with X2=1):This is the probability that X2=1 and X1 +1 is prime.Since X1 is picked uniformly, then:Probability = (1/4) * [Probability(X1 +1 is prime)]Because X2=1 has probability 1/4, and X1 is uniform.What's the probability that X1 +1 is prime?X1 can be 1,2,3,4.For X1=1: 1+1=2 (prime)X1=2: 2+1=3 (prime)X1=3:3+1=4 (not prime)X1=4:4+1=5 (prime)So primes when X1 +1: 1,2,4. So three out of four X1's.Therefore, Probability(X1 +1 is prime) = 3/4.Thus, Probability(stop at step2 with X2=1) = (1/4)*(3/4) = 3/16.Next, Probability(not stopping at step2):This is 1 - Probability(stop at step2). Probability(stop at step2) is the probability that X1 + X2 is prime. Since X1 and X2 are both uniform:Probability(stop at step2) = sum_{i=1}^4 sum_{j=1}^4 (1/16) * I_{i+j is prime}We can compute this by counting the number of pairs (i,j) where i + j is prime, divided by 16.From earlier, the prime sums are 2,3,5,7.Possible pairs:(1,1)=2, (1,2)=3, (1,4)=5,(2,1)=3, (2,3)=5,(3,2)=5, (3,4)=7,(4,1)=5, (4,3)=7.Total prime pairs:Let me count:i=1:j=1,2,4 → 3 pairsi=2:j=1,3 → 2 pairsi=3:j=2,4 → 2 pairsi=4:j=1,3 → 2 pairsTotal prime pairs: 3+2+2+2=9.Therefore, Probability(stop at step2) = 9/16.Thus, Probability(not stopping at step2) = 1 - 9/16 = 7/16.Now, E[P(X2) | not stopped at step2] is the expected value of P(X2) given that X1 + X2 is not prime.To compute this expectation, we need to find the distribution of X2 given that X1 + X2 is not prime.This requires calculating, for each j in {1,2,3,4}, the probability that X2 = j given that X1 + j is not prime.But since X1 is uniform and X2 is uniform and independent, the joint distribution is uniform over all pairs. The condition is that i + j is not prime. So for each j, the probability that X2=j and X1 + j is not prime, divided by the total probability of not stopping.Therefore,E[P(X2) | not stopped] = sum_{j=1}^4 P(j) * Probability(X2=j | not stopped)Where Probability(X2=j | not stopped) = [sum_{i=1}^4 (1/4 * 1/4) * I_{i+j not prime} ] / (7/16)Because the numerator is the probability that X2=j and X1 +j is not prime, which is sum_{i=1}^4 (1/16) * I_{i+j not prime}, and the denominator is 7/16.Therefore,Probability(X2=j | not stopped) = [sum_{i=1}^4 I_{i+j not prime}]/4 /7/16 * 1/16 ?Wait, wait. Let's clarify.The joint probability that X2=j and not stopped is sum_{i=1}^4 [1/16 * I_{i+j not prime} ]Which is equal to [number of i's such that i + j is not prime] * (1/16)Therefore, for each j, the number of i's where i + j is not prime:For j=1:i +1 is prime when i=1,2,4 (as before). Therefore, non-prime when i=3. So 1 i.For j=2:i +2 is prime when i=1,3 (sum=3,5). Non-prime when i=2,4 (sum=4,6). So 2 i's.For j=3:i +3 is prime when i=2,4 (sum=5,7). Non-prime when i=1,3 (sum=4,6). So 2 i's.For j=4:i +4 is prime when i=1,3 (sum=5,7). Non-prime when i=2,4 (sum=6,8). So 2 i's.Wait, let's verify:For j=1:i=1:1+1=2 (prime)i=2:2+1=3 (prime)i=3:3+1=4 (non-prime)i=4:4+1=5 (prime)So non-prime only at i=3. So 1.For j=2:i=1:1+2=3 (prime)i=2:2+2=4 (non-prime)i=3:3+2=5 (prime)i=4:4+2=6 (non-prime)Non-prime at i=2,4. So 2.For j=3:i=1:1+3=4 (non-prime)i=2:2+3=5 (prime)i=3:3+3=6 (non-prime)i=4:4+3=7 (prime)Non-prime at i=1,3. So 2.For j=4:i=1:1+4=5 (prime)i=2:2+4=6 (non-prime)i=3:3+4=7 (prime)i=4:4+4=8 (non-prime)Non-prime at i=2,4. So 2.Therefore, for each j:- j=1: 1 non-prime i- j=2:2 non-prime i's- j=3:2 non-prime i's- j=4:2 non-prime i'sThus, the number of non-prime i's for each j is [1,2,2,2].Therefore, the joint probability that X2=j and not stopped is:For j=1:1/16For j=2:2/16For j=3:2/16For j=4:2/16Therefore, the total is (1 +2 +2 +2)/16 =7/16, which matches our earlier result.Therefore, Probability(X2=j | not stopped) = (number of non-prime i's for j)/7So:- P(X2=1 | not stopped) =1/7- P(X2=2 | not stopped) =2/7- P(X2=3 | not stopped)=2/7- P(X2=4 | not stopped)=2/7Therefore, E[P(X2) | not stopped] = (1/7)P(1) + (2/7)P(2) + (2/7)P(3) + (2/7)P(4)Thus, the total probability Q is:Q = 3/16 + (7/16)[ (1/7)P(1) + (2/7)P(2) + (2/7)P(3) + (2/7)P(4) ]Simplify:Q = 3/16 + (1/16)[P(1) + 2P(2) + 2P(3) + 2P(4)]But wait, from earlier, we have equations for P(1), P(2), P(3), P(4). Let's recall their values from the previous (incorrect) approach:P(1) = 3/11 ≈0.2727P(2)=1/2=0.5P(3)=1/11≈0.0909P(4)=1/2=0.5But now, using the correct initial conditions, we need to compute Q as above.However, the P(i)'s were computed under the assumption that the initial state is uniform, which is not the case here. Therefore, the previous calculation of P(i)'s is incorrect because the actual initial state after step 2 (if not stopped) is not uniform, but skewed.Therefore, our earlier approach is invalid. We need to redefine P(j) correctly.Let’s redefine:Let P(j) be the probability that, starting from state j (i.e., the previous number is j), the process eventually stops with the last number being 1.Then, the equations for P(j) are as derived before:For each j:P(j) = sum_{k: j +k is prime} (1/4)δ_{k,1} + sum_{k: j +k not prime} (1/4)P(k)This part is correct because once you are in state j, the next number k is picked uniformly. If j +k is prime, you stop with k=1 or not; if not, you transition to state k.Therefore, the equations for P(1), P(2), P(3), P(4) are correct as derived earlier. However, the error was in the initial total probability calculation, where we averaged P(j) over the uniform distribution, which is incorrect because after step 2 (if not stopped), the distribution of j is not uniform.However, now that we have the correct expression for Q:Q = 3/16 + (1/16)(P(1) + 2P(2) + 2P(3) + 2P(4))But we already have the values of P(1), P(2), P(3), P(4) from solving the system earlier. Let's substitute them in:P(1) = 3/11, P(2)=1/2, P(3)=1/11, P(4)=1/2Therefore:Q = 3/16 + (1/16)(3/11 + 2*(1/2) + 2*(1/11) + 2*(1/2))Compute the numerator inside:3/11 + 2*(1/2) + 2*(1/11) + 2*(1/2)= 3/11 +1 + 2/11 +1= (3 +2)/11 + 2= 5/11 +2= 27/11Therefore,Q = 3/16 + (27/11)/16 = (3 + 27/11)/16 = (33/11 +27/11)/16 = (60/11)/16 = (60)/(11*16) = 60/176 = 15/44 ≈0.3409So the total probability is 15/44, which matches the result obtained earlier.But wait, this suggests that despite the initial mistake in reasoning about the distribution, the final result is the same. This is because the equations for P(j) are correctly derived based on the Markov chain transitions, and the Q expression correctly incorporates the initial step.Therefore, even though the intermediate step of averaging P(j) over uniform states was based on flawed reasoning, the correct expression for Q, which properly accounts for the non-uniform distribution after the first step, coincidentally leads to the same numerical result because of the way the equations are structured.Therefore, the final answer is 15/44.To confirm, let's do a sanity check. 15/44 is approximately 0.3409. Given that when you can stop at step 2 with probability 3/16 ≈0.1875, and the remaining probability comes from subsequent steps. Since the numbers are small and probabilities are cumulative, 0.34 seems reasonable.Another way to check is to simulate small cases.For example, consider stopping at step 2:Probability of stopping with X2=1 is 3/16.If not stopped at step2 (probability 7/16), then the expected contribution from there is (1/7)P(1) + 2/7 P(2) + 2/7 P(3) + 2/7 P(4).From earlier calculations, this is (1/7)(3/11) + (2/7)(1/2) + (2/7)(1/11) + (2/7)(1/2)= (3/77 + 1/7 + 2/77 + 1/7)= (5/77 + 2/7)= (5/77 + 22/77)= 27/77Multiply by 1/16:27/77 *1/16 =27/1232 ≈0.0219But total Q =3/16 +27/1232 ≈0.1875 +0.0219≈0.2094, which contradicts previous result. Wait, no, hold on, no:Wait, no. The total Q is 3/16 + (7/16)*(27/77). Let's compute correctly:Q=3/16 + (7/16)*(27/77)=3/16 + (27/(16*11))=3/16 +27/(176)Convert to common denominator:3/16 = 33/176So Q=33/176 +27/176=60/176=15/44≈0.3409. Correct.Yes, so that checks out.Therefore, despite the complexity, the answer is 15/44.Final AnswerThe probability that the last number selected is "1" is boxed{dfrac{15}{44}}.

✅:Given the set ( A = {1, 2, 3, 4} ), Ander randomly selects a number from ( A ) every second with replacement. The process stops when the sum of the last two selected numbers is a prime number. We need to find the probability that the last number selected is "1". Key Steps:1. Prime Sums Identification: - Possible sums of two numbers from ( A ) that are prime: 2, 3, 5, 7. - For each number ( i ) in ( A ), identify the numbers ( j ) such that ( i + j ) is prime.2. State Transitions and Equations: - Define ( P(i) ) as the probability that starting from state ( i ), the process stops with the last number being 1. - Set up equations for ( P(1), P(2), P(3), ) and ( P(4) ) based on transitions and prime checks.3. Solving the System of Equations: - For each state ( i ), the equation is: [ P(i) = sum_{substack{j=1 i+j text{ prime}}}^4 frac{1}{4} delta_{j,1} + sum_{substack{j=1 i+j text{ not prime}}}^4 frac{1}{4} P(j) ] - Solving these equations yields: [ P(1) = frac{3}{11}, quad P(2) = frac{1}{2}, quad P(3) = frac{1}{11}, quad P(4) = frac{1}{2} ]4. Total Probability Calculation: - The total probability is the average of ( P(1), P(2), P(3), ) and ( P(4) ) considering the initial uniform selection of the first number. - Correctly account for the non-uniform distribution after the first step and compute the expected value. Final Answer:The probability that the last number selected is "1" is boxed{dfrac{15}{44}}.

🔑:Firstly, let's denote the set as ( A = {1, 2, 3, 4} ) and define the events where sequences of numbers end with the sum of the last two numbers being a prime number, thereby halting the process. The process can also be interpreted as stopping when "1" is the last number drawn that fulfills this prime condition. Let's compute the probabilities in detail.# Definitions and Initial Probabilities:Let ( P_i ) be the probability that a sequence starting with ( i ) ends with "1". Thus, we need to find ( P_1 ).For a sequence starting with 1:1. If the next number is also 1: The sum is (1 + 1 = 2) (which is prime), hence the process stops. 2. If the next number is 2: The sum is (1 + 2 = 3) (which is prime), but this does not end with "1".3. If the next number is 3: The sum is (1 + 3 = 4) (not prime), the process continues.4. If the next number is 4: The sum is (1 + 4 = 5) (prime), but this does not end with "1".From the above, the recurrence relation for ( P_1 ) can be derived:[ P_1 = frac{1}{4} left( frac{1}{4} + P_3 right) ]Now, let's analyze other sequences.For sequences starting with 2:1. If the next number is 1: The sum is (2 + 1 = 3) (prime).2. If the next number is 2: The sum is (2 + 2 = 4) (not prime), the process continues.3. If the next number is 3: The sum is (2 + 3 = 5) (prime), the sequence continues if the next number ends in 1.4. If the next number is 4: The sum is (2 + 4 = 6) (not prime), the process continues.From the above, the recurrence relation for ( P_2 ) can be expressed:[ P_2 = frac{1}{4} left( frac{1}{4} + P_2 + P_4 right) ]For sequences starting with 3:1. If the next number is 1: The sum is (3 + 1 = 4) (not prime), the process continues.2. If the next number is 2: The sum is (3 + 2 = 5) (prime), the sequence continues if the next number ends in 1.3. If the next number is 3: The sum is (3 + 3 = 6) (not prime), the process continues.4. If the next number is 4: The sum is (3 + 4 = 7) (prime), the process continues.Thus:[ P_3 = frac{1}{4} left( P_1 + P_3 right) ]For sequences starting with 4:1. If the next number is 1: The sum is (4 + 1 = 5) (prime).2. If the next number is 2: The sum is (4 + 2 = 6) (not prime), the process continues.3. If the next number is 3: The sum is (4 + 3 = 7) (prime), the sequence continues if the next number ends in 1.4. If the next number is 4: The sum is (4 + 4 = 8) (not prime), the process continues.Hence:[ P_4 = frac{1}{4} left( frac{1}{4} + P_2 + P_4 right) ]# Solving the System of Equations:We now solve the obtained recurrence relations:1. ( P_1 = frac{1}{4}left(frac{1}{4} + P_3right) )2. ( P_2 = frac{1}{4} left( frac{1}{4} + P_2 + P_4 right) )3. ( P_3 = frac{1}{4} left( P_1 + P_3 right) )4. ( P_4 = frac{1}{4} left( frac{1}{4} + P_2 + P_4 right) )# Calculation:From equation (3):[ 4P_3 = P_1 + P_3 ][ 3P_3 = P_1 ][ P_1 = 3P_3 ]From (2) and (4), noting they are identical:[ P_2 = P_4 ]Rewriting (2) with P_2 as P_4:[ P_2 = frac{1}{4} left( frac{1}{4} + P_2 + P_2 right) ][ P_2 = frac{1}{4} left( frac{1}{4} + 2P_2 right) ][ P_2 = frac{1}{16} + frac{1}{2}P_2 ][ frac{1}{2}P_2 = frac{1}{16} ][ P_2 = frac{1}{8} ]Plugging ( P_2 = frac{1}{8} ) back into one of the other equations:[ P_4 = frac{1}{8} ]From equation 1:[ P_1 = frac{1}{4}left(frac{1}{4} + P_3right) ]Replacing ( P_3 = frac{1}{44} ) and solving similarly:[ P_1 = frac{3}{44} ]Summing up all ( P_i ):[ P_{1} + P_{2} + P_{3} + P_{4} = frac{3}{44} + frac{1}{8} + frac{1}{44} + frac{1}{8} = frac{15}{44} ] Conclusion:Thus, the probability of taking "1" being the last number in the sequence is:[boxed{frac{15}{44}}]

🧠:Okay, let's tackle this problem: we need to find the number of functions ( f: A_3 rightarrow A_3 ) such that applying the function three times gives us back the original function, that is, ( f^{(3)} = f ). Also, we have to list all such functions. Alright, let's start by understanding the problem step by step.First, what is ( A_3 )? The notation ( A_n ) typically refers to the alternating group on n elements, which consists of all even permutations of n elements. However, since the problem is about functions from ( A_3 ) to ( A_3 ), and not necessarily about group homomorphisms, I should confirm if ( A_3 ) here is meant to be the alternating group or perhaps another set. But given that ( A_3 ) in group theory is the set of even permutations of 3 elements, which has 3 elements (since the symmetric group S₃ has 6 elements, and the alternating group A₃ is the normal subgroup of index 2, so 3 elements). So ( A_3 ) is a set with 3 elements. Wait, but actually, the alternating group A₃ has order 3. The elements are the three even permutations: the identity permutation, (1 2 3), and (1 3 2). So yes, ( A_3 ) is a set with 3 elements. So, the problem is about functions from a 3-element set to itself, satisfying ( f^3 = f ).Wait, hold on. If ( A_3 ) is a 3-element set, then the number of functions from ( A_3 ) to itself is ( 3^3 = 27 ). But we need those functions where applying the function three times is equal to the function itself. So, ( f circ f circ f = f ). Let me note that ( f^{(3)} ) here means the third iterate of f, i.e., f composed with itself three times. So, ( f^{(3)}(x) = f(f(f(x))) ). Therefore, the condition is that for every x in ( A_3 ), ( f(f(f(x))) = f(x) ).So, the problem reduces to finding all functions f: {a, b, c} → {a, b, c} (assuming the three elements are a, b, c) such that f^3 = f. Then, count them and list them.First, let's recall some basics about function iterates. If we have a function f: S → S where S is a finite set, the behavior of f can be described in terms of cycles and chains leading into cycles. Each element is either part of a cycle or eventually maps into a cycle after some iterations.But our condition here is that f^3 = f. That is, applying f three times is the same as applying f once. Let's analyze what this implies for the structure of the function.First, for any element x in A₃, we have that f^3(x) = f(x). Let's consider the orbit of x under f. The orbit is the sequence x, f(x), f²(x), f³(x), ..., and since the set is finite, this must eventually cycle.But our condition is that f³(x) = f(x). Therefore, for each x, after three applications of f, we get back to f(x). So, starting from x:x → f(x) → f²(x) → f³(x) = f(x)So, after the third step, we loop back to f(x). This suggests that the orbit of x under f has a cycle of length dividing 2, because from f(x) onward, we have a cycle. Wait, let's see:Wait, if f³(x) = f(x), then starting from f(x), applying f two more times gets us back to f(x):f(x) → f²(x) → f³(x) = f(x). So, starting from f(x), the cycle length is 2. Because f²(f(x)) = f³(x) = f(x). Therefore, the cycle length is 2. Therefore, the function f, when restricted to its image, must be an involution, i.e., f²(y) = y for all y in the image of f. Because if y is in the image of f, then y = f(x) for some x, and then f²(y) = f³(x) = f(x) = y. So, f² acts as the identity on its image. Therefore, the image of f is a set on which f is an involution.Furthermore, for elements not in the image of f, but since f is a function from A₃ to itself, the image is some subset of A₃. However, since A₃ has 3 elements, the image can be of size 1, 2, or 3.But if the image has size 1, then f is a constant function. Let's check if a constant function satisfies f³ = f. If f(x) = c for all x, then f³(x) = c = f(x), so yes, constant functions satisfy the condition. So, there are 3 constant functions.If the image has size 2, then f maps A₃ onto a subset of two elements, and on those two elements, f acts as an involution (since f²(y) = y for y in the image). So, on the image {a, b}, f must satisfy f(a) = b and f(b) = a, or f(a) = a and f(b) = b. Wait, but if f is an involution on its image, then it can either fix both elements or swap them. However, since the image has two elements, an involution on two elements can be either the identity or the transposition (swap). But if f is the identity on its image, then each element in the image is a fixed point, which is allowed. However, in that case, the function f would map some elements to a and some to b, but both a and b are fixed by f.Wait, but if the image is {a, b}, then f maps all elements to either a or b, and once you're in the image, applying f again keeps you there, either fixed or swapped. However, if f is the identity on the image, then f²(y) = y, which would satisfy f³(y) = f(y) since f³(y) = f(f²(y)) = f(y). Wait, but if f is the identity on the image, then f²(y) = f(y) = y, so f³(y) = f(y) = y, which is equal to f(y). So, that works.Alternatively, if f swaps a and b on the image, then f²(y) = y, so again, f³(y) = f(f²(y)) = f(y). Therefore, whether f is the identity or the swap on the image {a, b}, it will satisfy f³ = f.But wait, the image is {a, b}, and f maps the entire domain to {a, b}. For the elements outside the image (but since the domain and codomain are both A₃, which has 3 elements, the third element, say c, must be mapped to either a or b as well. Wait, but if the image is {a, b}, then c must be mapped to a or b. So, the function f would have to map c to either a or b. Then, when we consider f³(c), it's f(f(f(c))). Since f(c) is in {a, b}, then f(f(c)) is f of an element in the image, which is either fixed or swapped. Then f³(c) = f(f(f(c))) = f(f(c)) if f²(f(c)) = f(c), which is true because on the image, f² is identity or swap. Wait, if f swaps a and b, then f² is the identity. So, if f is a swap on {a, b}, then f² is identity, so f³(c) = f(f(f(c))) = f(c). Similarly, if f is the identity on {a, b}, then f³(c) = f(f(f(c))) = f(f(c)) = f(c). So, in either case, if the image is {a, b}, then f³(c) = f(c), which is required.Therefore, for the image of size 2, we can have functions where the image is {a, b}, and on {a, b}, f is either the identity or the swap, and the third element c is mapped to either a or b. So, how many such functions are there?First, choose the image. There are C(3, 2) = 3 choices for the 2-element subset {a, b}, {a, c}, {b, c}. For each such subset, say {a, b}, we can define f on {a, b} either as the identity or the swap. So, 2 choices for the behavior on the image. Then, the third element c must be mapped to either a or b. So, 2 choices for f(c). Therefore, for each image subset, there are 2 * 2 = 4 functions. Since there are 3 image subsets, that would give 3 * 4 = 12 functions. Wait, but hold on: let's verify.Suppose the image is {a, b}. For the mapping on {a, b}, we have two possibilities: identity (f(a)=a, f(b)=b) or swap (f(a)=b, f(b)=a). Then, for the element c, which is not in the image, we map it to either a or b. So, for each of the two behaviors on {a, b}, we have two choices for c's image. Therefore, 2 * 2 = 4 functions per image subset. Since there are 3 such subsets, 3 * 4 = 12 functions. So, 12 functions with image size 2.Similarly, for image size 3, that is, f is a permutation of A₃. Since the image is the entire set, f must be a permutation. Now, permutations on 3 elements can be of two types: even permutations (since we're in A₃, but wait, the problem is just about functions, not necessarily group homomorphisms). Wait, no, A₃ here is just the set, not the group, unless specified otherwise. Wait, but the original question says ( f: A_3 rightarrow A_3 ). If A₃ is the alternating group, then the elements are permutations, but the function f is just any function, not necessarily a group homomorphism. However, since A₃ as a group has 3 elements: the identity, (1 2 3), and (1 3 2). So, if we consider A₃ as the set {e, σ, τ} where σ is a 3-cycle and τ is its inverse, then the function f is from this set to itself, with f^3 = f.But regardless of whether A₃ is considered as a group or just a set with 3 elements, the problem reduces to functions on a 3-element set satisfying f³ = f. So, perhaps the mention of A₃ is just to specify the set has 3 elements. So, we can treat A₃ as a 3-element set, say {0, 1, 2} for simplicity.So, returning to permutations. If f is a permutation of A₃ (i.e., bijective), then f^3 = f implies that f² = id, since if f is a permutation, composing it three times gives f again, so f³ = f ⇒ f² = id. Wait, because if we compose both sides with f⁻¹, we get f² = id. Therefore, permutations satisfying f³ = f are exactly the involutions (permutations equal to their own inverse). However, in the symmetric group S₃, the involutions are the transpositions and the double transpositions, but in A₃, since it's the alternating group, the only permutations are the identity and the 3-cycles. Wait, but 3-cycles are not involutions. The identity is an involution, but in S₃, the transpositions (which are involutions) are odd permutations, so they are not in A₃. Therefore, in A₃, the only involution is the identity? But the identity permutation composed three times is still the identity, which satisfies id³ = id. So, if f is a permutation (i.e., bijective) satisfying f³ = f, then in A₃, the only such permutation is the identity.Wait, this seems conflicting. Let's clarify: if f is a permutation of the 3-element set A₃, then f³ = f implies that f² = id. Therefore, f must be an involution. However, in the symmetric group S₃, the involutions are transpositions (which are not in A₃) and double transpositions (which are products of two transpositions, hence even permutations, so in A₃). Wait, but in S₃, a double transposition like (1 2)(3 4) is not possible because we have only 3 elements. Wait, actually, in S₃, the only involutions are the transpositions (1 2), (1 3), (2 3), and the identity permutation. Wait, no, the identity is also an involution, but it's trivial. However, the double transposition in S₃ would require four elements, which we don't have. So, in S₃, the only non-trivial involutions are the transpositions, which are odd permutations. Therefore, in A₃, the only involution is the identity permutation. Therefore, the only permutation in A₃ satisfying f² = id is the identity. Hence, the only bijective function satisfying f³ = f is the identity function.But wait, let's confirm. Suppose f is a 3-cycle. Then f³ would be the identity, but since f is a 3-cycle, f³ = id, which is not equal to f unless f is the identity. But the identity is a 0-cycle. Wait, no: if f is a 3-cycle, then f³ = id, which is not equal to f (unless f is id). Therefore, no non-identity permutation in A₃ satisfies f³ = f. Therefore, the only permutation that satisfies f³ = f is the identity permutation. So, there is 1 such function.But wait, hold on. If f is the identity permutation, then f³ is also the identity, which equals f. So, yes, the identity permutation satisfies f³ = f.Therefore, in the case where the image has size 3 (i.e., f is a permutation), there is only 1 function: the identity permutation.But wait, is that all? Let me check again. Suppose f is a permutation, so bijective. Then f³ = f ⇒ f³ = f ⇒ f² = id, as composing both sides with f⁻¹. Therefore, f must satisfy f² = id. Therefore, f is an involution. But in A₃, as a group, the only involution is the identity because A₃ consists of the identity and two 3-cycles. The 3-cycles are not involutions since their order is 3. So, indeed, only the identity permutation is an involution in A₃. Therefore, only the identity permutation satisfies f³ = f.But wait, but in S₃, there are three transpositions and the identity as involutions, but in A₃, only the identity is an involution. Therefore, only the identity permutation satisfies f² = id. Therefore, only the identity permutation satisfies f³ = f.Therefore, when the image is the entire set (size 3), there's only 1 function: the identity.So, combining the cases:- Image size 1: constant functions, 3 in total.- Image size 2: 12 functions as calculated.- Image size 3: 1 function, the identity.Therefore, total functions satisfying f³ = f are 3 + 12 + 1 = 16.But wait, wait a second. Let's verify this count because sometimes overlaps can occur. For example, could a constant function be counted in image size 1 and also somewhere else? No, because the image size is uniquely determined. Similarly, the identity function is a permutation, so image size 3. The image size 2 functions are those that map two elements to themselves or swap them, and the third element to one of them. So, the counts seem distinct.But let's cross-validate this with another approach.Another way to approach this problem is to consider that for each element x in A₃, f³(x) = f(x). So, for each x, either f(x) is a fixed point (i.e., f(f(x)) = f(x)), or f(x) is part of a 2-cycle (i.e., f(f(x)) = x, but then f³(x) = f(x) implies that f(x) is part of a 2-cycle. Wait, but if f(f(x)) = x, then f³(x) = f(x). Wait, yes:If f(f(x)) = x, then f³(x) = f(f(f(x))) = f(x). So, elements in 2-cycles satisfy f³(x) = f(x). Similarly, if f(x) is a fixed point, then f³(x) = f(x). Therefore, the function f can be composed of fixed points and 2-cycles. However, since f is a function, not necessarily a permutation, we have to consider that some elements may map into the cycles.Wait, this is similar to the structure of idempotent functions, but here it's f³ = f instead of f² = f. Let's think in terms of graph theory. The function f can be represented as a directed graph where each node has out-degree 1. The condition f³ = f means that after three steps, you end up where you would after one step. This implies that the graph consists of cycles of length 1 or 2, and trees of depth at most 2 leading into these cycles.Wait, more precisely, each element x either:- Is a fixed point: f(x) = x.- Is part of a 2-cycle: f(x) = y, f(y) = x, and x ≠ y.- Or maps to a fixed point or a 2-cycle in one or two steps.But since f³(x) = f(x), starting from x, after three steps, we must reach f(x). Let's see:If x maps to y, which maps to z, which maps to w, then f³(x) = w. But we need w = f(x) = y. Therefore, this chain must collapse such that after three steps, we end up at y. So, z must map to y, so w = y. Therefore, the chain x → y → z → y. Then, this would mean that f³(x) = y = f(x). But then, what about f²(x) = z, and f³(x) = f(z) = y. So, as long as f(z) = y, this holds. So, z can map to y, which is a fixed point or part of a cycle.Wait, but if z maps to y, then y must satisfy f³(y) = f(y). If y is a fixed point, then f(y) = y, so f³(y) = y = f(y). If y is part of a 2-cycle, then f(y) = w, f(w) = y, so f³(y) = f(f(f(y))) = f(f(w)) = f(y) = w. Wait, but f³(y) should equal f(y). If f(y) = w, then f³(y) must equal w, which it does. So, as long as the elements beyond the first step are in cycles of length 1 or 2, then f³(x) = f(x) will hold.Therefore, the structure of such functions is that each element is either in a cycle of length 1 or 2, or eventually leads into such a cycle in one or two steps.Therefore, for the entire function f, the graph can have:- Fixed points (1-cycles).- 2-cycles.- Trees of depth 1 or 2 rooted at the cycles.But in our case, since the domain and codomain are the same 3-element set, let's enumerate the possibilities.Case 1: All elements are fixed points. This is the identity function, which we already counted.Case 2: There is one 2-cycle and one fixed point. For example, f(a) = b, f(b) = a, f(c) = c. This function satisfies f³ = f because applying it three times swaps a and b twice (which is the identity) and then once, resulting in the same as f.Wait, let's check: f³(a) = f(f(f(a))) = f(f(b)) = f(a) = b = f(a). Similarly, f³(b) = f(f(f(b))) = f(f(a)) = f(b) = a = f(b). And f³(c) = f(f(f(c))) = f(f(c)) = f(c) = c = f(c). So, yes, functions that are permutations composed of a 2-cycle and a fixed point satisfy f³ = f.But wait, in A₃, which is a set, not a group, we can have such permutations. However, in the symmetric group S₃, a transposition (2-cycle) is an odd permutation, but here we are just considering functions, not group elements. So, regardless of parity, such a function exists. However, since our domain is A₃, which as a set has 3 elements, we can certainly have such functions.But earlier, we considered permutations and concluded that only the identity permutation is in A₃ (as a group). Wait, no. Wait, confusion arises if A₃ is considered as the alternating group or as a 3-element set. If the problem is purely about functions on a 3-element set, then A₃ is just a set with 3 elements, and the functions don't need to be group homomorphisms. Therefore, a 2-cycle plus a fixed point is a valid permutation on the set, regardless of whether it's even or odd. However, the problem states ( f: A_3 rightarrow A_3 ). If A₃ is the alternating group, then the elements are permutations, and the functions would be maps between permutations. However, the problem doesn't specify that f is a group homomorphism, so f can be any function from the set A₃ to itself. Therefore, even if A₃ is the alternating group, the functions are just set functions, not necessarily homomorphisms. Therefore, the 2-cycle plus fixed point permutation is a valid function in this context, even though it's not an element of the alternating group.But hold on, if A₃ is the alternating group, which is a group of three elements: the identity, σ = (1 2 3), and σ² = (1 3 2). Then, the set A₃ consists of these three permutations. So, a function f: A₃ → A₃ maps these three permutations to themselves. The function could be a transposition of two elements and fixing the third, but since transpositions are not in A₃, but here f is just a function, not a group element. So, for example, f could send σ to σ², σ² to σ, and fix the identity. That would be a valid function, and such a function would be a permutation (a transposition) on the set A₃. However, in the group theory sense, this is not a group automorphism unless it preserves the group operation, but the problem doesn't require that. So, as a function, it's allowed.But in our earlier analysis, when considering permutations of the 3-element set A₃, we need to consider all possible permutations, not just the group elements. Therefore, there are 3! = 6 permutations of the set A₃. Among these, the ones that satisfy f³ = f are the involutions (permutations with f² = id). In the symmetric group S₃, the involutions are the transpositions and the identity. There are three transpositions and the identity, totaling four involutions. However, in our case, since A₃ is the set, not the group, the answer depends on whether the problem treats A₃ as a set or as the alternating group.Wait, this is crucial. If A₃ is the alternating group, then its elements are the three even permutations: e, σ, σ². Then, a function f: A₃ → A₃ is a function between these three group elements. The condition f³ = f must hold for each element. So, for each element x in A₃, f³(x) = f(x). However, the function f does not need to be a group homomorphism. Therefore, f can be any function, and we can represent it as a mapping between the three elements.Therefore, in this case, treating A₃ as a 3-element set, regardless of its group structure, we need to find all functions f: {e, σ, σ²} → {e, σ, σ²} such that f³ = f. The analysis we did earlier for a 3-element set applies here, regardless of the nature of the elements.Therefore, the total number of such functions is 16, as previously calculated: 3 constant functions, 12 functions with image size 2, and 1 identity function. However, when we considered permutations, we thought only the identity permutation satisfies f³ = f, but in reality, if we consider all permutations of the 3-element set (not just the group elements), there are more involutions. Wait, this is a contradiction. Let's resolve this.If A₃ is a 3-element set, then the symmetric group on it is S₃, which has 6 elements: the identity, two 3-cycles (which are even permutations), and three transpositions (which are odd permutations). However, if we are considering functions f: A₃ → A₃ where A₃ is a set, then the fact that the elements might be permutations is irrelevant. The functions are just mappings between three elements.Therefore, the earlier analysis holds: the functions satisfying f³ = f are those where each element is either a fixed point or part of a 2-cycle, and other elements map into these in at most two steps. Therefore, the count would be as follows:1. Constant functions: 3.2. Functions with image size 2: For each pair of elements {a, b}, we can have: a. The function acts as the identity on {a, b} and maps the third element to a or b. b. The function swaps a and b and maps the third element to a or b. Each pair contributes 4 functions, and there are 3 pairs, totaling 12.3. Permutations: These are bijections. As we saw, permutations satisfying f³ = f must satisfy f² = id. In the symmetric group S₃, the involutions (elements of order 2) are the transpositions and the identity. However, the identity permutation is the only even permutation (i.e., element of A₃) that is an involution. The transpositions are odd permutations, so they are not in A₃. But wait, no, the set A₃ here is just a set, not a group. So, if we are talking about permutations of the set A₃ (which has three elements), then the involutions are the identity, and the three transpositions (which swap two elements and fix the third). However, if the problem is about functions on the set A₃, regardless of group structure, then these transpositions are valid permutations.Wait, I think there's confusion here between A₃ as a group and A₃ as a set. The problem states ( f: A_3 rightarrow A_3 ). If A₃ is the alternating group, then it's a group with three elements, but the functions are not required to be group homomorphisms. Therefore, we are simply dealing with functions from a 3-element set to itself. The elements of the set happen to be permutations, but the function f doesn't need to respect the group operation. Therefore, the analysis should proceed as for any 3-element set.Therefore, in this context, permutations of the set A₃ (considered as a set) are all bijections from the set to itself, which form the symmetric group S₃, regardless of the nature of the elements. Therefore, in this case, the involutions in S₃ are the identity and the three transpositions. Each transposition swaps two elements and fixes the third. Therefore, there are three non-identity involutions. Each of these satisfies f² = id, hence f³ = f. Therefore, these three transpositions, along with the identity permutation, make four permutations that satisfy f³ = f.But earlier, we considered only the identity permutation. That was a mistake because we were conflating A₃ as a group with A₃ as a set. If we treat A₃ as a set, then the permutations are just bijections on three elements, and the involutions include the transpositions. Therefore, there are four permutations satisfying f³ = f: the identity and the three transpositions.Therefore, correcting our earlier count:- Image size 3: There are four permutations (identity and three transpositions) satisfying f³ = f.Wait, but hold on: if f is a transposition, say swapping a and b and fixing c, then f² is the identity, so f³ = f² ∘ f = id ∘ f = f. Therefore, transpositions do satisfy f³ = f. Therefore, there are three such transpositions plus the identity, totaling four permutations.Therefore, revising the earlier count:- Image size 1: 3 functions.- Image size 2: previously 12 functions, but wait, if we have permutations with image size 3, which are four in total, then the image size 2 functions are different.Wait, no. If the function has image size 2, it's not a permutation. The permutations have image size 3. So, the previous count of 12 functions with image size 2 remains, and the permutations with image size 3 are four. Then, adding the constant functions (image size 1), the total is 3 + 12 + 4 = 19.But this contradicts the previous conclusion. Clearly, there's confusion here. Let's clarify.First, let's separate the concepts:1. Constant functions: Image size 1. There are 3 such functions.2. Non-surjective functions with image size 2: These are functions that map two elements to themselves or to each other and map the third element to one of the two. As analyzed earlier, for each pair of elements, we have 4 functions (identity on the pair + two choices for the third element; swap on the pair + two choices for the third element). So, 3 pairs × 4 functions = 12 functions.3. Permutations (image size 3): These are bijections. As we determined, the permutations satisfying f³ = f are the involutions, which are the identity and the three transpositions. So, 4 functions.Therefore, total functions: 3 + 12 + 4 = 19.But earlier, we thought permutations would only contribute 1 function (the identity), but that was under the mistaken assumption that A₃ is the alternating group, hence only containing even permutations. However, in reality, the problem is about functions on a 3-element set, regardless of the group structure. Therefore, the correct count should include all involutions, which are four permutations.But wait, there's a discrepancy here. Let's verify with a small example. Take the set {a, b, c}. Consider the transposition f that swaps a and b and fixes c. Then f² is the identity, so f³ = f. Indeed, f³(x) = f(f(f(x))) = f(f²(x)) = f(x). Therefore, this function satisfies the condition. Similarly, the other transpositions and the identity also satisfy it.Therefore, the correct total number of functions is 3 (constants) + 12 (image size 2) + 4 (permutations) = 19.But wait, this contradicts our previous calculation where we considered permutations in A₃ (the group) and thought there's only the identity. Therefore, the confusion arises from whether A₃ is a group or a set. The problem states ( f: A_3 rightarrow A_3 ). In standard mathematical notation, ( A_n ) usually denotes the alternating group on n elements. However, the alternating group ( A_3 ) is indeed a set of three elements (the even permutations), and the problem is about functions from this set to itself, not necessarily homomorphisms. Therefore, the elements of ( A_3 ) are three permutations, but the function f is just a map between these three elements, regardless of their group structure.Therefore, in this context, the set ( A_3 ) has three elements, say {e, σ, σ²}, where σ is a 3-cycle. The functions f: {e, σ, σ²} → {e, σ, σ²} can be any functions, and we need to count those satisfying f³ = f. Therefore, the analysis is the same as for any 3-element set. Hence, the total number of such functions is 19.But wait, this contradicts the standard result for functions on a 3-element set satisfying f³ = f. Let's verify this with another approach.For a set with n elements, the number of functions f satisfying f^k = f is known in combinatorics. For k=3, it's a specific case. However, I don't recall the exact formula, but for small n, we can compute it manually.For n=3, the total number of functions is 3^3 = 27. We need to subtract the functions that do not satisfy f³ = f. But it's easier to count directly.Each element x must satisfy f³(x) = f(x). This means that the sequence x, f(x), f²(x), f³(x), ... must stabilize after three steps. But since f³(x) = f(x), the sequence becomes periodic from f(x) onward with period 2 or 1.For each element x, the possible behaviors are:1. x is a fixed point: f(x) = x.2. x is part of a 2-cycle: f(x) = y, f(y) = x, and x ≠ y.3. x is a non-cyclic element that maps to a fixed point or a 2-cycle in one step.However, in a 3-element set, the possibilities are limited. Let's model the function as a directed graph where each node has out-degree 1.For the function to satisfy f³ = f, the graph must be such that after three steps, you reach the same node as after one step. This implies that each node is either in a cycle of length 1 or 2, or it points to a node in such a cycle within one step.So, possible components:- Fixed points (cycles of length 1).- 2-cycles (cycles of length 2).- Trees leading into these cycles with depth at most 2.But since there are only 3 elements, the possible structures are limited.Let's enumerate all possible functions:1. All elements are fixed points: This is the identity function. 1 function.2. Two fixed points and one 2-cycle: Impossible, because a 2-cycle requires two elements, leaving one fixed point. So, there are C(3,2) = 3 possible 2-cycles, each paired with one fixed point. Each 2-cycle corresponds to a transposition. So, 3 functions.Wait, but these are permutations (bijections). However, the previous analysis counted 4 permutations (identity + three transpositions). So, this accounts for 4 functions.3. Functions with a single 2-cycle and one element mapping into the 2-cycle:For example, suppose we have a 2-cycle between a and b, and c maps to a. Then, f(a)=b, f(b)=a, f(c)=a. This function is not a permutation because c maps to a, but a and b form a 2-cycle. This function satisfies f³ = f:- f³(a) = f(f(f(a))) = f(f(b)) = f(a) = b = f(a)- f³(b) = f(f(f(b))) = f(f(a)) = f(b) = a = f(b)- f³(c) = f(f(f(c))) = f(f(a)) = f(b) = a = f(c)So, yes, it satisfies the condition. Similarly, c could map to b instead. Therefore, for each 2-cycle, there are two such functions where the third element maps to either member of the 2-cycle. Since there are three 2-cycles, each contributing two functions, this gives 3 × 2 = 6 functions.4. Functions with all elements mapping to a single fixed point: These are the constant functions. There are 3 such functions (one for each element).5. Functions where two elements map to a fixed point, and the third maps to itself or the other:Wait, for example, suppose a and b map to c, and c maps to c. This is a constant function to c. Already counted in case 4.Alternatively, suppose a maps to b, b maps to c, and c maps to c. Let's check f³ = f:- f(a) = b- f²(a) = f(b) = c- f³(a) = f(c) = cBut f(a) = b ≠ c = f³(a). So, this does not satisfy f³ = f. Therefore, this function is invalid.Therefore, such chains longer than one step are not allowed because they would violate f³ = f. Therefore, elements not in cycles must map directly to a cycle.6. Functions with one fixed point and two elements mapping to it:For example, f(a)=a, f(b)=a, f(c)=a. This is a constant function to a, already counted in case 4.Another example: f(a)=a, f(b)=a, f(c)=b. Let's check f³ = f:- f(c)=b, f²(c)=f(b)=a, f³(c)=f(a)=a. But f(c)=b ≠ a=f³(c). So, this doesn't satisfy the condition. Therefore, invalid.Therefore, the only valid functions are those where elements not in cycles map directly to cycles in one step. Therefore, the valid functions are:- All constant functions: 3.- All permutations that are involutions (identity and three transpositions): 4.- For each transposition, two functions where the third element maps to either member of the transposition: 3 transpositions × 2 = 6.But wait, earlier I considered for each pair (which can form a transposition), two choices for the third element. So, 3 pairs × 2 (swap or identity on the pair) × 2 choices for the third element? Wait, no, let's clarify.Wait, in case 3 above, I considered that for each 2-cycle (transposition), we can have the third element map to either member of the cycle. That gives 3 × 2 = 6 functions. However, in case 2, the permutations (transpositions) are 3 functions where the third element is fixed. So, total functions with image size 2 are 6 (non-permutation) + 3 (permutation transpositions)? No, wait, the transpositions are permutations with image size 3. Wait, no: a transposition is a permutation, so it has image size 3. For example, swapping a and b and fixing c is a permutation; hence, the image is the entire set.Wait, this is where the confusion is. If we have a transposition permutation, it's a bijection, so all elements are in the image. However, the functions where two elements form a 2-cycle and the third maps to one of them are not permutations, since they're not injective. For example, f(a)=b, f(b)=a, f(c)=a. This is not injective, hence not a permutation. Therefore, these are distinct from the transposition permutations.Therefore, total functions:- Constant functions: 3.- Non-injective, non-surjective functions with image size 2: For each pair {a, b}, we can have: a. f(a)=a, f(b)=a, f(c)=a or b (but wait, if the image is {a, b}, then f(c) must be a or b. However, if we want the function to satisfy f³ = f, we need to ensure that on the image {a, b}, f behaves as an involution. So, for example, if we choose the image to be {a, b}, then f restricted to {a, b} can be: i. Identity: f(a)=a, f(b)=b. Then f(c) can be a or b. This gives 2 functions. ii. Swap: f(a)=b, f(b)=a. Then f(c) can be a or b. This gives 2 functions. Therefore, for each pair, 4 functions. With 3 pairs, 12 functions.- Permutations: 4 functions (identity and three transpositions).But wait, 3 + 12 + 4 = 19. But earlier, we thought the answer might be 16. However, according to this breakdown, it's 19. There's inconsistency here. Let's check with another method.Alternative approach: For each element x in A₃, f(x) can be any of the 3 elements. However, we have the constraint that f³(x) = f(x) for all x. Let's consider building such functions.For each element x, consider the sequence x, f(x), f²(x), f³(x). The constraint is that f³(x) = f(x). Therefore, the sequence must satisfy f³(x) = f(x), which implies that f²(f(x)) = f(x). Therefore, the element f(x) must be either a fixed point or part of a 2-cycle.Therefore, for each x, f(x) can be:1. A fixed point: f(f(x)) = f(x). Then, f²(x) = f(x), and f³(x) = f(f²(x)) = f(f(x)) = f(x), which satisfies the condition.2. Part of a 2-cycle: f(f(x)) = y and f(y) = f(x), forming a 2-cycle between f(x) and y. Wait, no: if f(x) is part of a 2-cycle, then we must have f(f(x)) = y and f(y) = f(x), but y must be f(f(x)). So, if f(x) = a, then f(a) can be b, and f(b) = a. Then, f³(x) = f(f(f(x))) = f(f(a)) = f(b) = a = f(x), which satisfies the condition.Therefore, each element x can map to either a fixed point or to an element involved in a 2-cycle. However, the choices are interdependent because the images must form a consistent function.This seems complex, but perhaps we can model this as follows:Each element must map to a fixed point or to one element of a 2-cycle. Additionally, if an element maps to a member of a 2-cycle, then the target of that element must map back to complete the 2-cycle.Therefore, the function can be constructed by:1. Selecting a set of fixed points F.2. Selecting a set of 2-cycles C.3. The remaining elements (if any) must map to either F or C.But since the domain and codomain are the same 3-element set, and we need to cover all elements, this complicates things.Alternatively, think in terms of graph components. Each component is either a fixed point or a 2-cycle with trees of height 1 feeding into them.Given there are 3 elements, the possible component structures are:1. All three elements are fixed points. (1 function)2. One 2-cycle and one fixed point. The 2-cycle uses two elements, and the third is a fixed point. (3 possible 2-cycles, each with one fixed point. As permutations, these are the three transpositions.)3. One 2-cycle and one element mapping into the 2-cycle. For example, two elements form a 2-cycle, and the third maps to one of them. As non-permutation functions, these are 3 pairs × 2 choices = 6 functions.4. All three elements map to a single fixed point. (3 constant functions)5. Two elements map to one fixed point, and the third maps to another fixed point. For example, f(a)=a, f(b)=a, f(c)=b. But this function doesn't satisfy f³ = f, as checked earlier. So, invalid.Therefore, valid structures are 1, 2, 3, and 4.Count:1. 1 function (identity).2. 3 functions (transpositions).3. 6 functions (non-injective with image size 2).4. 3 functions (constant).Total: 1 + 3 + 6 + 3 = 13.Wait, this is different again. Now we're getting 13. There's clearly confusion in the methodology. Let's try to systematically list all possible functions.Let’s denote the three elements as 0, 1, 2 for simplicity.Case 1: All elements are fixed points.Only one function: f(0)=0, f(1)=1, f(2)=2.Case 2: Exactly two elements are fixed points, and one forms a 2-cycle with another.Wait, but with three elements, a 2-cycle requires two elements. If two are fixed, the third must also be fixed, so this case is impossible. Therefore, the only permutations with fixed points are those with one fixed point and one 2-cycle (the transpositions).Case 3: One 2-cycle and one fixed point (transpositions).There are three such permutations:- f(0)=1, f(1)=0, f(2)=2.- f(0)=2, f(2)=0, f(1)=1.- f(1)=2, f(2)=1, f(0)=0.Case 4: Functions with a 2-cycle and one element mapping into the cycle.For example:- Cycle 0 ↔ 1, and 2 maps to 0. Then f(0)=1, f(1)=0, f(2)=0. Check f³(2) = f(f(f(2))) = f(f(0)) = f(1) = 0 = f(2).- Similarly, 2 could map to 1 instead: f(2)=1. Then f³(2) = f(f(f(2))) = f(f(1)) = f(0) = 1 = f(2).Similarly, for the other 2-cycles:- Cycle 0 ↔ 2, and 1 maps to 0 or 2.- Cycle 1 ↔ 2, and 0 maps to 1 or 2.Each 2-cycle has two such functions, so 3 × 2 = 6 functions.Case 5: Constant functions.Each element maps to a single value:- f(x)=0 for all x.- f(x)=1 for all x.- f(x)=2 for all x.Case 6: Functions where two elements map to a fixed point, and the third maps to another fixed point.For example, f(0)=0, f(1)=0, f(2)=1. Let's check f³(2) = f(f(f(2))) = f(f(1)) = f(0) = 0 ≠ f(2)=1. So, invalid.Similarly, any function where an element maps to a non-fixed point which is not part of a 2-cycle will fail.Case 7: Functions with all elements mapping to a 2-cycle element but not forming a cycle.For example, f(0)=1, f(1)=1, f(2)=1. This is a constant function, already counted.Another example: f(0)=1, f(1)=2, f(2)=1. Then f³(0) = f(f(f(0))) = f(f(1)) = f(2) = 1, but f(0)=1, so f³(0)=1=f(0). For x=1: f³(1)=f(f(f(1)))=f(f(2))=f(1)=2 ≠ f(1)=2? Wait, f(1)=2, f(f(1))=f(2)=1, f³(1)=f(1)=2. Wait, f³(1)=f(f(f(1)))=f(f(2))=f(1)=2. But f(1)=2, so f³(1)=2=f(1). Similarly, f³(2)=f(f(f(2)))=f(f(1))=f(2)=1=f(2). Therefore, this function satisfies f³ = f.But this function is f(0)=1, f(1)=2, f(2)=1. Let's analyze its structure:- 0 ↦ 1 ↦ 2 ↦ 1, so after three steps, we get back to 1, which is f(0)=1.- 1 ↦ 2 ↦ 1 ↦ 2, so f³(1)=2=f(1).- 2 ↦ 1 ↦ 2 ↦ 1, so f³(2)=1=f(2).So, this function does satisfy f³ = f. However, it's not accounted for in previous cases. It has image {1, 2}, where 1 and 2 form a 2-cycle, and 0 maps to 1. But wait, in this function, f(1)=2 and f(2)=1, forming a 2-cycle, and f(0)=1. This is similar to case 4, but here, the element 0 maps to 1, which is part of the 2-cycle. Therefore, this should be included in case 4. Wait, in case 4, we considered functions where two elements form a 2-cycle and the third maps to one of them. This function is exactly that: 1 and 2 form a 2-cycle, and 0 maps to 1. Similarly, 0 could map to 2. Therefore, this is already counted in case 4.Similarly, the function f(0)=2, f(1)=0, f(2)=0. Here, 0 and 1 form a 2-cycle (since f(0)=2, wait, no: f(0)=2, f(2)=0, so 0 and 2 form a 2-cycle, and f(1)=0. Wait, this function has a 2-cycle between 0 and 2, and 1 maps to 0. This is also covered in case 4.Therefore, case 4 indeed covers these functions.Case 8: Functions with a 3-cycle. For example, f(0)=1, f(1)=2, f(2)=0. Then f³(0)=f(f(f(0)))=f(f(1))=f(2)=0, which is not equal to f(0)=1. So, this function does not satisfy f³ = f. Similarly, the inverse 3-cycle also doesn't satisfy the condition. Therefore, 3-cycles are excluded.Case 9: Functions with a fixed point and two elements forming a chain leading to it. For example, f(0)=1, f(1)=0, f(2)=0. This is similar to case 4, but with 0 and 1 forming a 2-cycle and 2 mapping to 0. This is already covered in case 4.After considering all cases, the valid functions are:- 1 identity function.- 3 transposition permutations.- 6 functions with a 2-cycle and one element mapping into it.- 3 constant functions.Total: 1 + 3 + 6 + 3 = 13 functions.But wait, earlier I thought it was 19, but now according to this breakdown, it's 13. There's a discrepancy because of differing interpretations of the cases. Let's attempt to manually list all possible functions for a concrete example.Let’s take the set {0, 1, 2}.Constant functions (3):1. f(x)=02. f(x)=13. f(x)=2Transposition permutations (3):4. f(0)=1, f(1)=0, f(2)=25. f(0)=2, f(2)=0, f(1)=16. f(1)=2, f(2)=1, f(0)=0Identity permutation (1):7. f(x)=xFunctions with a 2-cycle and one element mapping into it (6):8. f(0)=1, f(1)=0, f(2)=09. f(0)=1, f(1)=0, f(2)=110. f(0)=2, f(2)=0, f(1)=011. f(0)=2, f(2)=0, f(1)=212. f(1)=2, f(2)=1, f(0)=113. f(1)=2, f(2)=1, f(0)=2So, that's a total of 13 functions. Therefore, the correct count is 13.But this contradicts our earlier count of 19. Where did we go wrong before? The error was in assuming that for each pair, we have 4 functions (identity on the pair and swap on the pair, each with two choices for the third element). However, when the function is a swap on the pair (transposition) and the third element maps to one of the pair, that is already counted in the transposition permutations if it's a bijection. However, in reality, the transposition permutations are bijections, so they require the third element to be fixed. Therefore, the functions where the third element maps to a different element are not permutations.Let me clarify:For each pair {a, b}:- If we define f as the identity on {a, b}, then f(a)=a, f(b)=b. The third element c can map to a or b. This gives two functions per pair: 3 pairs × 2 = 6 functions.- If we define f as a swap on {a, b}, then f(a)=b, f(b)=a. The third element c can map to a or b. This gives two functions per pair: 3 pairs × 2 = 6 functions.However, among these, the cases where the third element is mapped to a or b and the swap is performed are not permutations unless c maps to itself. But in this case, if we swap a and b and map c to a or b, the function is not injective, hence not a permutation. Therefore, these are distinct from the transposition permutations. The transposition permutations are bijections where c is fixed.Therefore, the 6 functions where we have a swap on {a, b} and c maps to a or b are non-injective, hence not permutations. The transposition permutations are a separate case where c is fixed. Therefore, total functions:- Constant functions: 3.- Identity permutation: 1.- Transposition permutations: 3.- Image size 2 functions (identity on pair): 3 pairs × 2 (choices for c) = 6.- Image size 2 functions (swap on pair): 3 pairs × 2 (choices for c) = 6.Total: 3 + 1 + 3 + 6 + 6 = 19.But this contradicts the manual count of 13. The issue arises because the image size 2 functions where we have a swap on the pair and map c to a or b are different from the transposition permutations. However, in the manual count, we considered that the transposition permutations fix the third element, whereas the non-injective functions map the third element into the cycle.Therefore, let's reconcile this:- Transposition permutations (bijections): 3 functions where two elements are swapped and the third is fixed.- Non-injective functions with swap on pair and third element mapped to pair: 3 pairs × 2 choices = 6 functions.- Non-injective functions with identity on pair and third element mapped to pair: 3 pairs × 2 choices = 6 functions.- Constant functions: 3.- Identity permutation: 1.Total: 3 + 6 + 6 + 3 + 1 = 19.But when we manually listed, we only found 13 functions. The discrepancy is because some of these functions are being double-counted or misclassified.Wait, in the manual list, we had:- 3 constants.- 3 transpositions.- 1 identity.- 6 non-injective functions with swap and third element mapped.That totals 13. The missing 6 functions are the non-injective functions with identity on the pair and third element mapped. For example, for pair {0,1}, identity on {0,1}, and 2 maps to 0 or 1. So:14. f(0)=0, f(1)=1, f(2)=015. f(0)=0, f(1)=1, f(2)=1Similarly for pairs {0,2} and {1,2}:16. f(0)=0, f(2)=2, f(1)=017. f(0)=0, f(2)=2, f(1)=218. f(1)=1, f(2)=2, f(0)=119. f(1)=1, f(2)=2, f(0)=2These are additional 6 functions where two elements are fixed, and the third maps to one of them. These functions satisfy f³ = f because:For example, function 14: f(2)=0, which is a fixed point. Therefore, f³(2) = f(f(f(2))) = f(f(0)) = f(0) = 0 = f(2). Similarly, the other elements are fixed, so they satisfy the condition.Therefore, these 6 functions are valid and were missing from our initial manual count. Therefore, the correct total is 19 functions.Therefore, the correct answer is 19 such functions.But let's verify one of these new functions. Take function 14: f(0)=0, f(1)=1, f(2)=0.Check f³ = f:- For 0: f³(0) = f(f(f(0))) = f(f(0)) = f(0) = 0 = f(0).- For 1: Similarly, f³(1) = 1 = f(1).- For 2: f³(2) = f(f(f(2))) = f(f(0)) = f(0) = 0 = f(2).Yes, it satisfies the condition. Similarly for the others.Therefore, the total number of functions is indeed 19.But this contradicts some of our previous counts. Why? Because initially, we forgot to consider the functions where two elements are fixed and the third maps to one of them, which are not constant functions but still satisfy f³ = f.Therefore, the breakdown is:- Constant functions: 3.- Identity permutation: 1.- Transposition permutations: 3.- Non-injective functions with a 2-cycle and third element mapped: 6.- Non-injective functions with identity on a pair and third element mapped: 6.Total: 3 + 1 + 3 + 6 + 6 = 19.Therefore, the correct number of functions satisfying f³ = f is 19.However, this answer seems high. Let's cross-validate with another approach. Each element must be either a fixed point or part of a 2-cycle, or map in one step to a fixed point or a 2-cycle. For a 3-element set, the possibilities are:1. All three elements are fixed points: 1 function.2. One 2-cycle and one fixed point: 3 permutations (transpositions).3. One 2-cycle and one element mapping to the cycle: 3 pairs × 2 choices = 6 functions.4. All elements map to a single fixed point: 3 constant functions.5. Two elements map to one fixed point, and the third maps to another: However, this creates a chain which may violate the condition, but in reality, if two elements map to a fixed point and the third also maps to a fixed point (possibly the same or different), this can be valid.Wait, for example, f(0)=0, f(1)=0, f(2)=0: constant function.f(0)=0, f(1)=0, f(2)=1: This maps 0→0, 1→0, 2→1. Let's check f³(2) = f(f(f(2))) = f(f(1)) = f(0) = 0 ≠ f(2)=1. So, invalid.f(0)=0, f(1)=0, f(2)=0: valid (constant).f(0)=0, f(1)=0, f(2)=0: same as above.Therefore, the only valid functions where two elements map to a fixed point are the constant functions.Therefore, the only valid non-constant, non-permutation functions are those with a 2-cycle and one element mapping into it (6 functions) and those with two fixed points and one element mapping to one of them (6 functions). However, the latter case requires that the third element maps to a fixed point, which would mean the function has image size 2 (if two elements are fixed and the third maps to one of them), but these functions are covered in case 5 above.Wait, for example, f(0)=0, f(1)=1, f(2)=0: image is {0,1}, which has size 2. This function is covered in the non-injective functions with identity on pair {0,1} and third element mapped to 0.Similarly, f(0)=0, f(1)=1, f(2)=1: image {0,1}.These are the 6 functions where a pair is fixed, and the third element maps to either member of the pair. Therefore, these are valid and satisfy f³ = f.Therefore, the count of 19 stands.But this contradicts standard results. Let's look for references or mathematical literature. For a set of size n, the number of functions f with f^k = f is known. For k=3, it's a specific case. However, I don't have the exact formula at hand, but for n=3, enumerating manually should be feasible.Given that we've systematically identified 19 functions and they all satisfy f³ = f, and cross-validated with multiple approaches, it seems that the correct answer is 19. However, this contradicts the initial intuition and requires careful verification.But let's consider the following: each element has three choices for its image, but with constraints. The constraints are that the image of each element must be a fixed point or part of a 2-cycle. Additionally, if an element maps to a member of a 2-cycle, the target must map back.This is complex, but the total number of such functions can be calculated using the formula for the number of idempotent functions, but here it's f³ = f instead of f² = f. However, the structure is more involved.Alternatively, for each element, we can determine the number of choices it has for its image, considering the constraints.An element x can map to:1. Itself (fixed point).2. Another element y, which must then map back to x or to another element z such that z maps to y or is a fixed point.But this quickly becomes recursive. Instead, let's consider that the entire function's graph must consist of cycles of length 1 or 2, and trees of depth at most 2 leading into these cycles.For 3 elements, the possible structures are:1. Three fixed points. (1 function)2. One fixed point and one 2-cycle. (3 permutations)3. One 2-cycle and one element mapping to the cycle. (6 functions)4. All elements mapping to a single fixed point. (3 functions)5. Two elements mapping to one fixed point and the third mapping to another fixed point. However, as seen earlier, this creates a chain that violates the condition unless all mappings are to the same fixed point. Hence, only constant functions.Wait, but we already considered two elements mapping to a fixed point and the third also mapping to a fixed point in case 5, which results in a constant function if all map to the same, but invalid if they map to different ones.Therefore, the only valid non-constant functions with image size 2 are those where the image set forms a fixed point and another element, or a 2-cycle with an element mapping into it.However, the case where two elements are fixed and the third maps to one of them (image size 2) are valid and contribute 6 functions. These are separate from the constant functions.Therefore, the total count is indeed 3 (constants) + 1 (identity) + 3 (transpositions) + 6 (2-cycle with third element) + 6 (two fixed points and third element mapped) = 19.Therefore, after careful consideration and verification, the correct number of functions satisfying f³ = f is 19.However, I must acknowledge that different sources may present conflicting information, and sometimes these counts are non-trivial. Given the systematic breakdown and manual verification, I'm confident the answer is 19.

✅:The number of functions ( f: A_3 rightarrow A_3 ) satisfying ( f^{(3)} = f ) is 19. These functions can be categorized as follows:1. Constant Functions (3): Each element maps to a single fixed element. - ( f(x) = a ) - ( f(x) = b ) - ( f(x) = c )2. Identity Permutation (1): Every element maps to itself. - ( f(a) = a, , f(b) = b, , f(c) = c )3. Transposition Permutations (3): Swaps two elements and fixes the third. - ( f(a) = b, , f(b) = a, , f(c) = c ) - ( f(a) = c, , f(c) = a, , f(b) = b ) - ( f(b) = c, , f(c) = b, , f(a) = a )4. Non-Injective Functions with a 2-Cycle (6): Two elements form a 2-cycle, and the third maps to one of them. - For cycle ( a leftrightarrow b ): ( f(a) = b, , f(b) = a, , f(c) = a ) or ( f(c) = b ) - For cycle ( a leftrightarrow c ): ( f(a) = c, , f(c) = a, , f(b) = a ) or ( f(b) = c ) - For cycle ( b leftrightarrow c ): ( f(b) = c, , f(c) = b, , f(a) = b ) or ( f(a) = c )5. Non-Injective Functions with Fixed Pair (6): Two elements are fixed, and the third maps to one of them. - For pair ( {a, b} ): ( f(a) = a, , f(b) = b, , f(c) = a ) or ( f(c) = b ) - For pair ( {a, c} ): ( f(a) = a, , f(c) = c, , f(b) = a ) or ( f(b) = c ) - For pair ( {b, c} ): ( f(b) = b, , f(c) = c, , f(a) = b ) or ( f(a) = c )Final Answer: The number of such functions is (boxed{19}).

🔑:Given the function ( f: A_{3} rightarrow A_{3} ) (where ( A_{3} ) is the set with elements {1, 2, 3}), we are required to find functions ( f ) such that ( f^{(3)} = f ). Here, ( f^{(3)} ) represents applying the function ( f ) three times.To solve the problem, we'll use the fact that ( f^{(3)} = f ). This indicates that ( f ) must be of a form that results in the same function after iterating three times. We shall use cycle notation, considering permutations.1. Identify the Divisors of ( 3 - 1 = 2 ): [ k_i mid 3-1 implies k_i mid 2 ] The divisors of 2 are 1 and 2.2. Check Each Divisor (Cycle Length): - For ( k_i = 1): [ x_{1} = 1, 2, 3 ] Each element maps to itself, which yields: [ begin{array}{ccc} 1 & 2 & 3 1 & 2 & 3 end{array} ] This gives us the following functions: [ f(1) = 1, f(2) = 2, f(3) = 3 ] - For ( k_i = 2): [ x_{1}, x_{2} in {1, 2, 3} ] Starting combinations: - (x_{1} = 1): - ( x_{2} = 2 ) yields: [ begin{array}{ccc} 1 & 2 & 3 1 & 2 & 1 end{array} ] - ( x_{2} = 3 ) yields: [ begin{array}{ccc} 1 & 2 & 3 1 & 3 & 1 end{array} ] - Other combinations similarly yield mappings. We thus need to enumerate all valid permutations.3. Account for and Calculate All Possible Functions: - Case (k = 1): [ text{Arrangements for } A(3,1) = 1 ] - Case (k = 2): [ A(3,2) = frac{2!}{1^2 cdot 2!} + frac{2!}{2^1 cdot 1!} = 2 ] - Case (k = 3): [ A(3,3) = frac{3!}{1^3 cdot 3!} + frac{3!}{1^1 cdot 2^1 cdot 1! cdot 1!} = 4 ]4. Combine Results: Using the formula for counting composite functions: [ B(3,3) = 1^{3-1} binom{3}{1} A(3,1) + 2^{3-2} binom{3}{2} A(3,2) + 3^{3-3} binom{3}{3} A(3,3) ] Calculating each term: [ B(3,3) = 1 cdot 3 cdot 1 + 2 cdot 3 cdot 2 + 1 cdot 1 cdot 4 = 3 + 12 + 4 = 19 ]5. Enumerate all 19 Functions: Listing functions, each case corresponds to the explicit mappings provided in the reference problem. Conclusion:Thus, there are 19 such functions ( f: A_{3} rightarrow A_{3} ) that satisfy ( f^{(3)} = f ).[boxed{19}]

🧠:Okay, so I need to figure out how many integers between 2020 and 2400 have digits in strictly increasing order with all digits distinct. Let me start by understanding the problem properly. First, strictly increasing digits mean that each digit is larger than the one before it. For example, 2357 is valid because 2 < 3 < 5 < 7. But something like 2335 wouldn't work because the digit 3 repeats, and 2353 is not in order. Also, all digits must be distinct. So, the number must be a four-digit number since we're dealing with numbers between 2020 and 2400, which are all four-digit numbers. Wait, hold on, 2020 is a four-digit number, and 2400 is also four digits. So all numbers in this range are four-digit numbers. Therefore, each number we consider will have four digits. Let's confirm that. The smallest number is 2020, which is four digits, and the largest is 2400, also four digits. So yes, all numbers in this range are four-digit numbers. Now, the problem is to count all four-digit numbers between 2020 and 2400 where each digit is strictly increasing and all digits are distinct. Let me break this down. First, let's consider the range. The numbers start at 2020 and go up to 2400. So, the thousands digit is either 2. Then, the hundreds digit ranges from 0 to 3 (since 2400 is the upper limit). Wait, but 2400 is the upper limit, so numbers up to 2399 would have thousands digit 2, hundreds digit 3, and then the tens and units digits can vary. But since the upper limit is 2400, we need to check if 2400 is included. The problem says "between 2020 and 2400," so does that include 2400? The wording is a bit ambiguous. But since 2400 has digits 2, 4, 0, 0, which are not strictly increasing and have repeating digits, it wouldn't count even if it were included. So maybe the upper bound is up to 2399? But let's confirm the exact range. The problem says "between 2020 and 2400," which in mathematical terms can be interpreted as [2020, 2400). That is, inclusive of 2020 but exclusive of 2400. However, sometimes "between" can be inclusive on both ends. But since 2400 does not satisfy the digit condition, whether it's included or not doesn't matter. So we can consider numbers from 2020 up to 2399. But let's get back to the digits. The thousands digit is fixed at 2 for all numbers in this range. So the first digit is 2. The next digit, the hundreds place, can be from 0 to 3 because 2020 has a hundreds digit of 0 and 2399 has a hundreds digit of 3. But wait, in the case of 2020, the hundreds digit is 0, but numbers like 2020 itself would have digits 2, 0, 2, 0 which are not strictly increasing. So we need to find numbers where each subsequent digit is larger than the previous one. So the first digit is 2. Then the hundreds digit (second digit) must be greater than 2? Wait, no. Wait, the digits must be strictly increasing. So if the first digit is 2, the second digit must be greater than 2. But wait, in the number 2020, the hundreds digit is 0, which is less than 2. Therefore, 2020 is not a strictly increasing number. So actually, for a four-digit number with strictly increasing digits starting with 2, the second digit must be greater than 2, the third digit greater than the second, and the fourth greater than the third. Therefore, any number in the range 2020-2399 that has strictly increasing digits must have the hundreds digit (second digit) greater than 2. So the hundreds digit can be 3, 4, ..., but since the upper limit is 2400, the hundreds digit can only go up to 3. Wait, because 2400 is the upper limit, so numbers up to 2399. So the hundreds digit can be 0 to 3. But if the digits have to be strictly increasing, starting with 2, then the hundreds digit must be at least 3? Wait, no. Wait, let's think again. If the thousands digit is 2, then the hundreds digit must be greater than 2 to have strictly increasing digits. So possible hundreds digits are 3, 4, 5, 6, 7, 8, 9. But since the upper limit is 2400, which is 2 4 0 0, but our numbers can only go up to 2399. So the hundreds digit can be 0, 1, 2, 3. However, if the hundreds digit is greater than 2, that is 3, then the number would be 23XY, where X and Y are digits greater than 3. But wait, 23XY would have to be between 2300 and 2399, but with X and Y such that 3 < X < Y. However, if the hundreds digit is 3, then the next digit (tens place) must be greater than 3. Similarly, if the hundreds digit was 0, 1, or 2, but since the thousands digit is 2, the hundreds digit must be greater than 2 to maintain strictly increasing order. Wait, this is conflicting. Let me clarify. If we have a four-digit number with strictly increasing digits, starting with 2, then the hundreds digit must be greater than 2. The tens digit must be greater than the hundreds digit, and the units digit greater than the tens digit. Therefore, the hundreds digit can be 3, 4, 5, 6, 7, 8, or 9. However, since our upper limit is 2400, the maximum number is 2399. Therefore, the hundreds digit cannot exceed 3. Because if the hundreds digit is 4, the number would be 24XY, which is already beyond 2400. Therefore, the hundreds digit can only be 3. Wait, that seems too restrictive. Let's verify. Numbers between 2020 and 2400: thousands digit is 2, hundreds digit can be from 0 to 3 (since 2400 is excluded). But to have strictly increasing digits, the hundreds digit must be greater than the thousands digit, which is 2. Therefore, the hundreds digit must be at least 3. But the hundreds digit in this range can only go up to 3, because if the hundreds digit is 4, the number becomes 24XX, which is beyond 2400. Therefore, the hundreds digit must be exactly 3. Wait, that makes sense. So the only possible hundreds digit is 3. Therefore, the number must be in the form 23XY, where X and Y are digits such that 3 < X < Y. But let's confirm. Suppose the hundreds digit is 3, so the number is 23XY. Then, X must be greater than 3, and Y must be greater than X. Since the number must be less than 2400, the hundreds digit is 3, and the tens and units digits can be from 4 to 9, but with X and Y such that 3 < X < Y. However, the original range is from 2020 to 2400, so 23XY is between 2300 and 2399. But 2020 is 2020, so numbers starting with 20XX, 21XX, 22XX, and 23XX. Wait, but if the hundreds digit must be greater than 2 (to have strictly increasing digits starting with 2), then the hundreds digit can only be 3. Because if the hundreds digit is 0, 1, or 2, then it's less than or equal to 2, which would not satisfy the strictly increasing condition. Therefore, the only possible hundreds digit is 3. Therefore, all valid numbers must start with 23, followed by two digits where each subsequent digit is higher. But wait, what about numbers like 2345? That's 2345, which is 2 < 3 < 4 < 5, so that works. Similarly, 2346, 2347, etc., up to 2389. But let's check if that's the case. So if the number is 23XY, where X and Y are digits greater than 3, and X < Y. So X can be from 4 to 8, and Y from X+1 to 9. Therefore, the problem reduces to counting how many two-digit combinations (X, Y) exist where 4 ≤ X < Y ≤ 9. This is equivalent to choosing 2 digits from the digits 4 to 9 inclusive. The number of such combinations is C(6, 2) since there are 6 digits (4,5,6,7,8,9). C(6,2) is 15. Therefore, there are 15 such numbers. But wait, the options given are (A) 9, (B)10, (C)15, (D)20, (E)28. So 15 is option C. But let's double-check to make sure we haven't missed anything. Wait, hold on. The thousands digit is 2, hundreds digit is 3, and then the tens and units digits are selected from 4-9 with X < Y. So the number of combinations is indeed C(6,2) = 15. That seems right. But let's also check if there are numbers in the range 2020-2399 that start with 2, have hundreds digit less than 3, but still have strictly increasing digits. Wait, if the hundreds digit is less than 3, like 0, 1, or 2, then since the thousands digit is 2, the hundreds digit would have to be greater than 2 to be strictly increasing. Therefore, hundreds digits less than 3 are invalid. Therefore, only hundreds digit 3 is valid. Therefore, all valid numbers are of the form 23XY where X and Y are digits from 4-9 with X < Y. Therefore, the total number is C(6,2) = 15. So the answer should be C)15. But let me confirm by enumerating some numbers. For example, starting with 2345, 2346, 2347, 2348, 2349, 2356, 2357, 2358, 2359, 2367, 2368, 2369, 2378, 2379, 2389. That's 15 numbers. Yes, that seems correct. So the answer should be 15, which is option C. But wait, the original question says "between 2020 and 2400". So does 2345 count? 2345 is between 2020 and 2400, yes. Similarly, 2389 is 2389, which is less than 2400, so that's included. But wait, could there be numbers like 2456? But 2456 is greater than 2400, so it's outside the range. So all numbers must be less than 2400, so the hundreds digit is 3, making the number 23XY, as we thought. Therefore, the calculation is correct. So the answer is 15, option C. Wait, but let me check once more. Let's think about the possible numbers. Possible tens digits (X) can be 4,5,6,7,8. For each X, Y can be from X+1 to 9. For X=4: Y=5,6,7,8,9 → 5 numbersFor X=5: Y=6,7,8,9 → 4 numbersFor X=6: Y=7,8,9 → 3For X=7: Y=8,9 → 2For X=8: Y=9 → 1Total: 5+4+3+2+1=15. Yes, that's 15 numbers. Therefore, the answer is C)15.Final Answerboxed{C}

✅:To determine how many integers between 2020 and 2400 have digits in strictly increasing order with all digits distinct, we need to consider four-digit numbers starting with 2. 1. Range Analysis: The numbers range from 2020 to 2400, so the thousands digit is fixed at 2. The hundreds digit must be greater than 2 to maintain strictly increasing order, but since the upper limit is 2400, the hundreds digit can only be 3. This restricts the numbers to the form 23XY where X and Y are digits such that 3 < X < Y.2. Digit Combinations: The tens digit (X) and units digit (Y) must be chosen from the digits 4 to 9. The number of valid combinations is equivalent to choosing 2 digits from 6 options (4, 5, 6, 7, 8, 9), which is calculated using the combination formula ( binom{6}{2} ).3. Calculation: [ binom{6}{2} = frac{6!}{2!(6-2)!} = frac{6 times 5}{2 times 1} = 15 ]4. Verification: Enumerating the valid numbers confirms there are 15 combinations, such as 2345, 2346, ..., 2389.Thus, the number of integers between 2020 and 2400 with strictly increasing distinct digits is (boxed{C}).

🔑:We are tasked with finding how many integers exist between 2020 and 2400 whose digits are strictly increasing and are composed of different digits.1. Understanding the Constraints: - The number must be between 2020 and 2400. - All digits should be strictly increasing. - All digits should be different. Let's start breaking down the digits place by place.2. Constraints on Digits: - The leading digit (thousands place) must be '2' since the integers must be within the range of 2020 to 2400. - The second digit (hundreds place) must be strictly greater than 2. Therefore, it cannot be 0, 1, or 2 (it also cannot be 4 to stay below 2400).3. Valid Digits for the Hundreds Place: - The valid options for the hundreds place are {3} because choosing 4 would make the number 2400 or more.4. Determining Subsequent Digits: - We now have '2' at the thousands place and '3' at the hundreds place. The digits to the right of '3' (the tens and units places) must be taken from the set {4, 5, 6, 7, 8, 9} and must be strictly increasing.5. Choosing Tens and Units Digits: - From the set {4, 5, 6, 7, 8, 9}, we need to select any 2 digits that are strictly increasing.6. Calculating the Number of Combinations: - The number of ways to select 2 digits out of 6 (with the order mattering) can be calculated using the binomial coefficient: [ mathrm{C}_6^2 = binom{6}{2} = frac{6 times 5}{2 times 1} = 15 ] Conclusion:Thus, the total number of valid integers within the given range is ( boxed{15} ).

🧠:Okay, so I have this problem here: On a 5x7 grid, 9 cells are marked. We define a pair of cells sharing a side as interesting if at least one cell in the pair is marked. The question is asking for the maximum number of interesting pairs that can be formed. Hmm, let me try to wrap my head around this.First, let's make sure I understand the problem correctly. We have a grid that's 5 rows by 7 columns, so 35 cells in total. Out of these, 9 cells are marked. An "interesting pair" is two cells that are adjacent (share a side) and at least one of them is marked. We need to find the maximum number of such pairs possible by optimally placing those 9 marked cells.So, the goal is to arrange the 9 marked cells in such a way that the number of adjacent pairs (including both horizontal and vertical adjacents) where at least one is marked is maximized. Let me think about how to approach this. Maybe I should first recall some similar problems or concepts. This seems related to graph theory, where the grid is a graph with cells as nodes and edges between adjacent cells. The marked cells would be like a subset of nodes, and we want to count the number of edges incident to at least one node in this subset. Maximizing this count would be equivalent to maximizing the number of edges in the edge boundary of the subset, or perhaps something similar.Alternatively, maybe think in terms of percolation or coverage. If I spread out the marked cells, perhaps I can cover more adjacent pairs. But if I cluster them together, maybe the overlapping adjacents would reduce the total count? Wait, but when you cluster marked cells together, the adjacent pairs between marked cells would still count, right? Because at least one is marked. Wait, no, if both are marked, then the pair is still interesting. So, actually, clustering marked cells might allow each marked cell to contribute more adjacent pairs, but some pairs would be overlapping. Hmm, maybe there's a balance here.Alternatively, maybe spreading out the marked cells as much as possible would cover more unique adjacent pairs. For example, if marked cells are not adjacent to each other, then each can contribute their four adjacent edges (up, down, left, right), but if they are on the edge of the grid, they have fewer. However, if two marked cells are adjacent, the shared edge is counted once, but each marked cell can still contribute their other edges. Hmm.Wait, let's formalize this a bit. Let's model the grid as a graph where each cell is a vertex, and edges connect adjacent cells. Then, the problem reduces to selecting a subset S of 9 vertices such that the number of edges incident to at least one vertex in S is maximized.In graph theory terms, this is equivalent to finding the maximum number of edges in the union of the edge sets incident to each vertex in S. Alternatively, the edge set is the union of the edges of the subgraph induced by S plus the edges connecting S to the rest of the graph.But perhaps another way to think about it is that each edge in the grid is either between two unmarked cells, or it's adjacent to at least one marked cell. So, the total number of edges in the grid is fixed. Let me calculate that first.In a 5x7 grid, the number of horizontal edges is 5 rows * 6 columns (since each row has 7 cells, there are 6 edges between them), so 5*6=30. The number of vertical edges is 4 rows (between the 5 rows) *7 columns=28. So total edges in the grid: 30+28=58.Our goal is to maximize the number of edges that are adjacent to at least one marked cell. Therefore, equivalently, we need to minimize the number of edges that are between two unmarked cells, because total edges minus edges between unmarked cells will give us the interesting pairs. So, maximum interesting pairs = 58 - (number of edges between unmarked cells).Therefore, the problem reduces to placing 9 marked cells such that the number of edges between the remaining 26 unmarked cells is minimized. So, if we can arrange the unmarked cells in a way that they form as few edges as possible, then the interesting pairs will be maximized.But how do you minimize the number of edges between unmarked cells? That would be equivalent to making the unmarked cells as disconnected as possible, i.e., spread out so they don't share edges. Wait, but since there are 26 unmarked cells, which is a majority of the grid (35 cells total), it's going to be difficult to spread them out completely. So perhaps the best way is to cluster the marked cells in such a way that they break up the unmarked regions into as many disconnected components as possible, thereby minimizing the number of edges between unmarked cells.Alternatively, since the unmarked cells are 26, which is more than half the grid, maybe arranging the marked cells to form a sort of barrier that splits the unmarked cells into separate regions, each with as few internal edges as possible. But this seems a bit vague.Alternatively, maybe use the concept of independent sets. If the unmarked cells form an independent set (no two adjacent), then there are zero edges between them, which would be ideal. But since we have 26 unmarked cells, is that possible? In a 5x7 grid, the maximum independent set is... let's see. For a bipartite graph (which a grid is), the maximum independent set is equal to the number of vertices minus the size of the minimum vertex cover. But perhaps more straightforwardly, in a chessboard coloring, the grid can be divided into black and white cells. For a 5x7 grid, there are 18 of one color and 17 of the other. So, the maximum independent set is 18. But we need 26 unmarked cells, which is way larger than 18. Therefore, it's impossible to have all unmarked cells be independent. Therefore, the edges between unmarked cells cannot be zero. So, we need to minimize the number of edges between them.Hence, the problem is equivalent to finding a subset S of 9 cells such that the induced subgraph on the complement of S (i.e., the unmarked cells) has as few edges as possible. To minimize the number of edges in the unmarked subgraph, we need to maximize the number of edges removed from the original grid by marking those 9 cells. Each marked cell can potentially remove edges adjacent to it. Wait, but edges are shared between two cells, so if we mark a cell, we remove all edges connected to it from the unmarked graph. Wait, no. If a cell is marked, any edge connected to it is considered in the interesting pairs. So the edges between a marked cell and an unmarked cell are included in the interesting pairs. The edges between two unmarked cells are not. So, to minimize the edges between unmarked cells, we need to arrange the marked cells such that they block as many edges between unmarked cells as possible.Alternatively, think of the marked cells as acting as "cuts" that break the grid into regions where the unmarked cells can't connect through those cuts. But since the marked cells are only 9, it's challenging to fully disconnect the grid.Alternatively, maybe use some formula. Let's denote U as the set of unmarked cells, and E(U) as the number of edges between cells in U. Then, we need to minimize E(U). The total edges in the grid is 58. The number of edges adjacent to marked cells is equal to the sum over all marked cells of their degrees, minus twice the number of edges between marked cells (since those edges are counted twice). So, if S is the set of marked cells, then the number of edges adjacent to S is sum_{v in S} deg(v) - 2 * E(S), where E(S) is the number of edges between cells in S. Therefore, the number of interesting pairs is equal to sum_{v in S} deg(v) - 2 * E(S). Hence, to maximize this, we need to maximize sum deg(v) - 2 * E(S). But since the total interesting pairs is 58 - E(U), and E(U) is equal to total edges - (sum deg(v) - 2 * E(S))). Wait, perhaps let me re-express this.Total edges = E_total = 58.Edges in E(U) = edges between unmarked cells.Edges adjacent to marked cells = E_total - E(U).Therefore, interesting pairs = E_total - E(U) = 58 - E(U). Therefore, to maximize interesting pairs, we need to minimize E(U).So, the problem reduces to choosing 9 cells to mark such that the remaining 26 cells have as few edges between them as possible.Therefore, we need to arrange the 9 marked cells in such a way that they break as many edges as possible between the unmarked cells.Alternatively, since the unmarked cells are 26, which is a large number, perhaps the minimal number of edges in U is achieved when the unmarked cells are as fragmented as possible, with each connected component being as small as possible.But how can we do that with only 9 marked cells?Alternatively, think of the marked cells as forming a sort of checkerboard pattern, but given that 9 is quite a number, perhaps arranging them in a way that they separate the grid into small regions.Alternatively, perhaps arrange the marked cells in a line, but maybe a snake-like pattern through the grid to divide it into parts.Alternatively, think of the grid as a 5x7 rectangle. If we place the 9 marked cells in a way that they form a barrier from top to bottom or left to right, splitting the grid into two parts, then the unmarked cells on each side of the barrier would have fewer edges between them. However, with 9 cells, can we form such a barrier?For example, in a 5x7 grid, a vertical barrier would require 5 cells (one in each row). Similarly, a horizontal barrier would require 7 cells (one in each column). So, if we use 5 cells to form a vertical barrier, we have 4 remaining marked cells. Maybe place them to form another partial barrier. But this might not be optimal.Alternatively, arrange the marked cells in a way that maximizes the number of edges they cover. Since each marked cell can cover up to 4 edges, but corner and edge cells have fewer.Wait, but perhaps the key is that when two marked cells are adjacent, they share an edge, which is then counted once instead of twice. So, if we can cluster the marked cells, we can cover more edges with fewer overlaps. Wait, but actually, if two marked cells are adjacent, the edge between them is still counted as an interesting pair. So, the edge is included in the interesting pairs regardless of whether one or both are marked. Therefore, clustering marked cells together would result in the edges between them being counted as interesting pairs, but those edges are still part of the total interesting pairs.But in terms of the edges between unmarked cells, clustering the marked cells might allow the unmarked cells to be in larger connected regions, which might have more edges between them. Whereas spreading out the marked cells might break the unmarked regions into smaller components with fewer internal edges.Wait, for example, if you have all 9 marked cells in a cluster, then the remaining unmarked cells form a single large area minus the cluster, which might still be connected and have many internal edges. But if you spread the marked cells out, you can divide the unmarked area into smaller regions, each with fewer edges.Therefore, to minimize E(U), we need to maximize the number of connected components in U and minimize the size of each connected component.But how to do that with 9 marked cells.Another thought: In graph theory, the number of edges in a planar graph is at most 3n - 6, where n is the number of vertices. But our grid is a planar graph, and each connected component of U is a planar graph. If we have multiple connected components, the total number of edges would be the sum over each component of (3n_i - 6), but this is a rough upper bound. However, since our grid is a subgraph of a planar graph, maybe the actual number of edges is less.But perhaps this is not the right approach. Maybe let's think about specific arrangements.Suppose we arrange the 9 marked cells in a way that they form a checkerboard pattern, i.e., no two marked cells are adjacent. Then, each marked cell can block the four edges around it. However, since they are not adjacent, their blocked edges don't overlap. But since we have 9 marked cells, each contributing up to 4 edges, that would be 9*4=36 edges. However, edge cells have fewer edges. For example, a corner cell has only 2 edges, a side cell (not corner) has 3 edges. So, if we place all 9 marked cells in the interior, each with 4 edges, that would maximize the edges covered. But since the grid is 5x7, the interior cells are those not on the borders.But 5x7 grid: the total interior cells would be (5-2)*(7-2)=3*5=15. So, 15 interior cells. If we place all 9 marked cells in the interior, each can block 4 edges. However, edges between marked cells would still be counted. Wait, but in this case, since they are not adjacent, right? If arranged in a checkerboard, then no two marked cells are adjacent, so the edges around each marked cell are all adjacent to unmarked cells. Therefore, each marked cell blocks 4 edges (if in the interior) from being between unmarked cells.But actually, if a marked cell is in the interior, then the four edges around it are adjacent to it, so those edges are included in the interesting pairs, meaning they are subtracted from the total edges. Therefore, the edges around the marked cells are not part of E(U). Therefore, by placing a marked cell in the interior, we remove 4 edges from E(U). If we place a marked cell on the edge, we remove 3 edges, and in the corner, 2 edges.Therefore, to minimize E(U), we should place marked cells in positions where they can block the most edges. Therefore, prefer interior cells over edge cells.But in a 5x7 grid, the number of interior cells is 3x5=15. So, if we place all 9 marked cells in the interior, each blocking 4 edges, that would remove 9*4=36 edges. However, if two marked cells are adjacent, their shared edge would have been counted twice, but in reality, the shared edge is only removed once. Wait, no. If two marked cells are adjacent, then the edge between them is adjacent to both marked cells. Therefore, that edge is part of the interesting pairs and is not in E(U). However, when we place two adjacent marked cells, we are removing that shared edge from E(U), but when calculating the total edges removed, we would have counted that edge once for each marked cell. But since it's a single edge, it's actually only removed once. Therefore, if two marked cells are adjacent, we have overcounted the edges removed.Therefore, the formula for edges removed is sum of degrees of marked cells minus twice the number of edges between marked cells. Because each edge between two marked cells is counted twice in the sum of degrees.Therefore, edges removed from E(U) = sum_{v in S} deg(v) - 2 * E(S).So, to maximize edges removed (which is equivalent to minimizing E(U)), we need to maximize sum deg(v) - 2 * E(S). Therefore, to maximize this, we need to have marked cells with high degrees (i.e., interior cells) but minimize the number of edges between them (E(S)).Therefore, the optimal arrangement would be to place as many marked cells as possible in high-degree positions (interior) while keeping them non-adjacent to each other to minimize E(S). So, a non-adjacent (independent set) placement of marked cells in the interior.But wait, in a grid, the maximum independent set in the interior (3x5 grid) is how much? In a 3x5 grid, which is bipartite, the maximum independent set is ceil(3*5 / 2) = 8 (since 3*5=15, 15/2=7.5, so 8). Therefore, we can place at most 8 non-adjacent marked cells in the interior. But we have 9 marked cells to place. So, after placing 8 in the interior, we need to place one more. That last one can be placed on the edge or somewhere else.But let's see. If we place 8 non-adjacent marked cells in the interior, each contributes 4 edges, so sum deg(v) = 8*4=32. Since they are non-adjacent, E(S) =0. Therefore, edges removed =32 -0=32. Then, placing the 9th marked cell. If we place it on an edge, its degree is 3, so edges removed becomes 32 +3 -0=35 (since the 9th cell is not adjacent to any marked cell). If we place it in the interior but adjacent to one of the existing marked cells, then sum deg(v)=32 +4=36, but E(S)=1 (the edge between the new marked cell and the existing one), so edges removed=36 -2*1=34. Therefore, it's better to place the 9th cell on the edge, contributing 3 edges removed without creating any adjacency. So total edges removed=35.Alternatively, maybe place the 9th cell in a corner, contributing 2 edges removed. But that's worse.Therefore, edges removed=35. Therefore, E(U)=58-35=23. Therefore, interesting pairs=58 -23=35. But wait, is that correct? Wait, no. Wait, edges removed from E(U) is equal to sum deg(v) - 2*E(S). But E(U)= total edges - edges removed. So edges removed = sum deg(v) - 2*E(S). Therefore, E(U)=58 - (sum deg(v) - 2*E(S)).But in this case, sum deg(v)=8*4 +3=35, E(S)=0, so edges removed=35. Therefore, E(U)=58-35=23. So interesting pairs=58-23=35. Wait, but that seems low. Let me check.Wait, but if all 9 marked cells are placed in the interior and edges as non-adjacent as possible, with 8 in the interior and 1 on the edge, then total edges removed=35. But is there a better way?Alternatively, maybe arrange some marked cells adjacent to each other but in a way that the total edges removed is higher. Let's consider an alternative arrangement.Suppose we place the 9 marked cells in a 3x3 block in the center. Let's see how many edges that removes.A 3x3 block in the center of a 5x7 grid. Each interior cell in the 3x3 block has 4 edges. However, edges between the marked cells are shared. So, sum deg(v) for the 3x3 block: The 3x3 block has 9 cells. Each corner cell of the block has 2 edges connecting to unmarked cells, each edge cell (not corner) has 3 edges connecting to unmarked cells, and the center cell has 4 edges connecting to unmarked cells. Wait, no. Wait, actually, each marked cell's degree counts all edges adjacent to it, whether they connect to marked or unmarked cells. But the edges between marked cells are subtracted twice in the formula.Wait, let's compute sum deg(v) for the 3x3 block. Each cell in the 3x3 block: the four corner cells of the block each have 2 edges going outside the block, the four edge cells (middle of each side) have 3 edges going outside, and the center cell has 4 edges going outside. Wait, no, actually, in the entire grid, each cell's degree is 4 if it's in the interior, 3 if on the edge, and 2 if in the corner. But if we have a 3x3 block of marked cells, then each marked cell's degree is as follows:Assuming the 3x3 block is placed such that it's entirely within the interior of the 5x7 grid. Wait, but a 5x7 grid only has 3 rows in the interior (rows 2-4) and 5 columns in the interior (columns 2-6). So a 3x3 block can fit there. Each cell in this 3x3 block is an interior cell of the grid, so each has degree 4. Therefore, sum deg(v)=9*4=36. The number of edges between marked cells: in a 3x3 block, there are 3 rows, each with 2 horizontal edges, so 3*2=6 horizontal edges, and 2 columns, each with 3 vertical edges, so 2*3=6 vertical edges. Total edges between marked cells: 6+6=12. Therefore, edges removed=sum deg(v) -2*E(S)=36 -2*12=36-24=12. Therefore, edges removed=12, which is way lower than the previous 35. Therefore, E(U)=58-12=46, so interesting pairs=12. That's much worse. So clustering marked cells is bad for this purpose. So definitely, spreading them out is better.So, going back, if we can place 8 non-adjacent marked cells in the interior (each contributing 4 edges) and 1 on the edge (contributing 3 edges), total edges removed=35, leading to E(U)=23 and interesting pairs=35. But wait, is there a way to place more marked cells in the interior without overlapping edges?Wait, the maximum independent set in the interior 3x5 grid is 8. So we can place 8 non-adjacent cells there. Then the 9th cell has to go on the edge. Alternatively, maybe place some on the edge but non-adjacent to the interior ones. Let's see.Alternatively, what if we place some marked cells on the edge but in positions where they don't interfere with the interior ones. For example, place 8 marked cells in the interior as an independent set, and then place the 9th marked cell on the edge such that it's not adjacent to any of the interior ones. Since the interior cells are in the 3x5 grid, their adjacent edge cells would be on the borders. So, to place a marked cell on the edge not adjacent to any interior marked cells, we need to find a cell on the border that is not adjacent (horizontally or vertically) to any of the interior marked cells.But if the interior marked cells are in an independent set, then their adjacent edge cells are all unmarked. Wait, no. If we have an interior marked cell at position (2,2), then the edge cell (1,2) is adjacent to it. So, if we want to place a marked cell on the edge not adjacent to any interior marked cells, we have to choose an edge cell that is not in the same column or row as any interior marked cell. But since the interior marked cells are in the 3x5 grid (rows 2-4, columns 2-6), their adjacent edge cells would be in rows 1 and 5, and columns 1 and 7. So, to place a marked cell on the edge not adjacent to any interior marked cell, we need to choose a cell in row 1 or 5, or column 1 or 7, such that it's not adjacent to any interior marked cell.But if the interior marked cells are in columns 2-6, then to avoid adjacency, the edge marked cells in column 1 or 7 must not be in the same row as any interior marked cell. Similarly, edge marked cells in row 1 or 5 must not be in the same column as any interior marked cell.But if the interior marked cells are placed in an independent set, they are spaced out. For example, in a checkerboard pattern, every other cell. Let's consider how an independent set in the interior 3x5 grid looks.An independent set in a grid is a set of cells with no two adjacent. In a 3x5 grid, the maximum independent set is indeed 8. For example, you can color the grid like a chessboard: alternate black and white. In a 3x5 grid, there are 15 cells. The number of black cells would be ceil(15/2)=8, and white cells=7. So the maximum independent set is 8.Therefore, if we place the 8 marked cells on the black cells of the interior 3x5 grid, then none of them are adjacent. Then, to place the 9th marked cell on the edge without adjacency, we need to pick an edge cell that is not adjacent to any of these black cells. Since each black cell in the interior is in a specific row and column, their adjacent edge cells would be in the outer rows and columns. For example, suppose we take the first row (row 1) of the grid. The cells in row 1, columns 1 to 7. If none of the interior marked cells are in row 2, then a marked cell in row 1 would not be adjacent to any interior marked cell. But if the interior marked cells are in rows 2-4, then a marked cell in row 1, column x is adjacent to the cell in row 2, column x. So, if the interior marked cells are placed in row 2, columns where the edge cell is in row 1, then those edge cells would be adjacent.Therefore, to avoid adjacency, the edge marked cell must be in a column where the corresponding interior cell (in row 2) is not marked. Similarly for row 5 and column 1,7.But since the interior marked cells are 8 in number, spread out in the 3x5 grid, it's likely that in some columns of row 2 and row 4, there are marked cells. Therefore, it might be difficult to find an edge cell that is not adjacent to any interior marked cell.Alternatively, perhaps place the 9th marked cell in a corner where it's only adjacent to one interior cell, which may or may not be marked. For example, if we place a marked cell at (1,1), it's adjacent to (2,1). If (2,1) is not marked (since the interior marked cells are in columns 2-6), then the edge cell (1,1) can be marked without adjacency. Wait, columns 2-6 are the interior columns. So, in row 2, columns 2-6 are the interior. Therefore, column 1 in row 2 is part of the edge. Wait, no. The interior is rows 2-4 and columns 2-6. Therefore, the cells in row 2, column 1 are still edge cells. So, the interior marked cells are in rows 2-4 and columns 2-6. Therefore, a marked cell at (1,1) is adjacent to (2,1), which is an edge cell (not in the interior). Therefore, if we place a marked cell at (1,1), it's only adjacent to unmarked cells (assuming the interior marked cells are in columns 2-6). Therefore, that's allowed. So, maybe we can place the 9th marked cell at (1,1), which is a corner, adjacent only to (1,2) and (2,1). If (1,2) and (2,1) are unmarked (since interior marked cells are in columns 2-6 and rows 2-4), then marking (1,1) would only contribute edges to (1,2) and (2,1), which are unmarked. Therefore, the edges removed would be 2 (the degree of the corner cell). But earlier, placing the 9th marked cell on an edge (non-corner) would remove 3 edges. For example, placing it at (1,2), but (1,2) is adjacent to (2,2), which is in the interior. If (2,2) is marked, then we cannot place a marked cell at (1,2) because they would be adjacent. But if (2,2) is unmarked (which it isn't, since in the interior marked set), then we could place it. But since in the interior marked set, (2,2) is marked (assuming the independent set is placed there), then we cannot place a marked cell at (1,2). Therefore, perhaps the best we can do for the 9th marked cell is place it on an edge where it's adjacent to only one interior cell, which is unmarked. Wait, but if the interior marked cells are in an independent set, maybe there are some edge cells adjacent to unmarked interior cells. For example, suppose in the interior 3x5 grid, we have an independent set of 8 cells. Let's visualize it:Let me label the grid rows 1-5 and columns 1-7. The interior is rows 2-4, columns 2-6. So a 3x5 grid. If we color this interior grid like a chessboard, alternating black and white. Suppose we choose the black cells as the marked ones. In a 3x5 grid, the number of black cells would be 8 (rows 2,3,4; columns 2,3,4,5,6). Starting at (2,2) black, (2,4) black, (2,6) black; then (3,3), (3,5); then (4,2), (4,4), (4,6). Wait, that's 8 cells. So in this arrangement, the marked cells in the interior are at positions:Row 2: columns 2,4,6Row 3: columns 3,5Row 4: columns 2,4,6Each of these is non-adjacent. Then, to place the 9th marked cell on the edge, we need to find an edge cell not adjacent to any of these.Looking at the edge cells, for example, in row 1, columns 1-7. The cells in row 1, column 1 is adjacent to (2,1), which is not in the interior marked set. Similarly, row 1, column 2 is adjacent to (2,2), which is marked. So we can't place a marked cell there. Column 3 in row 1 is adjacent to (2,3), which is unmarked (since in the interior marked set, row 2 has columns 2,4,6 marked). Similarly, column 4 in row 1 is adjacent to (2,4), which is marked. Column 5 in row 1 is adjacent to (2,5), which is unmarked (since row 2's marked are 2,4,6). Column 6 in row 1 is adjacent to (2,6), which is marked. Column 7 in row 1 is adjacent to (2,7), which is not in the interior.So in row 1, we can place marked cells in columns 1,3,5,7. However, these cells are on the edge. Let's pick one, say column 1. If we place a marked cell at (1,1), it's adjacent to (1,2) and (2,1). Neither of these are marked (since (1,2) is on the edge and (2,1) is not in the interior marked set). Therefore, this is safe. Similarly, placing a marked cell at (1,3) is adjacent to (1,4) and (2,3). (2,3) is unmarked. So that's safe too. Same with (1,5) and (1,7). So we can choose any of these. Each of these edge cells has degree 3 (since they are on the edge but not the corner). Wait, (1,1) is a corner cell, degree 2. (1,3), (1,5), (1,7) are edge cells, degree 3. So to maximize edges removed, we should place the 9th marked cell in an edge non-corner cell. For example, (1,3). Then, the degree is 3, and edges removed=3. Therefore, total edges removed=8*4 +3=35. Therefore, E(U)=58-35=23. Therefore, interesting pairs=35.But can we do better? Let's see.Suppose instead of placing the 9th marked cell on the edge, we try to place it somewhere else in the interior but adjacent to an existing marked cell. For example, suppose we place a 9th marked cell in the interior, adjacent to one of the existing ones. Then, sum deg(v)=8*4 +4=36, E(S)=1 (the edge between the new marked cell and the existing one). Therefore, edges removed=36 -2*1=34. Which is less than 35. Therefore, worse.Alternatively, place two marked cells on the edge. Wait, but we only have 9 marked cells. If we place 7 in the interior and 2 on the edge. Let's calculate. Suppose 7 interior non-adjacent marked cells. Then sum deg(v)=7*4=28. Place two marked cells on the edge, each with degree 3. Total sum deg(v)=28+3+3=34. If they are non-adjacent to each other and to the interior marked cells, then E(S)=0. Therefore, edges removed=34. Then E(U)=58-34=24. Which is worse than 23. So interesting pairs=58-24=34, which is less than 35. So not better.Alternatively, maybe place 8 in the interior and 1 on the edge as before, edges removed=35, E(U)=23. That's better.Alternatively, what if we place some marked cells on the edge in such a way that they block more edges. For example, if we place a marked cell at a position where it can block multiple edges adjacent to unmarked cells. Wait, but each edge is only blocked once. So maybe if we place a marked cell in a position where it can block edges that would otherwise be part of E(U). But this is similar to what we're already doing.Alternatively, maybe a different arrangement of marked cells. Suppose instead of an independent set in the interior, we arrange marked cells in a way that they form a cross or some other shape that fragments the unmarked cells more.For example, placing marked cells in a vertical line down the center. In a 5x7 grid, the center column is column 4. If we place marked cells in column 4, rows 1-5. But that's 5 marked cells. Then, we have 4 more to place. If we place them in adjacent columns to split the grid further. For example, placing marked cells in columns 3 and 5 as well. But this may create a larger E(S) and reduce the edges removed.Alternatively, this approach might not be optimal. Let's try to calculate.Suppose we create a vertical barrier with marked cells in column 4, rows 1-5. That's 5 marked cells. Each of these cells has degree 4 except the ones at the ends (rows 1 and 5), which have degree 3. So sum deg(v)=3+4+4+4+3=18. E(S) is the edges between marked cells: between each consecutive row in column 4, so 4 edges. Therefore, edges removed=18 -2*4=18-8=10. Then, place the remaining 4 marked cells. Suppose we place them in columns 2 and 6, rows 2-4. Each of these cells is in the interior, degree 4. So 4 cells *4=16. Edges between them: in columns 2 and 6, rows 2-4, each column has 3 cells, with 2 edges each, so 2 columns *2 edges=4 edges. Also, edges between columns 2 and 4 and columns 6 and 4? No, because column 4 is already marked. So no horizontal edges between columns 2 and 4 or 6 and 4. So total E(S) for these 4 cells is 4. Therefore, edges removed=16 -2*4=8. Then total edges removed so far=10+8=18. Plus the remaining edges from the vertical and horizontal barriers. Wait, maybe I'm complicating it. This seems like a worse approach since edges removed=18 is much less than the previous 35. Therefore, this method is not good.Therefore, the initial approach of placing as many marked cells as possible in the interior in an independent set and the remaining on the edge seems better.But let's confirm with another arrangement. Suppose we place all 9 marked cells on the edge. How many edges would that remove?Each corner marked cell removes 2 edges, each edge (non-corner) marked cell removes 3 edges. Suppose we place 4 corners (but there are only 4 corners in a 5x7 grid: (1,1), (1,7), (5,1), (5,7)). Each corner has degree 2. If we mark all 4 corners, that's 4 cells, sum deg(v)=4*2=8. Then, we have 5 remaining marked cells. Place them on the edges, non-corner. Each has degree 3. So 5*3=15. Total sum deg(v)=8+15=23. Edges between marked cells: corners are not adjacent to each other. The edge cells might be adjacent. For example, if we place marked cells along the top edge, (1,2) to (1,6), but they are adjacent to each other. If we place them non-adjacent, but on the edge. However, on the top edge (row 1), there are 7 cells. If we place marked cells alternately, we can place 3 non-adjacent marked cells (e.g., (1,1), (1,3), (1,5), (1,7)), but we already used the corners. So maybe on the top edge, we have 4 corners and then some. But this is getting messy. Assume that we place the 5 non-corner edge cells such that none are adjacent. On each edge, the maximum independent set is roughly half the length. For the top edge (7 cells), maximum independent set is 4. But we already have the corners marked. So between the corners, we can place 2 more on the top edge without adjacency. Similarly for other edges. But this is complicated. However, even if we manage to place all 5 non-corner edge cells without adjacency, sum deg(v)=23, E(S)=0 (since no edges between marked cells), edges removed=23. Therefore, E(U)=58-23=35, interesting pairs=23. Which is worse than the previous 35. So placing all marked cells on the edge is worse than placing some in the interior.Therefore, the best arrangement seems to be placing as many marked cells as possible in the interior (non-adjacent) and the rest on the edges (non-adjacent to the interior ones). That gives us edges removed=35, interesting pairs=35.But wait, let me think again. If edges removed=35, then E(U)=58-35=23. Therefore, interesting pairs=58-23=35. But can we remove more edges? Let's think.Suppose we manage to place 9 marked cells in the interior 3x5 grid, but since the maximum independent set is 8, we have to place one more cell adjacent to an existing one. So sum deg(v)=8*4 +4=36, E(S)=1. Therefore, edges removed=36 -2*1=34. So E(U)=58-34=24, interesting pairs=34. Which is worse than 35. So no.Alternatively, what if we place 7 marked cells in the interior and 2 on the edge, but arrange them such that the edge cells are adjacent to unmarked cells. For example, 7 interior non-adjacent marked cells, sum deg(v)=7*4=28. Two edge cells, each with degree 3, non-adjacent. So sum deg(v)=28+3+3=34. E(S)=0. Edges removed=34. E(U)=58-34=24. Interesting pairs=34. Worse than 35.Alternatively, what if we place some marked cells in the interior and some on the edge, but overlapping their coverage. For example, a marked cell on the edge adjacent to an interior marked cell. Then, the edge cell's adjacency to the interior cell is already counted, but the edge cell can block additional edges. Let's see.Suppose we have 8 interior marked cells (sum deg(v)=32) and 1 edge marked cell adjacent to one of them. The edge cell has degree 3, but one of its edges is adjacent to the interior marked cell. Therefore, edges removed from E(U) would be sum deg(v)=32+3=35, but E(S)=1 (the edge between the edge marked cell and the interior one). Therefore, edges removed=35 -2*1=33. So E(U)=58-33=25, interesting pairs=33. Worse than 35.Therefore, the previous calculation of 35 edges removed leading to 35 interesting pairs seems better.Wait, but is this possible? If we have 8 marked cells in the interior (non-adjacent), and 1 marked cell on the edge (non-adjacent to any interior marked cells), then sum deg(v)=8*4 +3=35. E(S)=0. Therefore, edges removed=35. Therefore, E(U)=23. So interesting pairs=35. But let's verify with an example. Let's try to construct such a configuration.Interior grid: rows 2-4, columns 2-6. Place 8 marked cells in an independent set. For example, in a checkerboard pattern where we select every other cell. Let's color the interior grid in black and white. Suppose we choose the black cells:Row 2: columns 2,4,6Row 3: columns 3,5Row 4: columns 2,4,6That's 3+2+3=8 cells. None of these are adjacent. Then, place the 9th marked cell on the edge, say at (1,3). This cell is on the top edge, column 3. It is adjacent to (2,3), which in the interior is unmarked (since in row 2, column 3 is white). Therefore, (1,3) is not adjacent to any marked cell. The degree of (1,3) is 3 (up, right, down), but since it's on the top edge, up is outside the grid. So actually, degree is 3? Wait, (1,3) has neighbors (1,2), (1,4), and (2,3). But (1,3) is on the edge, but not a corner. So degree 3. Yes. So edges removed by marking (1,3) are 3: the edges to (1,2), (1,4), and (2,3). Therefore, total edges removed=8*4 +3=35. E(U)=58-35=23. Therefore, interesting pairs=35.But let's count E(U). The unmarked cells are 35-9=26. The edges between them are 23. Is this plausible? Let's see. If the unmarked cells are mostly in the remaining parts of the grid, but with 23 edges between them, which is quite low.Alternatively, perhaps I made a miscalculation. Let's think differently. If we have 9 marked cells arranged to block 35 edges, then the remaining edges are 58-35=23. But the unmarked cells have 26 cells. In a planar graph, the number of edges is at most 3n -6. For n=26, 3*26 -6=72. But 23 is way less than that, so it's possible. But how?Actually, the unmarked cells would form several disconnected components. For example, the interior marked cells block certain areas, and the edge marked cell further blocks. The unmarked cells would be fragmented into small components with few edges each.But to verify, let's consider that the 8 interior marked cells are spread out in the 3x5 interior, and the 9th marked cell is on the edge. The unmarked cells would include the entire periphery except the marked edge cell, and the interior except the 8 marked cells. However, the periphery has many cells, so unless the marked cells fragment the unmarked areas, they might still have many edges.Wait, maybe I need to visualize this. If we have 8 marked cells in the interior in a checkerboard pattern, they are spaced out such that each is not adjacent. Then, the unmarked cells in the interior are the remaining 7 cells (since 15-8=7). These 7 cells are also in a checkerboard pattern (the white cells). Then, the periphery has 35-15=20 cells, minus 1 marked cell on the edge, so 19 unmarked cells. Therefore, total unmarked cells=7+19=26. The edges between unmarked cells would include:1. Edges within the interior unmarked cells (7 cells). Since they are in a checkerboard pattern, each unmarked interior cell is adjacent to marked cells, but not necessarily to other unmarked cells. Wait, no. In the checkerboard pattern, each white cell is adjacent to black cells (marked), but white cells are not adjacent to each other. Because in a checkerboard, no two white cells are adjacent. Therefore, the 7 interior unmarked cells form an independent set. Therefore, there are no edges between them. So E(U) from the interior unmarked cells is 0.2. Edges within the periphery unmarked cells (19 cells). The periphery includes the outer layer of the grid. These 19 cells are on the edge, except for the 1 marked cell. The periphery is a 5x7 grid with the interior 3x5 removed, so it's like a frame. The periphery has 5*7 -3*5=35-15=20 cells. Minus 1 marked cell, 19 unmarked. The periphery is a 5x7 frame, which is a grid of 5 rows and 7 columns, with the inner 3x5 removed. So, it's like a loop. The number of edges within the periphery unmarked cells would be the edges between these 19 cells. But calculating this is complicated. However, since we have a marked cell on the edge, say at (1,3), this cell is marked, so its adjacent edges are removed from E(U). The periphery includes the entire outer edge except the inner 3x5. So the periphery has two horizontal bars (top and bottom rows) and two vertical bars (left and right columns) minus the corners which are part of the horizontal bars.But the periphery's edges include the connections between these outer cells. For example, in the top row (row 1), columns 1-7, except the inner columns 2-6. So, row 1 has cells (1,1), (1,7). Similarly, row 5 has cells (5,1), (5,7). The columns 1 and 7 have rows 2-4. So, the periphery is:- Top row: (1,1), (1,7)- Bottom row: (5,1), (5,7)- Left column: (2,1), (3,1), (4,1)- Right column: (2,7), (3,7), (4,7)But wait, no. Actually, the periphery is all cells not in the interior 3x5. The interior 3x5 is rows 2-4, columns 2-6. Therefore, the periphery is:- Rows 1 and 5: all columns 1-7- Columns 1 and 7: rows 1-5- Rows 2-4, columns 1 and 7But the intersection of rows 1 and 5 with columns 1 and 7 are the corners. So total periphery cells:- Row 1: 7 cells- Row 5: 7 cells- Columns 1 and 7, rows 2-4: 3 cells each, total 6- Total: 7+7+6=20 cells.But we marked one of these periphery cells, say (1,3), so unmarked periphery cells are 19.The edges between these 19 cells include:- Horizontal edges in row 1: between (1,1)-(1,2), (1,2)-(1,3), ..., (1,6)-(1,7). But (1,3) is marked, so the edges (1,2)-(1,3) and (1,3)-(1,4) are removed. Similarly, in row 5, if no cells are marked there, all edges remain.- Vertical edges between row 1 and row 2: for columns 1-7, except columns 2-6.Wait, this is getting too complex. Let's think about how many edges are in the periphery.The periphery has 20 cells. Each periphery cell is on the edge of the grid. The number of edges within the periphery:- Horizontal edges in row 1: 6 edges (between 7 cells)- Horizontal edges in row 5: 6 edges- Vertical edges in column 1: 4 edges (between rows 1-5)- Vertical edges in column 7: 4 edges- Horizontal edges in columns 2-6, row 1 and row 5: but these are part of the periphery?Wait, no. The periphery is rows 1 and 5 (all columns), and columns 1 and 7 (rows 1-5). Therefore, the horizontal edges in row 1 are all within the periphery. Similarly, row 5. The vertical edges in columns 1 and 7 are within the periphery.But between the periphery and the interior, there are edges connecting row 1 to row 2 (columns 2-6), and row 4 to row 5 (columns 2-6), and columns 1 to 2 (rows 2-4), and columns 6 to 7 (rows 2-4). But these edges are between periphery and interior, which are already handled by the marked cells.But in our case, the periphery unmarked cells are 19. The edges between them would be:- In row 1: originally 6 edges, but with a marked cell at (1,3), two edges are removed: (1,2)-(1,3) and (1,3)-(1,4). So 6-2=4 edges remaining.- In row 5: 6 edges.- In column 1: vertical edges between (1,1)-(2,1), (2,1)-(3,1), (3,1)-(4,1), (4,1)-(5,1). That's 4 edges.- In column 7: similarly, 4 edges.But the marked cell is at (1,3), which doesn't affect columns 1 and 7. So total edges in the periphery unmarked cells:4 (row 1) +6 (row 5) +4 (column 1) +4 (column 7) =18 edges.But wait, but some edges are counted twice. For example, the cell (1,1) is part of row 1 and column 1. The edge (1,1)-(2,1) is in column 1, and the edge (1,1)-(1,2) is in row 1. So there's no overlap in edge counts.Therefore, total edges in the periphery unmarked cells: 4+6+4+4=18 edges.Additionally, the interior unmarked cells have 7 cells, but they form an independent set, so no edges between them.But wait, there's also edges between the periphery and the interior unmarked cells. Wait, no. The interior unmarked cells are in the interior 3x5 grid, rows 2-4, columns 2-6. The periphery unmarked cells are in rows 1,5 and columns 1,7, rows 2-4. The edges between periphery and interior are:- Between row 1 and row 2 (columns 2-6). These edges are adjacent to the marked cell in row 1,3, but row 2 columns 2-6 include marked cells. For example, in row 2, columns 2,4,6 are marked. Therefore, the edges between row 1 and row 2 for columns 2,4,6 are adjacent to marked cells (in row 2). Therefore, those edges are already counted as edges removed. The edges between row 1 and row 2 for columns 3,5 are adjacent to unmarked cells (since row 2, columns 3,5 are unmarked). But since the periphery unmarked cells are in row 1, columns 1-7 except the marked one. So for example, in row 1, column 3 is marked, so edges from row 1, columns 2,4,5,6,7 to row 2. But columns 2,4,6 in row 2 are marked, so edges (1,2)-(2,2), (1,4)-(2,4), (1,6)-(2,6) are adjacent to marked cells and thus removed. The edges (1,3)-(2,3), (1,5)-(2,5), (1,7)-(2,7) are adjacent to unmarked cells. But (1,3) is marked, so (1,3)-(2,3) is removed. The other edges (1,5)-(2,5) and (1,7)-(2,7) are between unmarked cells. Wait, but (2,5) is unmarked? In our interior marked set, row 2 has columns 2,4,6 marked. So (2,5) is unmarked. Similarly, (2,7) is part of the periphery. Therefore, the edges (1,5)-(2,5) and (1,7)-(2,7) are between unmarked cells. So these edges are part of E(U). Similarly, edges between row 4 and row 5.This is getting very complex. It seems that my initial calculation might have been too optimistic. If the periphery unmarked cells have 18 edges among themselves, and there are additional edges between the periphery and the interior unmarked cells, then E(U) could be higher than 23. For example, let's consider edges between periphery and interior unmarked cells. The interior unmarked cells are 7 in number, located at rows 2-4, columns 3,5. Each of these cells is adjacent to periphery cells. For example, (2,3) is adjacent to (1,3) [marked], (3,3) is adjacent to periphery cells in column 1 or 7? No, (3,3) is in the interior. Wait, the interior unmarked cells are at:Row 2: columns 3,5Row 3: columns 2,4,6Row 4: columns 3,5Wait, no. Wait, in the interior marked set, we had:Row 2: 2,4,6Row 3: 3,5Row 4: 2,4,6Therefore, the interior unmarked cells are:Row 2: 1,3,5,7 (but columns 1 and 7 are periphery). Wait, no. The interior is only columns 2-6. So row 2, columns 3,5 are unmarked. Row 3, columns 2,4,6 are unmarked. Row 4, columns 3,5 are unmarked. So total interior unmarked cells: 2 (row 2) +3 (row 3) +2 (row 4)=7. These cells are all in the interior. Therefore, their adjacent periphery cells are in rows 1 and 5, and columns 1 and 7. For example, (2,3) is adjacent to (1,3) [marked], so that edge is removed. (2,5) is adjacent to (1,5) [unmarked]. So edge (2,5)-(1,5) is between unmarked cells. Similarly, (3,2) is adjacent to (3,1) [periphery unmarked]. (3,4) is adjacent to (3,5) [interior unmarked]. (3,6) is adjacent to (3,7) [periphery unmarked]. (4,3) is adjacent to (5,3) [periphery unmarked]. (4,5) is adjacent to (5,5) [periphery unmarked]. Therefore, edges between interior unmarked cells and periphery unmarked cells are:- (2,5)-(1,5)- (3,2)-(3,1)- (3,6)-(3,7)- (4,3)-(5,3)- (4,5)-(5,5)That's 5 edges. Additionally, edges between interior unmarked cells:- (3,2)-(3,4) [but (3,2) and (3,4) are both unmarked? No, in row 3, columns 2,4,6 are unmarked. So (3,2)-(3,3), but (3,3) is marked. (3,2)-(3,4): are these adjacent? (3,2) and (3,4) are two apart, so not adjacent. Similarly, other interior unmarked cells are not adjacent to each other because of the checkerboard pattern. So there are no edges between interior unmarked cells.Therefore, total edges in E(U):- Periphery unmarked edges:18- Periphery-interior unmarked edges:5- Total E(U)=18+5=23.Yes! Therefore, the calculation holds. So E(U)=23, interesting pairs=58-23=35.Therefore, the maximum number of interesting pairs is 35.But wait, let's check with another arrangement. Suppose we place the 9 marked cells not just in the interior and one edge, but spread out differently. For example, place marked cells in both the interior and edges in a way that they are non-adjacent but cover more edges.But as we saw earlier, the formula gives us edges removed=sum deg(v) -2*E(S). To maximize this, we need high deg(v) and low E(S). Placing as many high-degree (interior) cells as possible, non-adjacent, gives the maximum edges removed.Given that we can place 8 non-adjacent interior cells (sum deg(v)=32) and one non-adjacent edge cell (sum deg(v)=3), total edges removed=35, which gives E(U)=23. Any other arrangement would result in lower edges removed due to either lower sum deg(v) or higher E(S).For example, if we place 7 interior non-adjacent marked cells (sum deg(v)=28) and 2 non-adjacent edge cells (sum deg(v)=6), total edges removed=34, which is less than 35. Or if we place 9 marked cells all on the edge, sum deg(v) is at most 23, which is much lower.Therefore, the maximum number of interesting pairs is 35.But wait, let me confirm once more. If E(U)=23, then interesting pairs=58-23=35. But let's think of another way: each marked cell can contribute edges. If they are non-adjacent, each marked cell's edges are all to unmarked cells. Therefore, each interior marked cell contributes 4 edges, and each edge marked cell contributes 3 edges. So total interesting pairs=8*4 +1*3=35. But this counts each edge only once, even if it's adjacent to two marked cells. But in our case, the marked cells are non-adjacent, so no edges are shared between them. Therefore, the total interesting pairs=35. Which matches the previous result.Therefore, the answer should be 35.But to be thorough, let's consider if there's a way to place 9 marked cells such that they cover more edges. For example, if we place some marked cells adjacent to each other but in a way that they cover more unique edges.Suppose we have two marked cells adjacent vertically. Each has degree 4. The shared edge is counted once. So total edges covered=4+4 -2=6. Whereas if they are non-adjacent, they cover 4+4=8 edges. Therefore, adjacency reduces the total edges covered. Therefore, it's better to keep them non-adjacent.Similarly, if we have a marked cell on the edge adjacent to an interior marked cell, as before, it reduces the total edges covered.Therefore, the optimal configuration is indeed placing as many non-adjacent marked cells in the interior as possible, and the rest on the edge non-adjacent to any marked cells.Therefore, the maximum number of interesting pairs is 35.

✅:To determine the maximum number of interesting pairs on a 5 times 7 grid with 9 marked cells, we need to maximize the edges adjacent to at least one marked cell. This is equivalent to minimizing the number of edges between unmarked cells. Here's the systematic approach:1. Grid Analysis: The grid has 35 cells and 58 edges (30 horizontal, 28 vertical). Marking cells affects edges adjacent to them.2. Edge Removal: Each marked cell removes edges from the unmarked set. Interior cells (degree 4) remove more edges than edge cells (degree 3) or corners (degree 2).3. Optimal Placement: - Place as many marked cells as possible in the interior (higher degree) to maximize edge removal. - Use an independent set (non-adjacent) arrangement to avoid overlapping edge removals.4. Interior Maximization: The interior (3x5) can hold 8 non-adjacent cells (maximum independent set). Each removes 4 edges, totaling 8 times 4 = 32.5. Edge Cell Addition: The 9th cell is placed on the edge (non-adjacent to interior cells), removing 3 more edges. Total edges removed: 32 + 3 = 35.6. Result Calculation: Total edges (58) minus edges between unmarked cells (35 removed) gives interesting pairs: 58 - 23 = 35.Answer: boxed{35}

🔑:1. Define the Concept of Interesting Pairs: In a 5 times 7 grid, a pair of cells sharing a common side is called "interesting" if at least one of the cells is marked.2. Calculate Maximum Possible Interesting Pairs: Each cell can be adjacent to a maximum of 4 other cells (up, down, left, right). Hence, marking a cell can contribute up to 4 interesting pairs. Since there are 9 marked cells, the maximum number of interesting pairs is: [ 9 times 4 = 36 ]3. Adjustment for Boundary Cells: - Boundary cells (corner cells and edge cells) have fewer than 4 neighbors. - Corner cells have at most 2 neighbors. - Edge cells (not corners) have at most 3 neighbors. We must take into account that some cells might be on the boundary, reducing their potential to form the maximum number of pairs.4. Consider Overlapping Interesting Pairs: If among the marked cells, two are neighbors, the interesting pair they share will be counted twice in the initial tally. To avoid double-counting, subtract any overlap.5. Guarantee of Neighboring Cells: Within any rectangular grid of 3 times 5, with 15 cells total, marking 9 cells ensures that at least 2 marked cells will be neighbors (Pigeonhole Principle).6. Maximum Pairs Considering Constraints: If we have an arrangement in which marking 9 cells ensures at least one of these neighboring cells interactions, and each cell pair is only counted once, we adjust the total: If each marking generates some boundary cases and shared pairs: [ text{Total pairs} leq 36 - 1 = 35 ]7. Optimal Distribution: To achieve exactly 35 interesting pairs without overlaps or missing potential pairs, there exists an arrangement to showcase this maximum.Thus, the number of interesting pairs in a 5 times 7 grid with 9 marked cells, taking into account boundary constraints and overlaps, is:[boxed{35}]

🧠:Okay, so I have this problem about a parallelogram ABCD. The diagonal AC is twice the length of side AB. On side BC, there's a point K such that angle KDB is equal to angle BDA. I need to find the ratio BK : KC. Hmm, let's start by drawing a diagram to visualize this. First, let me recall that in a parallelogram, opposite sides are equal and parallel, and the diagonals bisect each other. But here, they mention that diagonal AC is twice the length of side AB. So, if AB is, say, length x, then AC is 2x. That might be important for using the Law of Cosines or something later.Now, point K is on BC, and angle KDB is equal to angle BDA. Let me note that angle BDA is the angle at D between BD and DA. Since ABCD is a parallelogram, DA is congruent to BC, and AB is congruent to CD. Also, BD is another diagonal. Wait, diagonals in a parallelogram bisect each other, but they aren't necessarily equal unless it's a rectangle. So BD might not be equal to AC here.Given that angle KDB = angle BDA, maybe there's some similar triangles involved here. Let me try to mark the angles. Let me denote angle BDA as θ, so angle KDB is also θ. If I can find triangles that have these angles, maybe they are similar, which would let me set up proportions.Let me label the parallelogram. Let's say AB is the base, so AB is horizontal, and AD is vertical, but since it's a parallelogram, AD and BC are both vertical but maybe not. Wait, actually, in a general parallelogram, sides are just parallel, not necessarily horizontal or vertical. Maybe assigning coordinates could help here. That might be a good approach. Let's assign coordinates to the points to make it easier.Let me place point A at the origin (0,0). Since ABCD is a parallelogram, let's let AB be along the x-axis. Let me denote AB as vector a, so AB has length |a|. Then AD is another vector, let's say vector b. So coordinates would be: A(0,0), B(a,0), D(0,b), and C(a,b). Wait, no. Wait, in a parallelogram, the coordinates would be A(0,0), B(a,0), C(a + c, d), D(c,d). But actually, if AB is from (0,0) to (a,0), then AD is from (0,0) to (c,d), so D is (c,d), then C would be (a + c, d). But maybe to make it simpler, since it's a parallelogram, if I set AB as (a,0) and AD as (p,q), then coordinates would be:A(0,0), B(a,0), D(p,q), C(a + p, q). But maybe I can set coordinates such that calculations are easier. Since the diagonal AC is twice the length of AB. Let me set AB to be length 1 for simplicity, so AC is length 2. Then I can assign coordinates accordingly.Let me try this. Let me set point A at (0,0). Let me let AB be along the x-axis with length 1, so point B is at (1,0). Since ABCD is a parallelogram, point D can be at (0, d), so AD is vertical? Wait, but a parallelogram doesn't have to have vertical sides. Maybe I should assign D as (p, q) so that AD is a vector (p, q). Then point C would be B + AD vector, so (1 + p, q). The diagonal AC is from A(0,0) to C(1 + p, q), so its length is sqrt((1 + p)^2 + q^2). According to the problem, AC is twice AB. Since AB is length 1, AC must be 2. Therefore, sqrt((1 + p)^2 + q^2) = 2. So that's one equation.But we have another diagonal BD, which goes from B(1,0) to D(p, q). Its length is sqrt((p - 1)^2 + q^2). But we might not need that yet.We need to find point K on BC such that angle KDB = angle BDA. Let's figure out angle BDA first. Angle BDA is the angle at point D between BD and DA. Let me find vectors for BD and DA. BD is from D to B: vector BD = B - D = (1 - p, -q). DA is from D to A: vector DA = A - D = (-p, -q). The angle between BD and DA is angle BDA. So the angle between vectors BD and DA can be found using the dot product:cos(theta) = (BD · DA) / (|BD| |DA|)BD · DA = (1 - p)(-p) + (-q)(-q) = -p + p^2 + q^2|BD| = sqrt((1 - p)^2 + q^2)|DA| = sqrt(p^2 + q^2)So cos(theta) = (-p + p^2 + q^2) / (sqrt((1 - p)^2 + q^2) * sqrt(p^2 + q^2))Hmm, this is getting complicated. Maybe there's a better approach.Alternatively, let's use coordinates to find point K. Since K is on BC, which goes from B(1,0) to C(1 + p, q). So the parametric equation of BC is (1 + tp, tq) where t ranges from 0 to 1. Wait, no. Wait, BC is from B(1,0) to C(1 + p, q). So the vector from B to C is (p, q). So any point K on BC can be represented as B + t*(BC vector) = (1 + tp, 0 + tq) where t is between 0 and 1. So K is (1 + tp, tq). Then, we need to find t such that angle KDB = angle BDA.Angle KDB is the angle at D between DK and DB. So angle between vectors DK and DB. Similarly, angle BDA is the angle at D between BD and DA, which is the angle between vectors DB and DA.So maybe if the angles are equal, the triangles involved might be similar or there's some trigonometric relation.Alternatively, using the Law of Sines in the triangles. Let me consider triangle DKB and triangle D something.Wait, angle KDB = angle BDA = theta. So in triangle DKB, angle at D is theta, and in triangle BDA, angle at D is theta.Wait, perhaps triangle DKB is similar to triangle D something?Alternatively, maybe using coordinates to set up the angle equality. Let's compute the slopes or the vectors.First, let's compute angle BDA. As before, vectors BD and DA. Vector BD is (1 - p, -q), vector DA is (-p, -q). The angle between them is theta.Now, angle KDB is the angle between vectors DK and DB. Vector DK is K - D = (1 + tp - p, tq - q) = (1 - p + tp, tq - q) = (1 - p(1 - t), q(t - 1)). Vector DB is B - D = (1 - p, -q).So the angle between DK and DB is theta. So using the dot product:cos(theta) = (DK · DB) / (|DK| |DB|)Similarly, we have for angle BDA:cos(theta) = (BD · DA) / (|BD| |DA|)Since both equal to cos(theta), we can set them equal:(BD · DA) / (|BD| |DA|) = (DK · DB) / (|DK| |DB|)Simplify this equation. Let's compute each part.First, BD · DA = (1 - p, -q) · (-p, -q) = (1 - p)(-p) + (-q)(-q) = -p + p^2 + q^2|BD| = sqrt((1 - p)^2 + q^2)|DA| = sqrt(p^2 + q^2)Now, DK · DB = (1 - p(1 - t), q(t - 1)) · (1 - p, -q) = [1 - p(1 - t)](1 - p) + q(t - 1)(-q)Let me expand that:First term: [1 - p + pt](1 - p) = (1)(1 - p) - p(1 - p) + pt(1 - p) = (1 - p) - p + p^2 + pt(1 - p)Wait, maybe better to compute directly:= [1 - p + pt]*(1 - p) + [q(t - 1)]*(-q)= (1 - p)(1 - p) + pt(1 - p) - q^2(t - 1)= (1 - 2p + p^2) + pt(1 - p) - q^2 t + q^2Similarly, |DK| = sqrt([1 - p(1 - t)]^2 + [q(t - 1)]^2)= sqrt([1 - p + pt]^2 + [q(t - 1)]^2)And |DB| is sqrt((1 - p)^2 + q^2), same as |BD|.So putting all together:(-p + p^2 + q^2) / (sqrt((1 - p)^2 + q^2) * sqrt(p^2 + q^2)) = [ (1 - 2p + p^2) + pt(1 - p) - q^2 t + q^2 ] / (sqrt([1 - p + pt]^2 + [q(t - 1)]^2) * sqrt((1 - p)^2 + q^2))This looks really complicated. Maybe there's a better way. Let me think.Alternatively, since ABCD is a parallelogram, diagonals bisect each other. But AC is twice AB. Wait, if AB is 1, AC is 2. So maybe we can find the coordinates based on this.Earlier, we set AB to be length 1, so AB is from (0,0) to (1,0). Then AC is from (0,0) to (1 + p, q) with length 2. So (1 + p)^2 + q^2 = 4.But we also have the condition that ABCD is a parallelogram, so AD must be equal to BC. AD is from (0,0) to (p, q), so its length is sqrt(p^2 + q^2). BC is from (1,0) to (1 + p, q), which is the same vector as AD, so length is also sqrt(p^2 + q^2). So that's consistent.But we need another condition. Wait, in a parallelogram, the sum of the squares of the diagonals equals twice the sum of the squares of the sides. Wait, that's a theorem. So:AC² + BD² = 2(AB² + AD²)Given that AC = 2 AB, and AB = 1, so AC = 2. Then:2² + BD² = 2(1² + AD²)So 4 + BD² = 2 + 2 AD²Therefore, BD² = 2 AD² - 2But BD is the other diagonal. Wait, but maybe this is not helpful here. Let me check.Alternatively, let's assign coordinates such that point A is (0,0), point B is (1,0), point D is (p, q), then point C is (1 + p, q). Then AC has length 2, so:(1 + p)^2 + q^2 = 4.We need to find p and q. But we have only one equation. Maybe we need to express the angle condition to get another equation.But the angle condition is about angle KDB = angle BDA. Let's see if we can relate this.Alternatively, maybe using vectors. Let me consider vectors.Vector DB is B - D = (1 - p, -q)Vector DA is A - D = (-p, -q)Angle between DB and DA is theta.Vector DK is K - D. Since K is on BC, which is from B(1,0) to C(1 + p, q). So K can be parameterized as K = B + t*(C - B) = (1,0) + t*(p, q) = (1 + tp, tq). So vector DK = K - D = (1 + tp - p, tq - q) = (1 - p + tp, q(t - 1)).We need the angle between DK and DB to be equal to the angle between DB and DA.So using the dot product formula for the angle between two vectors:cos(theta) = (DB · DA) / (|DB| |DA|) = (DB · DK) / (|DB| |DK|)Therefore:(DB · DA) / |DA| = (DB · DK) / |DK|Because |DB| cancels out.So:(DB · DA) / |DA| = (DB · DK) / |DK|Let me compute DB · DA:DB · DA = (1 - p)(-p) + (-q)(-q) = -p + p² + q²|DA| = sqrt(p² + q²)DB · DK = (1 - p)(1 - p + tp) + (-q)(q(t - 1))Let me expand this:= (1 - p)(1 - p + tp) - q²(t - 1)= (1 - p)(1 - p) + (1 - p)(tp) - q² t + q²= (1 - 2p + p²) + tp(1 - p) - q² t + q²|DK| = sqrt( (1 - p + tp)^2 + (q(t - 1))^2 )So the equation becomes:(-p + p² + q²) / sqrt(p² + q²) = [ (1 - 2p + p²) + tp(1 - p) - q² t + q² ] / sqrt( (1 - p + tp)^2 + q²(t - 1)^2 )This is a complicated equation, but maybe we can find t in terms of p and q. However, we have another equation from AC = 2 AB, which is (1 + p)^2 + q² = 4.Perhaps we can assign specific values to p and q that satisfy (1 + p)^2 + q² = 4, and then solve for t. Let me see if choosing specific coordinates can simplify this.Let me try to set p = 0. Then AC length would be sqrt( (1 + 0)^2 + q^2 ) = sqrt(1 + q²) = 2. Therefore, 1 + q² = 4 => q² = 3 => q = sqrt(3). So point D would be (0, sqrt(3)), and C would be (1, sqrt(3)). But in this case, the parallelogram would be a rhombus because all sides would be equal? Wait, no. If p = 0 and q = sqrt(3), then AB is from (0,0) to (1,0), so length 1. AD is from (0,0) to (0, sqrt(3)), length sqrt(3). So sides are 1 and sqrt(3), so it's a rhombus only if all sides are equal, which they aren't. So it's just a regular parallelogram.But let's see if with p = 0, q = sqrt(3), can we find t such that angle KDB = angle BDA.So coordinates:A(0,0), B(1,0), D(0, sqrt(3)), C(1, sqrt(3)).So BC is from (1,0) to (1, sqrt(3)). Wait, p = 0, so BC is vertical line x=1 from y=0 to y=sqrt(3). So point K is somewhere on BC, so coordinates (1, t sqrt(3)), where t is between 0 and 1.So K is (1, t sqrt(3)).Now, angle BDA is the angle at D between BD and DA.Point D is (0, sqrt(3)), B is (1,0), A is (0,0).So vector DB is B - D = (1 - 0, 0 - sqrt(3)) = (1, -sqrt(3))Vector DA is A - D = (0 - 0, 0 - sqrt(3)) = (0, -sqrt(3))So angle between vectors DB and DA. Let's compute this angle.cos(theta) = (DB · DA) / (|DB| |DA|)DB · DA = (1)(0) + (-sqrt(3))( -sqrt(3)) = 0 + 3 = 3|DB| = sqrt(1² + (sqrt(3))²) = sqrt(1 + 3) = 2|DA| = sqrt(0² + (sqrt(3))²) = sqrt(3)Therefore, cos(theta) = 3 / (2 * sqrt(3)) = 3 / (2 sqrt(3)) = sqrt(3)/2Thus theta = 30 degrees, since cos(theta) = sqrt(3)/2 implies theta = 30°.Now, angle KDB is also 30°, which is the angle at D between DK and DB.Point K is (1, t sqrt(3)), so vector DK is K - D = (1 - 0, t sqrt(3) - sqrt(3)) = (1, sqrt(3)(t - 1))Vector DB is (1, -sqrt(3))We need the angle between DK and DB to be 30°, so:cos(30°) = (DK · DB) / (|DK| |DB|)We know cos(30°) = sqrt(3)/2Compute DK · DB = (1)(1) + (sqrt(3)(t - 1))(-sqrt(3)) = 1 - 3(t - 1) = 1 - 3t + 3 = 4 - 3t|DK| = sqrt(1² + [sqrt(3)(t - 1)]²) = sqrt(1 + 3(t - 1)^2 )|DB| = 2, as before.Thus:sqrt(3)/2 = (4 - 3t) / ( sqrt(1 + 3(t - 1)^2 ) * 2 )Multiply both sides by 2:sqrt(3) = (4 - 3t) / sqrt(1 + 3(t - 1)^2 )Multiply both sides by sqrt(1 + 3(t - 1)^2 ):sqrt(3) * sqrt(1 + 3(t - 1)^2 ) = 4 - 3tSquare both sides:3(1 + 3(t - 1)^2 ) = (4 - 3t)^2Expand left side:3 + 9(t - 1)^2 = 16 - 24t + 9t²Expand (t - 1)^2:3 + 9(t² - 2t + 1) = 16 - 24t + 9t²3 + 9t² - 18t + 9 = 16 - 24t + 9t²Combine like terms on left:(3 + 9) + 9t² - 18t = 12 + 9t² - 18tSo left side: 12 + 9t² - 18tRight side: 16 - 24t + 9t²Subtract right side from left side:12 + 9t² - 18t - 16 + 24t - 9t² = 0Simplify:(12 - 16) + (9t² - 9t²) + (-18t + 24t) = 0-4 + 6t = 06t = 4t = 4/6 = 2/3Therefore, t = 2/3. So point K is at (1, (2/3)sqrt(3)). Therefore, BK is the distance from B(1,0) to K(1, (2/3)sqrt(3)) which is (2/3)sqrt(3). KC is the distance from K to C(1, sqrt(3)) which is sqrt(3) - (2/3)sqrt(3) = (1/3)sqrt(3). Therefore, BK : KC = (2/3)sqrt(3) : (1/3)sqrt(3) = 2:1.Wait, that seems straightforward once I assigned specific coordinates. But does this hold in general? Because I chose p = 0 for simplicity. But in the original problem, we are not given any specific coordinates. However, in a parallelogram, the ratios should be preserved regardless of the specific dimensions, as long as the given condition AC = 2 AB holds. So by choosing coordinates that satisfy AC = 2 AB, the ratio BK : KC comes out as 2:1. Therefore, the answer is 2:1.But let me verify this with another coordinate system to ensure it's not dependent on my choice of p = 0. Let's take another set of coordinates where p ≠ 0.Suppose AB is still length 1, from (0,0) to (1,0). Let me choose p = 1, then AC length is sqrt((1 + 1)^2 + q²) = sqrt(4 + q²) = 2. So 4 + q² = 4 => q² = 0 => q = 0. But that would make ABCD a degenerate parallelogram with all points colinear. So that's not possible. So p cannot be 1. Let me pick p = 0.5. Then AC length sqrt((1 + 0.5)^2 + q²) = sqrt(2.25 + q²) = 2. So 2.25 + q² = 4 => q² = 1.75 => q = sqrt(1.75) ≈ 1.322.So point D is (0.5, sqrt(1.75)), point C is (1.5, sqrt(1.75)). Then BC is from (1,0) to (1.5, sqrt(1.75)). Parametric equations for BC: x = 1 + 0.5t, y = 0 + sqrt(1.75) t, where t ∈ [0,1]. So point K is (1 + 0.5t, sqrt(1.75) t).Vector DK = K - D = (1 + 0.5t - 0.5, sqrt(1.75) t - sqrt(1.75)) = (0.5 + 0.5t, sqrt(1.75)(t - 1))Vector DB = B - D = (1 - 0.5, 0 - sqrt(1.75)) = (0.5, -sqrt(1.75))Compute angle between DK and DB. Let's use the dot product formula.First, compute cos(theta) for angle BDA:Vectors BD and DA. BD = D to B = (0.5, -sqrt(1.75)), DA = D to A = (-0.5, -sqrt(1.75))Dot product BD · DA = (0.5)(-0.5) + (-sqrt(1.75))(-sqrt(1.75)) = -0.25 + 1.75 = 1.5|BD| = sqrt(0.5² + (sqrt(1.75))²) = sqrt(0.25 + 1.75) = sqrt(2) ≈ 1.414|DA| = sqrt(0.5² + (sqrt(1.75))²) = sqrt(0.25 + 1.75) = sqrt(2) ≈ 1.414Therefore, cos(theta) = 1.5 / (sqrt(2)*sqrt(2)) = 1.5 / 2 = 0.75, so theta ≈ 41.41 degrees.Now, angle KDB should also be ≈41.41 degrees. Let's compute DK · DB and |DK|.DK · DB = (0.5 + 0.5t)(0.5) + (sqrt(1.75)(t - 1))(-sqrt(1.75))= 0.25 + 0.25t - 1.75(t - 1)= 0.25 + 0.25t - 1.75t + 1.75= (0.25 + 1.75) + (0.25t - 1.75t)= 2 - 1.5t|DK| = sqrt( (0.5 + 0.5t)^2 + (sqrt(1.75)(t - 1))^2 )= sqrt(0.25(1 + t)^2 + 1.75(t - 1)^2 )= sqrt(0.25(t² + 2t + 1) + 1.75(t² - 2t + 1))= sqrt(0.25t² + 0.5t + 0.25 + 1.75t² - 3.5t + 1.75)= sqrt( (0.25t² + 1.75t²) + (0.5t - 3.5t) + (0.25 + 1.75) )= sqrt(2t² - 3t + 2)|DB| = sqrt(0.5² + (sqrt(1.75))²) = sqrt(0.25 + 1.75) = sqrt(2)So cos(theta) = (2 - 1.5t) / ( sqrt(2t² - 3t + 2) * sqrt(2) )We know this should equal 0.75.So:(2 - 1.5t) / ( sqrt(2t² - 3t + 2) * sqrt(2) ) = 0.75Multiply both sides by sqrt(2):(2 - 1.5t) / sqrt(2t² - 3t + 2) = 0.75 * sqrt(2) ≈ 1.06066But let's do it symbolically:(2 - 1.5t) = 0.75 * sqrt(2) * sqrt(2t² - 3t + 2)Square both sides:(2 - 1.5t)^2 = (0.75)^2 * 2 * (2t² - 3t + 2)Compute left side:(4 - 6t + 2.25t²) = 0.5625 * 2 * (2t² - 3t + 2)Simplify right side:0.5625 * 2 = 1.1251.125*(2t² - 3t + 2) = 2.25t² - 3.375t + 2.25Left side: 4 - 6t + 2.25t²Set equation:4 - 6t + 2.25t² = 2.25t² - 3.375t + 2.25Subtract right side from left side:4 - 6t + 2.25t² - 2.25t² + 3.375t - 2.25 = 0Simplify:(4 - 2.25) + (-6t + 3.375t) + (2.25t² - 2.25t²) = 01.75 - 2.625t = 0-2.625t = -1.75t = (-1.75)/(-2.625) = 1.75 / 2.625 = (7/4) / (21/8) = (7/4)*(8/21) = (56/84) = 2/3So t = 2/3 again. Therefore, regardless of the specific coordinates (as long as AC = 2 AB), the parameter t is 2/3, so BK : KC = t : (1 - t) = 2/3 : 1/3 = 2:1.Therefore, the ratio is 2:1.I think this confirms that the ratio is indeed 2:1, regardless of the specific parallelogram configuration, as long as AC is twice AB. Therefore, the final answer is boxed{2:1}.Final Answerboxed{2:1}

✅:To solve the problem, we start by assigning coordinates to the vertices of the parallelogram (ABCD). We place point (A) at ((0,0)), point (B) at ((1,0)), and point (D) at ((p, q)). Since (AC) is twice the length of (AB), we know the coordinates of point (C) are ((1 + p, q)) and the length of (AC) is 2. This gives us the equation:[(1 + p)^2 + q^2 = 4]Next, we need to find point (K) on (BC) such that (angle KDB = angle BDA). We parameterize point (K) on (BC) as (K = (1 + tp, tq)) where (t) is a parameter between 0 and 1. We calculate the vectors involved in the angles:- Vector (BD) is ((1 - p, -q))- Vector (DA) is ((-p, -q))- Vector (DK) is ((1 - p + tp, q(t - 1)))Using the dot product to find the cosine of the angles, we set up the equation for the angles being equal:[frac{(BD cdot DA)}{|BD| |DA|} = frac{(BD cdot DK)}{|BD| |DK|}]Simplifying this, we find:[frac{-p + p^2 + q^2}{sqrt{(1 - p)^2 + q^2} sqrt{p^2 + q^2}} = frac{(1 - 2p + p^2) + pt(1 - p) - q^2 t + q^2}{sqrt{(1 - p + tp)^2 + (q(t - 1))^2} sqrt{(1 - p)^2 + q^2}}]By choosing specific coordinates that satisfy (AC = 2AB), we simplify the problem. For example, setting (p = 0) and (q = sqrt{3}), we find the coordinates of (K) and verify the ratio (BK : KC). Through detailed calculations, we find that the parameter (t = frac{2}{3}) consistently, leading to the ratio (BK : KC = 2:1).Thus, the final answer is:[boxed{2:1}]

🔑:1. Given a parallelogram (ABCD) with diagonal (AC) twice the length of side (AB). We need to find the ratio (BK:KC), where (K) is a point on (BC) such that (angle KDB = angle BDA).2. Note that in any parallelogram, the diagonals bisect each other. Let (O) be the intersection point of the diagonals (AC) and (BD).3. Therefore, (AO = OC) and (BO = OD). Given (AC = 2AB), we have: [ AO = OC = AB = CD ]4. Observing the angle condition (angle KDB = angle BDA), we notice that this implies (angle KDB = angle DBK), making (triangle BDK) isosceles with (BK = KD).5. Since (K) is on segment (BC) such that (K) divides (BC), and (BK = KD), the triangle (triangle BDK) formed has (K) as the midpoint of (BD). But since (O) is already known to be the midpoint of (BD), we conclude (K) coincides with point (O), making (BO) median of (triangle BKC).6. As diagonals in a parallelogram bisect each other, (O) is the midpoint of (BD) and since (K) divides (BC) internally in the same ratio (2:1): [ BK:KC = 2:1 ] Conclusion:[boxed{2:1}]