🧠:Okay, let's see. The problem is about whether the Little Boy can transform the fraction 10/97 into either 1/2 or 1 using two specific operations: adding any natural number to both numerator and denominator, or multiplying both by the same natural number. Hmm, interesting. Let me try to break this down step by step for both parts a) and b).Starting with part a) first. The goal is to get to 1/2. So we need to see if there's a sequence of those operations that can take 10/97 to 1/2. Let me recall the operations allowed. First operation: add a natural number 'k' to both numerator and denominator. So starting with 10/97, adding k would make it (10 + k)/(97 + k). Second operation: multiply numerator and denominator by a natural number 'm', resulting in (10*m)/(97*m). But multiplying by m just scales the fraction, but since 10/97 is already in reduced form (since 10 and 97 are coprime, as 97 is a prime number), multiplying numerator and denominator by any m would keep the fraction the same. Wait, so that operation is actually not changing the fraction's value. So that might not help unless we use it in combination with addition. Hmm. So the key operations here are adding k and then maybe scaling? But scaling alone doesn't change the fraction. So maybe we need to first add some k to get a fraction that can be scaled to 1/2. Let's think.Alternatively, maybe we can first multiply and then add. Since multiplying by m first, then adding k. Let me formalize this.Let me denote the original fraction as N/D = 10/97. The operations allowed are:1. N → N + k, D → D + k for some natural number k.2. N → m*N, D → m*D for some natural number m.We need to reach 1/2. Let's suppose that after some operations, we can get (m*(10 + k) + k')/(m*(97 + k) + k'') = 1/2. Wait, maybe that's complicating. Let me think step by step.Alternatively, perhaps we can represent the problem as equations. Let's suppose that after some additions and multiplications, we can reach 1/2. Let me see. Suppose after adding k several times and multiplying several times. But the order matters. For example, if we first add k, then multiply by m, then add l, then multiply by n, etc. But this can get complex.Alternatively, maybe the operations can be combined in a certain way. Let's think of each operation as a transformation. Let's consider that after some number of additions and multiplications, the fraction becomes 1/2.Let me consider that the operations can be applied in any order. Let's suppose that we first add a number k to numerator and denominator, making it (10 + k)/(97 + k). Then multiply by m, making it (m*(10 + k))/(m*(97 + k)) which simplifies back to (10 + k)/(97 + k). So multiplying doesn't help here. So perhaps the key is to use addition first, then multiplication, but since multiplication doesn't change the value, unless we use addition again after multiplication. Wait, but if you multiply first, then add, that might have a different effect.Wait, if we first multiply numerator and denominator by m, getting 10m/97m, which is still 10/97. Then add k to both, resulting in (10m + k)/(97m + k). So that's a different fraction. So perhaps the sequence is important. So maybe first multiply, then add. Let's consider that.Suppose we first multiply by m, then add k. So starting with 10/97, multiply by m: 10m/97m. Then add k: (10m + k)/(97m + k). We want this to equal 1/2. So:(10m + k)/(97m + k) = 1/2Cross multiplying:2*(10m + k) = 1*(97m + k)20m + 2k = 97m + k20m + 2k = 97m + kSubtract 20m and k from both sides:2k - k = 97m - 20mk = 77mSo k must be 77m. Since k and m are natural numbers, we can choose m = 1, which gives k = 77. So then:First multiply numerator and denominator by 1 (which does nothing), then add 77. Wait, but if m =1, then after multiplying by 1, we have 10/97, then add 77 to both numerator and denominator: (10 +77)/(97 +77) = 87/174. Simplify that: 87 divides by 3: 29/58. Wait, 29/58 simplifies to 29/(2*29) = 1/2. Oh! So actually, adding 77 to numerator and denominator gives 87/174, which reduces to 1/2. Wait, but does the Little Boy need the fraction to be exactly 1/2 without reducing? Or is reducing allowed?Wait, the problem says "obtain a fraction equal to 1/2". So equality in value, not necessarily the same representation. So 87/174 is equal to 1/2 because 87*2 = 174. So yes, if you add 77 to both numerator and denominator, you get 87/174 = 1/2. Therefore, the answer to part a) is yes.Wait, but let me verify this. Starting with 10/97. Add 77 to numerator and denominator: 10 +77=87, 97 +77=174. 87/174 = 1/2. Yes, exactly. So that works. So part a) is possible just by adding 77. So the answer is yes.Wait, but the problem allows multiple operations. So the Little Boy can use any combination of adding and multiplying. But in this case, just adding 77 once suffices. So that's straightforward. Therefore, part a) is possible.Now moving to part b). Can he obtain 1? Which is the fraction 1/1. So we need to see if through adding numbers and multiplying, we can get from 10/97 to 1/1.Again, operations allowed: add k to numerator and denominator, or multiply numerator and denominator by m. Let's think. Let's try similar approach as part a). Suppose we want to reach 1/1. Let's see if adding k would work. Suppose we add k to numerator and denominator: (10 +k)/(97 +k). To make this equal to 1, we need 10 +k = 97 +k, which implies 10 = 97, which is impossible. So adding any k cannot make the fraction equal to 1, because numerator and denominator differ by 87 (since 97 -10 =87), and adding the same k to both preserves the difference. So 97 +k - (10 +k) =87. So the difference remains 87. So numerator will always be 87 less than denominator. So the fraction can never be 1, which requires numerator and denominator to be equal. So addition operations cannot make the fraction equal to 1.What about multiplication? If we multiply numerator and denominator by m, the fraction remains 10/97. So multiplying alone doesn't help. But perhaps combining multiplication and addition? Let's think. Let's suppose we first multiply by m, then add k. So:Starting with 10/97, multiply by m: 10m/97m. Then add k: (10m +k)/(97m +k). We want this to be 1. So:(10m +k)/(97m +k) = 1Which implies 10m +k = 97m +kSubtract k: 10m =97mWhich implies 87m=0. But m is a natural number, so this is impossible. So even after multiplying and then adding, we can't reach 1.Alternatively, if we first add k and then multiply. Let's see. Start with 10/97, add k: (10 +k)/(97 +k). Then multiply by m: (m*(10 +k))/(m*(97 +k)) = (10 +k)/(97 +k). Again, same as before. So multiplying after adding doesn't change the fraction's value. So that also can't help.Alternatively, maybe multiple additions and multiplications. Let's try adding k1, then multiplying by m1, then adding k2, etc. Let's attempt that. Suppose we first add k1:(10 +k1)/(97 +k1)Then multiply by m1:(m1*(10 +k1))/(m1*(97 +k1)) = (10 +k1)/(97 +k1)Same as before. Then add k2:(10 +k1 +k2)/(97 +k1 +k2)Wait, but adding k1 first and then k2 is equivalent to adding (k1 +k2) in one step. So essentially, multiple additions can be combined into a single addition. Similarly, multiplications can be combined into a single multiplication. Therefore, any combination of additions and multiplications can be represented as a sequence where first we add some k, then multiply by some m, then add some more k', then multiply by m', etc. But because multiplication cancels the effect of previous additions if followed by another addition. Wait, let's formalize this.Suppose we start with N/D =10/97.1. Add k: (N +k)/(D +k)2. Multiply by m: (m(N +k))/(m(D +k)) = (N +k)/(D +k)3. So multiplying after adding does nothing. So the multiplication is redundant.Alternatively, if we first multiply then add:1. Multiply by m: (mN)/(mD)2. Add k: (mN +k)/(mD +k)Which is different from the original. So in this case, multiplying first and then adding can lead to a different fraction. So the order matters here. Let's explore this path.So if we first multiply by m, then add k, resulting in (10m +k)/(97m +k). Let's set this equal to 1:(10m +k)/(97m +k) =1Which implies 10m +k =97m +kAgain, 10m =97m, leading to 87m=0. Impossible. So this approach also doesn't work.Alternatively, maybe multiple operations. Let's try multiplying, adding, multiplying, adding, etc. Let's see.Suppose we first multiply by m1, then add k1, then multiply by m2, then add k2, etc. But let's see:Starting with 10/97.Multiply by m1: 10m1/97m1Add k1: (10m1 +k1)/(97m1 +k1)Multiply by m2: (m2*(10m1 +k1))/(m2*(97m1 +k1)) = (10m1 +k1)/(97m1 +k1)So again, multiplying after adding doesn't help. So if we alternate multiply and add, the multiplications after additions are redundant. So the only effective operations would be the additions and the multiplications done before any additions. Therefore, the only effective sequences are:1. Multiply by m, then add k, resulting in (10m +k)/(97m +k)2. Add k, then multiply by m, which is same as adding k first, then multiplying (but since multiplying doesn't change the fraction, it's equivalent to just adding k)3. Any combination where multiplications are interleaved with additions, but as above, the multiplications after additions don't contribute anything.Thus, the only viable way to change the fraction beyond simple addition is to first multiply and then add. So let's focus on that.So, we need to find m and k such that (10m +k)/(97m +k) =1. But as before, this leads to 10m +k =97m +k => 10m=97m, which is impossible. So this approach cannot yield 1.Alternatively, perhaps multiple multiply-add steps. Let's suppose we first multiply by m1, add k1, then multiply by m2, add k2, etc. Let's try two steps:First step: multiply by m1, add k1: (10m1 +k1)/(97m1 +k1)Second step: multiply by m2: (m2*(10m1 +k1))/(m2*(97m1 +k1)) = same as beforeThen add k2: (m2*(10m1 +k1) +k2)/(m2*(97m1 +k1) +k2)Set this equal to 1:[m2*(10m1 +k1) +k2]/[m2*(97m1 +k1) +k2] =1Which implies:m2*(10m1 +k1) +k2 = m2*(97m1 +k1) +k2Cancel k2:m2*(10m1 +k1) = m2*(97m1 +k1)Divide both sides by m2 (since m2 is natural number, cannot be zero):10m1 +k1 =97m1 +k1Which again gives 10m1 =97m1 => 87m1=0, impossible.Same result. So no matter how many times we do multiply-add steps, we end up with an impossible equation. Therefore, it's impossible to reach 1 via these operations.Another angle: The key difference between part a) and b) is that in a) we can exploit the difference between numerator and denominator. The original difference is 87. When we add k to both, the difference remains 87. But when we multiply first and then add, we can manipulate the difference. Wait, no. Let's see:If we first multiply by m and then add k, the difference becomes (97m +k) - (10m +k) =87m. So the difference scales by m. For part a), when we set (10m +k)/(97m +k) =1/2, we found k=77m. So the difference is 87m, and the required fraction is 1/2. That worked because 10m +77m =87m, and 97m +77m =174m, which is 87m*2. Hence, 87m/174m=1/2. So in that case, scaling allowed us to reach the desired fraction.But for part b), the difference after multiply-add is 87m. To reach 1, the numerator and denominator must be equal, so their difference must be zero. But 87m is always 87 times a natural number, which can't be zero. Therefore, regardless of m, the difference remains non-zero, making it impossible to reach 1.Alternatively, let's consider invariant properties. The original fraction is 10/97. Let's see if there's an invariant that prevents it from reaching 1.One possible invariant is the difference between denominator and numerator. Originally, it's 87. When we add k to both, the difference remains 87. When we multiply both by m, the difference becomes 87m. If we then add k after multiplying, the difference becomes 87m. So the difference is always a multiple of 87. For the fraction to equal 1, the difference must be zero, but 87m can't be zero. Therefore, it's impossible. Hence, part b) is impossible.Yes, that makes sense. The difference is preserved as a multiple of 87 through the operations, so it can never be zero. Therefore, it's impossible to reach 1.So summarizing:a) By adding 77 to both numerator and denominator of 10/97, we get 87/174 =1/2. So yes.b) The difference between denominator and numerator is always a multiple of 87, which can't be zero. Hence, impossible.

✅:a) YesTo transform ( frac{10}{97} ) into ( frac{1}{2} ), the Little Boy can add ( 77 ) to both the numerator and denominator:[frac{10 + 77}{97 + 77} = frac{87}{174} = frac{1}{2}.]This operation directly achieves the target fraction.b) NoThe difference between the denominator and numerator of ( frac{10}{97} ) is ( 87 ). Both operations preserve this difference or scale it by a natural number ( m ):- Adding ( k ): The difference remains ( 87 ).- Multiplying by ( m ) then adding ( k ): The difference becomes ( 87m ).To reach ( 1 ), the numerator and denominator must be equal, requiring a difference of ( 0 ). However, ( 87m ) (for ( m in mathbb{N} )) is never zero. Thus, it is impossible to obtain ( 1 ).Final Answers a) (boxed{text{Yes}}) b) (boxed{text{No}})

🔑:Given fraction: frac{10}{97}.Actions allowed:1. Add any natural number x to both the numerator and the denominator.2. Multiply both the numerator and the denominator by any natural number.We need to determine whether the following fractions can be obtained:(a) frac{1}{2}1. Addition Insight: Adding the same natural number x to both the numerator and the denominator, the new fraction becomes: [ frac{10 + x}{97 + x} ]2. Equation Setup: We want this new fraction to equal frac{1}{2}: [ frac{10 + x}{97 + x} = frac{1}{2} ]3. Cross Multiplication: [ 2(10 + x) = 1(97 + x) ]4. Simplification: Expanding and solving for x: [ 20 + 2x = 97 + x ] [ 20 + 2x - x = 97 ] [ x + 20 = 97 ] [ x = 97 - 20 ] [ x = 77 ]5. Conclusion: Yes, it's possible to obtain the fraction frac{1}{2} by adding 77 to both the numerator and the denominator: [ frac{10 + 77}{97 + 77} = frac{87}{174} = frac{1}{2} ](b) frac{1}{1}1. Analyzing the Condition for 1: For a fraction to be equal to 1, its numerator must be equal to its denominator.2. Possible Fractions: Starting from frac{10}{97}, to make the numerator equal the denominator: - Addition: frac{10 + x}{97 + x} requires 10 + x = 97 + x, which is impossible since 10 neq 97. - Multiplication does not change the equality condition: frac{10k}{97k} = frac{10}{97} for any natural number k.3. Conclusion: It is impossible to make frac{10}{97} equal frac{1}{1} because the numerator and the denominator of the initial fraction are inherently unequal. Conclusion:- Part (a): It is possible to obtain the fraction (frac{1}{2}).- Part (b): It is impossible to obtain the fraction (frac{1}{1}).(boxed{text{a) Yes; b) No.}})

🧠:Okay, let's try to tackle this problem. It's about permutations of numbers from 1 to 105 with certain divisibility conditions. The problem states that for any m in {3, 5, 7}, and for all n such that 1 ≤ n < n + m ≤ 105, the difference between the (n + m)-th term and the n-th term in the permutation should be divisible by m. We need to find how many such distinct permutations exist.First, let's parse the problem carefully. The permutation is a rearrangement of numbers 1 through 105. The condition is that for each modulus m = 3, 5, 7, every pair of terms separated by m positions must have a difference divisible by m. So, for example, if m = 3, then for any n where n + 3 ≤ 105, the difference a_{n+3} - a_n must be divisible by 3. Similarly for m = 5 and m = 7.This seems like a problem involving modular arithmetic constraints on the permutation. Maybe each position in the permutation has to satisfy certain congruence conditions modulo 3, 5, and 7. Since 3, 5, and 7 are coprime, their least common multiple is 105, which is the length of the permutation. That might be a clue.Let me think. If we have conditions modulo 3, 5, and 7, then perhaps each number in the permutation must be congruent to a certain residue modulo each of these primes. Because the differences a_{n+m} - a_n must be divisible by m, that implies that a_{n+m} ≡ a_n mod m. So, if we fix the residue of a_n mod m, then all terms at positions differing by m must have the same residue modulo m. But since the permutation is of numbers from 1 to 105, each number is unique, so their residues mod 3, 5, 7 must be unique combinations? Wait, but residues mod 3, 5, 7 can repeat, but in a permutation, each number is unique, so their residues mod 3, 5, 7 must form a complete residue system?Wait, no. For example, mod 3, there are three residues: 0, 1, 2. Since 105 is divisible by 3 (105 = 3*35), there are 35 numbers congruent to 0 mod 3, 35 congruent to 1, and 35 congruent to 2. Similarly for mod 5 and mod 7.But how do the conditions affect the permutation? Let's take m = 3 first. For every n, a_{n+3} ≡ a_n mod 3. This recurrence relation suggests that the residues mod 3 repeat every 3 positions. So, positions 1, 4, 7, ..., 103 must all have the same residue mod 3. Similarly, positions 2, 5, 8, ..., 104 must have the same residue mod 3, and positions 3, 6, 9, ..., 105 must have the same residue mod 3.Similarly, for m = 5, the residues mod 5 must repeat every 5 positions. So, positions 1, 6, 11, ..., 101 have the same residue mod 5; positions 2, 7, 12, ..., 102; and so on up to positions 5, 10, 15, ..., 105.The same logic applies for m = 7, with residues mod 7 repeating every 7 positions.But since 3, 5, and 7 are pairwise coprime, the combination of these conditions would partition the permutation into residue classes modulo each of these primes, but overlapping in a way that each position is part of a residue cycle for each modulus.This seems similar to the concept of the Chinese Remainder Theorem (CRT). Since the positions are from 1 to 105, each position can be uniquely identified by its residues modulo 3, 5, and 7. For example, position 1 is (1 mod 3, 1 mod 5, 1 mod 7), position 2 is (2 mod 3, 2 mod 5, 2 mod 7), and so on. Wait, actually, position numbers themselves can be considered modulo 3, 5, 7. So each position i can be represented by (i mod 3, i mod 5, i mod 7). Because 3, 5, 7 are coprime, the CRT tells us that each combination of residues corresponds to a unique position modulo 105. Hence, each position has a unique triplet of residues (r3, r5, r7) where r3 = i mod 3, r5 = i mod 5, r7 = i mod 7.Given the problem's conditions, for each modulus m, the residue of a_i mod m must be the same for all positions i that are congruent modulo m. For example, for m=3, all positions congruent to 1 mod 3 must have a_i ≡ c1 mod 3, positions congruent to 2 mod 3 must have a_i ≡ c2 mod 3, and positions congruent to 0 mod 3 must have a_i ≡ c0 mod 3, where c0, c1, c2 are distinct residues mod 3. Similarly for m=5 and m=7.But since the permutation consists of numbers 1 to 105, each residue class mod 3 must contain exactly 35 numbers, as 105/3=35. Similarly, mod 5, each residue class has 21 numbers, and mod 7, each residue class has 15 numbers.Therefore, the permutation must be such that the numbers are partitioned into residue classes mod 3, 5, 7 according to their position's residue mod 3, 5, 7. That is, for each position i, which has a residue (r3, r5, r7), the number in that position must be congruent to some fixed residue mod 3, mod 5, and mod 7. Moreover, each residue class mod 3 must contain exactly 35 numbers, mod 5 exactly 21, and mod 7 exactly 15.Wait, but how do these constraints interact? Since the residues mod 3, 5, 7 of the numbers in the permutation are determined by their positions. For example, all numbers in positions congruent to 1 mod 3 must be congruent to a fixed residue mod 3, say c1. Similarly, all numbers in positions congruent to 1 mod 5 must be congruent to a fixed residue mod 5, say d1, and so on.But since each number in the permutation is unique, the combination of residues mod 3, 5, 7 for each position must be unique. However, the CRT tells us that each number from 1 to 105 has a unique combination of residues mod 3, 5, 7. Therefore, if we can assign to each position i (which has a unique (r3, r5, r7)) a number with a unique (s3, s5, s7), such that:1. For each m in {3,5,7}, and for each residue rm mod m, all positions with rm mod m must have numbers with a fixed residue sm mod m.In other words, the assignment of residues to positions is a bijection between the residue triplets of positions and the residue triplets of numbers. But since each position's residue triplet is unique, and each number's residue triplet is unique, the only way this can happen is if the mapping from position residues to number residues is a permutation of the residues for each modulus.For example, for modulus 3, the positions have residues 0,1,2 mod 3. The numbers assigned to positions with residue r mod 3 must all have a fixed residue s_r mod 3, where s_0, s_1, s_2 are a permutation of 0,1,2. Similarly for modulus 5 and 7.Therefore, the problem reduces to permuting the residue classes for each modulus. That is, we can independently choose a permutation of the residues for mod 3, mod 5, and mod 7, and then assign numbers to positions based on these permutations. Once the residues are assigned, the numbers in each residue class can be permuted among themselves.Let me elaborate. For each modulus m (3,5,7), we can permute the residues 0,1,...,m-1. For example, for m=3, we can choose a permutation (s0, s1, s2) of (0,1,2). Then, all positions congruent to r mod 3 will be assigned numbers congruent to s_r mod 3. Similarly for m=5 and m=7. Since these permutations are independent, the number of ways to choose these permutations is 3! for m=3, 5! for m=5, and 7! for m=7. Therefore, the total number of residue assignments would be 3! * 5! * 7!.However, once the residue classes are assigned, we need to assign the actual numbers within each residue class. For example, for mod 3, each residue class has 35 numbers. Once we fix that positions with residue r mod 3 get numbers congruent to s_r mod 3, we have 35 numbers in each of the three residue classes for mod 3. The numbers within each residue class can be permuted freely among their assigned positions. Similarly, for mod 5, each residue class has 21 numbers, and they can be permuted within their positions, and for mod 7, 15 numbers per residue class.However, we have to consider the overlap. Since a number is uniquely determined by its residues mod 3, mod 5, and mod 7, each number belongs to exactly one residue class for each modulus. Therefore, the assignments for each modulus are independent, and the total number of permutations is the product of the permutations within each residue class for all moduli.Wait, but the problem is that when we permute within residue classes for different moduli, we have to ensure that we don't overcount. Wait, perhaps not. Since the residue assignments for each modulus are independent, the total number of permutations would be the product of the permutations for each modulus. However, since the residue classes for different moduli interact via the Chinese Remainder Theorem, maybe the permutations are constrained across moduli.Wait, let's think differently. Suppose that for each modulus m, we fix a permutation of the residues mod m. For example, for m=3, we choose a permutation π_3 of {0,1,2}; for m=5, π_5 of {0,1,2,3,4}; and for m=7, π_7 of {0,1,2,3,4,5,6}. Then, for each position i, which has residues (r3, r5, r7) mod 3,5,7, we assign a number that has residues (π_3(r3), π_5(r5), π_7(r7)) mod 3,5,7. Since each number is uniquely determined by its residues mod 3,5,7, this assignment is well-defined and bijective. Therefore, such a mapping would define a unique permutation of the numbers 1 to 105.However, in addition to permuting the residues, we might also have to permute the numbers within each combined residue class. Wait, no, because once we fix the permutation of residues for each modulus, the mapping from positions to numbers is fixed via the CRT. Each position's residue triplet (r3, r5, r7) is mapped to a number's residue triplet (π_3(r3), π_5(r5), π_7(r7)), and by CRT, this number is unique. Therefore, this gives us exactly one permutation for each combination of π_3, π_5, π_7. Hence, the total number of such permutations would be 3! * 5! * 7!.But wait, the problem says "distinct permutations". Is this the final answer? Let me verify.Suppose we fix permutations π_3, π_5, π_7. Then, the permutation of numbers is determined by mapping each position i (with residues (i mod 3, i mod 5, i mod 7)) to the number with residues (π_3(i mod 3), π_5(i mod 5), π_7(i mod 7)). This is a bijection because of the Chinese Remainder Theorem. Therefore, each combination of π_3, π_5, π_7 gives a unique permutation. Hence, the total number is indeed 3! * 5! * 7! = 6 * 120 * 5040.Wait, hold on. Let's compute this. 3! is 6, 5! is 120, 7! is 5040. Multiplying these together: 6 * 120 = 720, 720 * 5040 = 3,628,800. But that seems very large. Is this the correct answer?Wait, but maybe I made a mistake here. Let's think again. If each permutation π_3, π_5, π_7 leads to a unique permutation of the numbers, then the total number is 3! * 5! * 7!. However, is that the case? Let's take a small example.Suppose we have a permutation of numbers 1 to 15 (instead of 105), and mod 3, 5. Let's see if the same logic applies. For each modulus, permuting residues, and then numbers are assigned via CRT. In that case, the number of permutations would be 3! * 5! = 6 * 120 = 720. But if we try to construct such permutations, is each one unique? For example, permuting residues mod 3 and mod 5 independently would lead to different permutations. However, maybe there are overlaps or some constraints I'm missing.Wait, in the original problem, the conditions are that for each m in {3,5,7}, the difference a_{n+m} - a_n is divisible by m. This is equivalent to saying that a_{n+m} ≡ a_n mod m. Therefore, the sequence of residues mod m must be periodic with period m. Since the permutation is a single cycle of length 105, the residues mod m must be constant on each congruence class mod m. Therefore, for each residue r mod m (0 ≤ r < m), all positions congruent to r mod m must have the same residue s_r mod m in the permutation. Since we have 105/m numbers in each residue class mod m, and each residue s_r must be assigned to 105/m positions.But in a permutation, each residue mod m must appear exactly 105/m times. Therefore, the mapping from position residues mod m to number residues mod m must be a bijection. Wait, no. Since we have m residue classes for positions (0 to m-1 mod m), and m residue classes for numbers (0 to m-1 mod m), each with 105/m numbers. Therefore, to satisfy the counts, the mapping from position residues to number residues must be a permutation. For example, for m=3, we have 35 numbers in each residue class mod 3. The positions mod 3 also have 35 positions each. Therefore, if we permute the residues, mapping position residue r to number residue π(r), then each number residue π(r) will be assigned to 35 positions, which is exactly the number of numbers available for that residue. Therefore, this works out.Therefore, indeed, the number of such permutations is the product of the number of permutations for each modulus, which is 3! * 5! * 7!.But let's check with a smaller case. Suppose we have n = 3*5 = 15, and mod 3 and 5. Then the number of permutations would be 3! * 5! = 6*120=720. Let's see if this holds.Take a permutation of 1 to 15 where for m=3 and m=5, the differences a_{n+m} - a_n are divisible by m. Then, similar logic applies. The residues mod 3 must be constant on positions congruent to each residue mod 3, and similarly for mod 5. Each position has a unique (mod 3, mod 5) pair. The mapping would be permuting residues mod 3 and mod 5 independently, leading to 3! * 5! permutations. Each such permutation defines a unique permutation of the numbers 1 to 15 via CRT. This seems correct.Therefore, scaling up to 3,5,7 with n=105, the answer should indeed be 3! * 5! * 7! = 6 * 120 * 5040.Calculating that:First, 6 * 120 = 720.Then, 720 * 5040.Calculate 720 * 5000 = 3,600,000.720 * 40 = 28,800.Total: 3,600,000 + 28,800 = 3,628,800.So, 3,628,800.But wait, the problem states "distinct permutations". However, the mapping via CRT as described gives exactly 3! * 5! * 7! distinct permutations. Each choice of π_3, π_5, π_7 gives a unique permutation. Therefore, the answer should be 3! * 5! * 7! = 6 * 120 * 5040 = 3,628,800.But let me confirm once more.Suppose we fix permutations π_3, π_5, π_7. Then, each position i (from 1 to 105) has residues (i mod 3, i mod 5, i mod 7). We map this to the number with residues (π_3(i mod 3), π_5(i mod 5), π_7(i mod 7)). By the Chinese Remainder Theorem, this uniquely defines a number between 1 and 105. Since π_3, π_5, π_7 are permutations, this mapping is bijective. Therefore, each combination of π_3, π_5, π_7 gives a unique permutation.Moreover, these permutations satisfy the given conditions. For any m in {3,5,7}, the difference a_{n+m} - a_n ≡ π_m((n+m) mod m) - π_m(n mod m) mod m. But since (n + m) mod m = n mod m, we have π_m((n + m) mod m) = π_m(n mod m). Therefore, a_{n+m} ≡ a_n mod m, which implies that m divides (a_{n+m} - a_n). Hence, the conditions are satisfied.Therefore, the total number of such permutations is indeed 3! * 5! * 7! = 3,628,800.But wait, the problem statement says "distinct permutations". However, in our reasoning, we considered permutations of the residues for each modulus. Is there a possibility that different combinations of π_3, π_5, π_7 could lead to the same overall permutation? For example, if two different residue permutations result in the same mapping. But since the CRT uniquely determines the number based on the triplet of residues, and each π_3, π_5, π_7 are independent permutations, I think each combination gives a distinct permutation. Because even if two different π_3 permutations could affect different residue triplets, changing any π_m would alter the residue triplet and hence the number assigned to each position.Therefore, the total number is indeed 3! * 5! * 7! = 6 * 120 * 5040 = 3,628,800.But wait a second. Let me think again. When we permute the residues for each modulus, are there any overlaps or constraints that we haven't considered? For example, if permuting residues mod 3 affects the assignments mod 5 or 7. However, since the permutations for each modulus are independent, and the Chinese Remainder Theorem ensures that the combination of residues uniquely defines a number, there should be no overlap. Each permutation of residues for each modulus operates independently, leading to distinct overall permutations.Therefore, I think the answer is indeed 3! * 5! * 7! = 3,628,800.But to be thorough, let's consider a tiny example. Let’s take m=2 and m=3, n=6. Then, permutations of 1-6 where for m=2 and m=3, the differences a_{n+2} -a_n divisible by 2, and a_{n+3} -a_n divisible by 3. Following the same logic, the number of permutations would be 2! * 3! = 2*6=12. Let's verify.Each position i in 1-6 has residues (i mod 2, i mod 3). For each modulus, permute the residues. For m=2: permute (0,1). For m=3: permute (0,1,2). Each combination defines a permutation. For example, if π_2 is identity and π_3 is identity, then position 1 (1 mod 2, 1 mod 3) maps to number with (1 mod 2, 1 mod 3), which is 1. Position 2 (0 mod 2, 2 mod 3) maps to (0 mod 2, 2 mod 3) which is 2. Etc., giving the identity permutation.If we swap π_2, then even positions become odd and vice versa. Similarly, permuting π_3 would rearrange the residues mod 3. Each combination gives a different permutation. For example, swapping π_2 and reversing π_3 would result in a different permutation. Since there are 2! * 3! = 12 permutations, this seems correct. Each of these permutations satisfies the divisibility conditions. Therefore, the logic holds, and scaling up to 3,5,7 with 105 elements gives 3! * 5! * 7!.Hence, the answer should be 3! * 5! * 7! = 6 * 120 * 5040 = 3,628,800.But just to be 100% sure, let's check another small case. Take m=2 and n=2. Then, permutations of [1,2] where a_{n+2} -a_n is divisible by 2. But since n+2 exceeds 2, the condition is trivial. However, if we consider m=2 and n=2, then the permutation must satisfy that a_2 - a_1 is divisible by 2. But since it's a permutation of [1,2], the only possibilities are [1,2] and [2,1]. Checking divisibility: for [1,2], 2 - 1 =1, not divisible by 2. For [2,1], 1 -2 = -1, also not divisible by 2. Wait, so there are no such permutations? But according to our formula, it would be 2! = 2 permutations. But in reality, there are none. This suggests that our reasoning might be flawed.Wait, hold on. This discrepancy indicates a problem with the earlier reasoning. Let's analyze this.In the case of n=2 and m=2, the condition is that for all n such that 1 ≤ n < n + m ≤ 2. Here, m=2, so n can only be 1, since 1 ≤ n < n + 2 ≤ 2. But n=1: n + m =3, which exceeds 2. Therefore, there are no such n. Hence, the condition is vacuously true. Therefore, any permutation of [1,2] would satisfy the condition. Therefore, the number of permutations is 2. But according to the earlier formula, it would be 2! = 2. Which matches. The previous mistake was in assuming that there was a condition when there isn't.So in the n=2, m=2 case, the number of permutations is 2! = 2, which is correct. If we have m=3 and n=3, similarly, there are no pairs, so all 6 permutations would satisfy the condition. Wait, let's check with n=3 and m=3. The permutation of [1,2,3] must satisfy that for m=3, any n where 1 ≤ n < n +3 ≤3. But n can only be 1, n+3=4 >3, so vacuous. Therefore, all 6 permutations are valid. But according to the formula, it would be 3! =6. Which is correct.But if we have m=2 and n=4. Let's see. n=4, m=2. The conditions are for n=1,2,3: check if a_{n+2} - a_n divisible by 2. So for permutation [1,2,3,4], check a3 - a1 and a4 - a2. For example, permutation [1,3,2,4]: a3 -a1=2 -1=1, not divisible by 2. So invalid. Permutation [2,4,1,3]: a3 -a1=1 -2=-1, not divisible by 2. Invalid. Wait, but according to our formula, the number of valid permutations should be 2! * ... but n=4, which is 2*2. Wait, no. Wait, for n=4, m=2. The positions are 1,2,3,4. For m=2, we need residues mod 2 to be constant on even and odd positions. So positions 1,3 must have the same parity, and positions 2,4 must have the same parity. Therefore, possible permutations are those where odd positions are all even or all odd, and similarly for even positions. But since it's a permutation, we need exactly two even and two odd numbers. So possible assignments: assign even numbers to odd positions and odd numbers to even positions, or vice versa. Then, within each, permute the numbers. So number of permutations: 2! * 2! * 2 (for swapping even and odd). Wait, let's see.If we fix that positions 1 and 3 are even, and 2 and 4 are odd. There are 2 even numbers (2,4) and 2 odd numbers (1,3). Assign 2 and 4 to positions 1 and 3: 2! ways. Assign 1 and 3 to positions 2 and 4: 2! ways. Total: 2! * 2! =4. Similarly, if we assign odd numbers to positions 1 and 3, and even to 2 and 4: another 4 permutations. Total: 8. But the total number of permutations is 24. Therefore, 8 valid permutations.But according to our formula, it would be 2! (for m=2) times ... but n=4 is not a multiple of 3,5,7. Wait, in this case, the formula was for when n is the product of the moduli. If n=4, which is 2^2, but our earlier reasoning was for n= product of distinct primes. Wait, maybe the formula needs adjustment.But in our original problem, n=105=3*5*7, which is square-free and the product of distinct primes. Therefore, the earlier reasoning applies. But in the case of n=4, m=2, which is not square-free, the formula might not apply.Therefore, the initial reasoning works when the length of the permutation is the product of distinct primes (or coprime moduli). In such cases, the CRT applies directly, and each position can be uniquely identified by its residues modulo each prime.Hence, returning to the original problem, since 105=3*5*7, and each modulus is prime and distinct, the CRT applies, and the number of permutations is indeed 3! * 5! * 7! = 3,628,800.Therefore, after careful consideration and checking with smaller examples (adjusting for different cases), I believe the correct answer is 3! * 5! * 7! = 6 * 120 * 5040 = 3,628,800.

✅:The number of distinct permutations is calculated by considering the independent permutations of residue classes modulo 3, 5, and 7. Each modulus requires permuting its residues, leading to factorials of the moduli. Multiplying these gives the total count:[3! times 5! times 7! = 6 times 120 times 5040 = 3,!628,!800]Answer: (boxed{3628800})

🔑:1. Let (a_1, a_2, ldots, a_{105}) be a permutation of the numbers (1, 2, ldots, 105).2. The condition is that for any (m in {3, 5, 7}) and for every (n) such that (1 leq n < n+m leq 105), it holds that (m mid left(a_{n+m} - a_nright)).3. To satisfy the condition (m mid left(a_{n+m} - a_nright)), the values of (a_{n}) mod (m) must be in a complete residue system modulo (m).4. Thus, the sequences (a_1, a_2, ldots, a_{105}) can be divided into groups of (m) that each contains a complete set of residues modulo (m).5. Specifically, for (m = 3), the numbers can be split into sequences of 3 numbers each forming complete residues modulo 3. This means for each sequence of 3 numbers, they must cover all residues: {0, 1, 2} modulo 3.6. Similarly, for (m = 5), the numbers can be split into sequences of 5 numbers each forming complete residues modulo 5. For each sequence of 5 numbers, they must cover all residues: {0, 1, 2, 3, 4} modulo 5.7. And for (m = 7), the numbers can be split into sequences of 7 numbers each forming complete residues modulo 7. For each sequence of 7 numbers, they must cover all residues: {0, 1, 2, 3, 4, 5, 6} modulo 7.8. In each case, (a_{i}) is uniquely determined by the residues modulo 3, 5, and 7 respectively. 9. To find the total number of such permutations, we calculate the factorials of the sizes of these groups: - For (m = 3), we have sequences of 3 numbers, so: (3!) - For (m = 5), we have sequences of 5 numbers, so: (5!) - For (m = 7), we have sequences of 7 numbers, so: (7!) Combining these, we have: [ (3!)(5!)(7!) = 6 times 120 times 5040 ]10. Calculating this product: [ 6 times 120 = 720 ] [ 720 times 5040 = 3628800 ]11. Thus, the number of such permutations that satisfy the given condition is: ( 3628800 ) Conclusion:(boxed{3628800})

🧠:Okay, let's see. I need to find the minimum value of ( ab ) given that ( a ) and ( b ) are positive numbers satisfying ( ab = a + b + 3 ). Hmm, okay, so the problem is giving me an equation involving ( a ) and ( b ), and I have to minimize their product. First, maybe I should write down the given equation again to make sure I have it right: ( ab = a + b + 3 ). Yeah, that's correct. Both ( a ) and ( b ) are positive. So, I need to find the minimum possible value of ( ab ) under this constraint. I remember that when dealing with optimization problems (like finding minima or maxima) with two variables, one common approach is to use calculus, specifically using partial derivatives or maybe substituting one variable in terms of the other and then taking the derivative. Alternatively, since this involves a product and sum, maybe the AM-GM inequality could be useful here. Let me think about both approaches.Starting with the substitution method. Let me try to express one variable in terms of the other using the given equation. Let's solve for ( b ) in terms of ( a ). So, starting with ( ab = a + b + 3 ). Let's rearrange the equation to isolate ( b ). Subtract ( a ) and ( b ) from both sides: ( ab - a - b = 3 ). Hmm, maybe factor this equation. Let me try adding 1 to both sides to make it factorable. So, ( ab - a - b + 1 = 4 ). Then, the left side can be factored as ( (a - 1)(b - 1) = 4 ). Oh, that's clever! So, by adding 1 to both sides, the equation becomes a product of ( (a - 1) ) and ( (b - 1) ) equals 4. That seems like a useful factorization. So, if I let ( x = a - 1 ) and ( y = b - 1 ), then ( x ) and ( y ) are positive numbers because ( a ) and ( b ) are positive (since ( a > 1 ) and ( b > 1 ), because ( x = a - 1 > 0 ) implies ( a > 1 ), same with ( b )). Then, the equation becomes ( xy = 4 ). Now, the original goal is to find the minimum value of ( ab ). Let's express ( ab ) in terms of ( x ) and ( y ). Since ( a = x + 1 ) and ( b = y + 1 ), then:( ab = (x + 1)(y + 1) = xy + x + y + 1 ).But we know that ( xy = 4 ), so substituting that in:( ab = 4 + x + y + 1 = x + y + 5 ).Therefore, the problem reduces to minimizing ( x + y + 5 ) given that ( x ) and ( y ) are positive real numbers such that ( xy = 4 ).Now, minimizing ( x + y ) with ( xy = 4 ). Hmm, this is a classic optimization problem. By the AM-GM inequality, the arithmetic mean of two positive numbers is at least their geometric mean. So, ( frac{x + y}{2} geq sqrt{xy} ). Plugging in ( xy = 4 ), we get ( frac{x + y}{2} geq sqrt{4} = 2 ), so ( x + y geq 4 ). Equality holds when ( x = y = 2 ). Therefore, the minimum of ( x + y ) is 4, so the minimum of ( ab = x + y + 5 ) would be ( 4 + 5 = 9 ). Wait, but let me verify this with calculus to make sure. Let's see. If we have ( x + y ) with ( xy = 4 ), then we can write ( y = frac{4}{x} ). Then, the expression becomes ( x + frac{4}{x} ). Taking the derivative with respect to ( x ):( f(x) = x + frac{4}{x} ).( f'(x) = 1 - frac{4}{x^2} ).Set derivative equal to zero:( 1 - frac{4}{x^2} = 0 Rightarrow x^2 = 4 Rightarrow x = 2 ). Since ( x > 0 ), we take ( x = 2 ), then ( y = frac{4}{2} = 2 ). So, indeed, the minimum occurs at ( x = y = 2 ), giving ( x + y = 4 ), so ( ab = 9 ). That checks out.Therefore, the minimum value of ( ab ) is 9.But wait a minute, let me check if substituting back into the original equation works. Let me take ( a = x + 1 = 3 ), ( b = y + 1 = 3 ). Then, ( ab = 9 ), and ( a + b + 3 = 3 + 3 + 3 = 9 ). So, yes, it satisfies the equation. But let me see if there are other possible points where the minimum could be lower. Suppose ( x ) is very large, then ( y = 4/x ) becomes very small. Then, ( x + y + 5 ) would approach infinity as ( x ) approaches infinity, and similarly if ( x ) approaches zero, ( y ) approaches infinity, so again ( x + y +5 ) approaches infinity. So, the function has a minimum at ( x = y = 2 ), which gives the minimal value of 9. Therefore, the answer is 9.Alternatively, I could have approached the original equation without substitution. Let's try that method as well to confirm.Starting with ( ab = a + b + 3 ). Let me try to express this as ( ab - a - b = 3 ). Then, adding 1 to both sides gives ( ab - a - b + 1 = 4 ), which factors as ( (a - 1)(b - 1) = 4 ), as before. So, same result. Therefore, the substitution seems valid. Alternatively, maybe using Lagrange multipliers. Let's try that for practice. We need to minimize ( ab ) subject to the constraint ( ab - a - b - 3 = 0 ). Let me set up the Lagrangian:( mathcal{L}(a, b, lambda) = ab - lambda(ab - a - b - 3) ).Wait, actually, the standard form is to have the constraint equal to zero. So, the function to minimize is ( f(a, b) = ab ), and the constraint is ( g(a, b) = ab - a - b - 3 = 0 ).Then, the Lagrangian is ( mathcal{L} = ab - lambda(ab - a - b - 3) ).Taking partial derivatives:( frac{partial mathcal{L}}{partial a} = b - lambda(b - 1) = 0 )( frac{partial mathcal{L}}{partial b} = a - lambda(a - 1) = 0 )( frac{partial mathcal{L}}{partial lambda} = -(ab - a - b - 3) = 0 )So, from the first equation: ( b = lambda(b - 1) )From the second equation: ( a = lambda(a - 1) )Let me solve for ( lambda ) from both equations. From the first equation:( lambda = frac{b}{b - 1} )From the second equation:( lambda = frac{a}{a - 1} )Therefore, ( frac{b}{b - 1} = frac{a}{a - 1} )Cross-multiplying: ( b(a - 1) = a(b - 1) )Expanding both sides: ( ab - b = ab - a )Subtract ( ab ) from both sides: ( -b = -a Rightarrow a = b )So, the critical point occurs when ( a = b ). Let's substitute ( a = b ) into the constraint equation ( ab = a + b + 3 ). Since ( a = b ), this becomes ( a^2 = 2a + 3 ), which simplifies to ( a^2 - 2a - 3 = 0 ). Solving this quadratic equation:( a = frac{2 pm sqrt{4 + 12}}{2} = frac{2 pm sqrt{16}}{2} = frac{2 pm 4}{2} )Since ( a > 0 ), we take the positive root: ( a = frac{2 + 4}{2} = 3 ). Therefore, ( a = b = 3 ), and ( ab = 9 ). This matches the previous result.Therefore, using Lagrange multipliers also leads us to the conclusion that the minimum value of ( ab ) is 9 when ( a = b = 3 ).Another approach could be to express ( ab ) in terms of a single variable. Let's try that. From the original equation ( ab = a + b + 3 ), we can solve for one variable in terms of the other. Let's solve for ( b ):( ab = a + b + 3 )( ab - b = a + 3 )( b(a - 1) = a + 3 )( b = frac{a + 3}{a - 1} )Now, substitute this expression for ( b ) into the product ( ab ):( ab = a times frac{a + 3}{a - 1} = frac{a(a + 3)}{a - 1} )Now, define ( f(a) = frac{a(a + 3)}{a - 1} ). We need to find the minimum value of ( f(a) ) for ( a > 1 ) (since ( a - 1 ) must be positive because ( b ) is positive, so denominator ( a - 1 > 0 Rightarrow a > 1 )).To find the minimum, take the derivative of ( f(a) ) with respect to ( a ):First, simplify ( f(a) ):( f(a) = frac{a^2 + 3a}{a - 1} )Using the quotient rule: ( f'(a) = frac{(2a + 3)(a - 1) - (a^2 + 3a)(1)}{(a - 1)^2} )Simplify numerator:( (2a + 3)(a - 1) - (a^2 + 3a) )First expand ( (2a + 3)(a - 1) ):( 2a(a) + 2a(-1) + 3(a) + 3(-1) = 2a^2 - 2a + 3a - 3 = 2a^2 + a - 3 )Then subtract ( a^2 + 3a ):( 2a^2 + a - 3 - a^2 - 3a = (2a^2 - a^2) + (a - 3a) - 3 = a^2 - 2a - 3 )Therefore, ( f'(a) = frac{a^2 - 2a - 3}{(a - 1)^2} )Set numerator equal to zero to find critical points:( a^2 - 2a - 3 = 0 )Solving the quadratic equation:( a = frac{2 pm sqrt{4 + 12}}{2} = frac{2 pm sqrt{16}}{2} = frac{2 pm 4}{2} )Again, since ( a > 1 ), take the positive root:( a = frac{2 + 4}{2} = 3 )So, critical point at ( a = 3 ). To confirm this is a minimum, check the second derivative or test intervals around ( a = 3 ).Let's test values around ( a = 3 ). Let's pick ( a = 2 ):( f'(2) = frac{(4) - 4 - 3}{(1)^2} = frac{-3}{1} = -3 ). Negative slope.Pick ( a = 4 ):( f'(4) = frac{16 - 8 - 3}{9} = frac{5}{9} ). Positive slope.Therefore, the function decreases before ( a = 3 ) and increases after ( a = 3 ), so ( a = 3 ) is indeed a minimum. Calculating ( ab ) at ( a = 3 ):( b = frac{3 + 3}{3 - 1} = frac{6}{2} = 3 ), so ( ab = 9 ), same as before.So, all methods lead to the conclusion that the minimal value of ( ab ) is 9.Another way to think about this is using symmetry. Since the equation ( ab = a + b + 3 ) is symmetric in ( a ) and ( b ), it's reasonable to guess that the minimum occurs when ( a = b ). Let me check if that's the case.Assuming ( a = b ), then the equation becomes ( a^2 = 2a + 3 ), which as before gives ( a = 3 ). Therefore, ( ab = 9 ). So, yes, the symmetric case gives the minimum. However, it's essential to confirm that this is indeed the minimum and that no asymmetric cases (where ( a neq b )) can give a lower product. But from earlier analysis using substitution and calculus, we saw that the minimal occurs at ( a = b = 3 ), so symmetry here gives the correct result.Additionally, considering the AM-GM inequality approach after substitution (( x + y geq 4 )) confirms that the minimal sum is achieved when ( x = y ), hence ( a = b ).In summary, all approaches—substitution with AM-GM, calculus using derivatives, Lagrange multipliers, and symmetry considerations—converge to the same minimum value of ( ab = 9 ).Just to make sure there's no oversight, let me test another pair of values. Suppose ( a = 4 ), then ( b = frac{4 + 3}{4 - 1} = frac{7}{3} approx 2.333 ). Then ( ab = 4 * 7/3 ≈ 9.333 ), which is higher than 9. If ( a = 2 ), then ( b = frac{2 + 3}{2 - 1} = 5 ), so ( ab = 10 ), which is also higher. If ( a = 5 ), ( b = (5 + 3)/(5 - 1) = 8/4 = 2 ), so ( ab = 10 ). So, these examples confirm that when ( a ) and ( b ) are not equal, the product is higher, reinforcing that the minimum occurs at ( a = b = 3 ).Therefore, I'm confident that the minimum value of ( ab ) is indeed 9.

✅:The minimum value of ( ab ) is found by transforming the given equation ( ab = a + b + 3 ) through substitution and applying optimization techniques. Here's the step-by-step solution:1. Factor the Equation: Rewrite ( ab = a + b + 3 ) as: [ (a - 1)(b - 1) = 4 ] by adding 1 to both sides and factoring.2. Substitute Variables: Let ( x = a - 1 ) and ( y = b - 1 ), so ( xy = 4 ). Express ( ab ) in terms of ( x ) and ( y ): [ ab = (x + 1)(y + 1) = xy + x + y + 1 = 4 + x + y + 1 = x + y + 5 ]3. Apply AM-GM Inequality: Minimize ( x + y ) given ( xy = 4 ). By AM-GM: [ frac{x + y}{2} geq sqrt{xy} implies x + y geq 4 ] Equality holds when ( x = y = 2 ).4. Compute Minimum ( ab ): Substitute ( x = y = 2 ): [ ab = 2 + 2 + 5 = 9 ]5. Verification: - Original Equation: If ( a = b = 3 ), then ( ab = 9 ) and ( a + b + 3 = 9 ), satisfying the equation. - Calculus: Expressing ( ab ) as a function of ( a ), taking derivatives confirms the minimum at ( a = b = 3 ). - Symmetry: Assuming ( a = b ) leads directly to the solution ( a = b = 3 ).Final Answer: The minimum value of ( ab ) is (boxed{9}).

🔑:1. We are given the equation ( ab = a + b + 3 ) for positive integers ( a ) and ( b ). 2. We can rearrange the given equation to solve for ( b ): [ ab = a + b + 3 implies ab - b = a + 3 implies b(a - 1) = a + 3. ] 3. Solve for ( b ): [ b = frac{a + 3}{a - 1}. ] 4. Since ( a ) and ( b ) are both positive, ( b = frac{a + 3}{a - 1} ) must be positive, and thus, the denominator ( a - 1 > 0 ): [ a > 1. ] 5. We need to express ( ab ) in terms of ( a ): [ ab = a cdot frac{a + 3}{a - 1}. ] 6. Simplify ( ab ): [ ab = a cdot frac{a + 3}{a - 1} = frac{a(a + 3)}{a - 1} = frac{a^2 + 3a}{a - 1}. ]7. To find the minimum value of ( ab ), let's rewrite ( frac{a^2 + 3a}{a - 1} ) by splitting and simplifying it differently: [ frac{a^2 + 3a}{a - 1} = frac{a^2 - a + 4a}{a - 1} = frac{(a^2 - a) + (4a)}{a - 1} = frac{a(a - 1) + 4a}{a - 1} = a + frac{4a}{a - 1}. ] 8. Continue to simplify the expression: [ ab = a + frac{4a}{a - 1}. ]9. Factor and separate the terms: [ ab = (a - 1) + 1 + frac{4}{a - 1} implies ab = (a - 1) + frac{4}{a - 1} + 1 + 4 = (a - 1) + frac{4}{a - 1} + 5. ]10. According to the AM-GM inequality, for any positive number ( x ), [ x + frac{4}{x} geq 2 sqrt{x cdot frac{4}{x}} = 2 sqrt{4} = 4. ] Thus, [ ab geq (a - 1) + 4 + 5. ]11. Attain the equality in AM-GM, we need [ x = frac{4}{x} implies x^2 = 4 implies x = 2. ], So ( a - 1=2implies a=3 ). 12. Check value of ab: [ a= 3,b = {frac{3+3}{3-1}}=3, ab = 3 cdot 3 = 9. ]Conclusion: The minimum value of ( ab ) is ( boxed{9} ).

🧠:Okay, so I have this geometry problem here. Let me try to parse it step by step. We have triangle ABC. A median AF is drawn. A median connects a vertex to the midpoint of the opposite side, right? So AF is a median, which means F is the midpoint of BC. So, BF = FC. Then, point D is the midpoint of segment AF. So AF is the median, and D splits it into two equal parts. So AD = DF. E is the intersection point of line CD with side AB. So we draw a line from C to D, and where it crosses AB is point E. Given that BD = BF, we need to prove that AE = DE. Hmm. Let me sketch this out mentally. Maybe drawing a diagram would help. Let me imagine triangle ABC, with AF as the median. D is the midpoint of AF, so halfway between A and F. Then CD is drawn, intersecting AB at E. And BD equals BF. So BD is a segment from B to D, and BF is from B to F. Since F is the midpoint of BC, BF is half of BC. So BD is equal in length to BF, which is half BC. Our goal is to show that AE equals DE. So, on side AB, the point E is such that the distance from A to E is the same as from E to D. Wait, but D is on AF. So DE is a segment from E to D, which is on the median. This seems like a problem where using coordinate geometry might be helpful. Assign coordinates to the points and compute the necessary lengths. Alternatively, maybe using mass point geometry or similar triangles. Let's consider different approaches.First, coordinate geometry. Let's assign coordinates to the triangle. Let's place point A at (0, 0) for simplicity. Let’s let point B be at (2b, 0) so that AB is along the x-axis. Then point C can be at some point (c, h). But since AF is a median, F is the midpoint of BC. So coordinates of F would be ((2b + c)/2, h/2). But maybe to make calculations easier, we can choose specific coordinates. Let’s set A at (0, 0), B at (2, 0), and C at (0, 2). Wait, but then BC would be from (2,0) to (0,2), so midpoint F would be (1,1). Then AF is from (0,0) to (1,1). The midpoint D of AF would be (0.5, 0.5). Then line CD is from C (0,2) to D (0.5, 0.5). Let me find the equation of line CD. Slope of CD: (0.5 - 2)/(0.5 - 0) = (-1.5)/0.5 = -3. So equation is y - 2 = -3(x - 0), which is y = -3x + 2. Now, find intersection E with AB. AB is from (0,0) to (2,0), so it's the x-axis. So set y=0 in the equation of CD: 0 = -3x + 2 ⇒ x = 2/3. So E is at (2/3, 0). Now, check BD and BF. First, BD is from B (2,0) to D (0.5, 0.5). Let's compute BD's length. Distance formula: sqrt[(2 - 0.5)^2 + (0 - 0.5)^2] = sqrt[(1.5)^2 + (-0.5)^2] = sqrt[2.25 + 0.25] = sqrt[2.5] ≈ 1.581. BF is from B (2,0) to F (1,1). Distance: sqrt[(2 - 1)^2 + (0 - 1)^2] = sqrt[1 + 1] = sqrt[2] ≈ 1.414. But in this case, BD ≈ 1.581 and BF ≈ 1.414, which are not equal. But the problem states BD = BF. So my coordinate system here doesn't satisfy the given condition. Hmm, so maybe assigning coordinates arbitrarily might not work unless we take into account the condition BD = BF. So perhaps we need to set up coordinates such that BD = BF. Alternatively, maybe using vectors or coordinate geometry with variables and then impose the condition BD = BF. Let me try that.Let’s denote coordinates with variables. Let’s place A at (0,0), B at (2b,0), and C at (0,2c). Then F, the midpoint of BC, is at ((2b + 0)/2, (0 + 2c)/2) = (b, c). Then AF is from (0,0) to (b, c). The midpoint D of AF is at (b/2, c/2). Line CD connects C (0,2c) to D (b/2, c/2). Let's find the parametric equation of line CD. Parametric equations: x = 0 + t*(b/2 - 0) = t*b/2, y = 2c + t*(c/2 - 2c) = 2c + t*(-3c/2), where t ranges from 0 to 1. This line intersects AB at y=0. So set y=0:2c - (3c/2)t = 0 ⇒ 2c = (3c/2)t ⇒ t = (2c) / (3c/2) = (2c) * (2/(3c)) )= 4/(3). Wait, but t=4/3 is greater than 1, which would be outside the segment CD. Hmm, that can't be. So perhaps in this coordinate system, the intersection point E is beyond D? But the problem states that E is the intersection of CD with AB. If in our coordinate system, line CD extended beyond D intersects AB at E, then E is outside of CD. But according to the problem, E is the intersection point, which might just be on line CD, not necessarily on segment CD. So in this case, E would be at t=4/3. Thus, the coordinates of E would be x = (4/3)*(b/2) = (2b/3), y=0. So E is at (2b/3, 0). Now, let's compute BD and BF. First, BD is the distance from B (2b, 0) to D (b/2, c/2). BD = sqrt[(2b - b/2)^2 + (0 - c/2)^2] = sqrt[(3b/2)^2 + (-c/2)^2] = sqrt[(9b² + c²)/4] = (1/2)sqrt(9b² + c²)BF is the distance from B (2b, 0) to F (b, c). BF = sqrt[(2b - b)^2 + (0 - c)^2] = sqrt[b² + c²]Given BD = BF, so:(1/2)sqrt(9b² + c²) = sqrt(b² + c²)Square both sides:(1/4)(9b² + c²) = b² + c²Multiply both sides by 4:9b² + c² = 4b² + 4c²Bring terms to left side:9b² + c² -4b² -4c² = 0 ⇒ 5b² -3c² =0 ⇒ 5b² = 3c² ⇒ c² = (5/3)b² ⇒ c = b*sqrt(5/3)So the coordinates must satisfy c = b*sqrt(5/3). So if we take b=1, then c= sqrt(5/3). Let's choose b=1 for simplicity. Then coordinates:A(0,0), B(2,0), C(0, 2*sqrt(5/3)), F(1, sqrt(5/3)), D(0.5, sqrt(5/3)/2). E is at (2/3, 0) as computed before. Now, need to check AE and DE. Coordinates:A(0,0), E(2/3, 0). So AE is the distance from (0,0) to (2/3,0), which is 2/3.DE is the distance from D(0.5, sqrt(5/3)/2) to E(2/3, 0). Compute coordinates:x-coordinates: 0.5 (which is 1/2) to 2/3. Difference: 2/3 - 1/2 = (4/6 - 3/6) = 1/6y-coordinates: sqrt(5/3)/2 - 0 = sqrt(5/3)/2Thus, DE = sqrt[(1/6)^2 + (sqrt(5/3)/2)^2]Compute each term:(1/6)^2 = 1/36(sqrt(5/3)/2)^2 = (5/3)/4 = 5/12So DE = sqrt(1/36 + 5/12) = sqrt(1/36 + 15/36) = sqrt(16/36) = 4/6 = 2/3So DE = 2/3, which is equal to AE = 2/3. Hence, AE = DE. Therefore, in this coordinate system with the condition BD=BF satisfied, we indeed get AE=DE. So the problem is proven. But wait, this is using coordinate geometry. Maybe there's a more geometric proof without coordinates. Let me think.Alternative approach: Using mass point geometry. In mass point, we can assign masses to the points to balance the segments. Given that AF is a median, so F is midpoint of BC. So masses at B and C are equal. Let's assign mass 1 to B and C each, so mass at F is 1+1=2. D is the midpoint of AF, so masses at A and F must be equal for D to be the midpoint. Since F has mass 2, assign mass 2 to A. Then D, being the midpoint of AF, will have mass 2+2=4. Now, line CD intersects AB at E. Let's analyze the masses. When considering line CD, the masses at C and D are 1 and 4. So the ratio CE:ED would be inverse of masses, so 4:1. Wait, mass at C is 1, mass at D is 4, so the ratio is mass at D : mass at C = 4:1. Hence, CE:ED = 4:1. But wait, is this correct? Mass point assigns the ratio based on the masses. So if we have point E on CD, then the masses at C and D would determine the ratio CE:ED. Since mass at C is 1 and mass at D is 4, the ratio CE:ED is 4:1. So CE is four times ED. But E is also on AB. Let's see. Mass point for AB. Since we assigned mass 2 to A and mass 1 to B (from the median AF). Wait, earlier we assigned mass 2 to A and 1 to B? Wait, no. Let me recap:In mass point, when dealing with medians, the masses are determined by the midpoints. So since F is the midpoint of BC, masses at B and C are equal, say mass 1 each. Then mass at F is 1+1=2. Then D is the midpoint of AF. Since D is the midpoint, masses at A and F must be equal. Since F has mass 2, A must also have mass 2. Then mass at D is 2+2=4. Now, looking at line CD. Point C has mass 1, point D has mass 4. So the ratio CE:ED is mass D : mass C = 4:1. So CE is 4 parts, ED is 1 part. But E is also on AB. Let's analyze AB. Point A has mass 2, point B has mass 1. So the ratio AE:EB is mass B : mass A = 1:2. So AE is 1 part, EB is 2 parts. Hence, AE:EB = 1:2, meaning E divides AB in the ratio 1:2. But wait, AE:EB = 1:2. So if AB is of length, say 3 units, AE is 1 unit, EB is 2 units. Hence, AE = (1/3)AB. But from the coordinate system earlier, E was at 2/3 from A, but that was when AB was from 0 to 2. Wait, maybe there's confusion here. Wait, mass point assigns the ratio inversely to the masses. If mass at A is 2 and mass at B is 1, then the ratio AE:EB is mass B : mass A = 1:2. So E divides AB such that AE:EB = 1:2. So AE is 1 part, EB is 2 parts. Therefore, AE is (1/3) of AB. So if AB is length L, AE = L/3 and EB = 2L/3. But in our coordinate example, AE was 2/3. Wait, that contradicts. Wait, in the coordinate system, AB was from (0,0) to (2,0), so length 2. Then AE was 2/3, which is (2/3)/2 = 1/3 of AB. Wait, no. If AB is length 2, then AE = 2/3, which is (2/3)/2 = 1/3 of AB? Wait, no. If AB is length 2, then AE = 2/3, so AE is (2/3)/2 = 1/3 of AB? No, that's not correct. Wait, actually, the ratio AE:EB is 2/3 : (2 - 2/3) = 2/3 : 4/3 = 1:2. So AE:EB = 1:2. So AE is 1 part, EB is 2 parts, making AB 3 parts. But in reality, AE is 2/3, EB is 4/3. So 2/3 : 4/3 = 1:2. So yes, the ratio is 1:2. Therefore, mass point gives us that E divides AB in ratio 1:2, which matches the coordinate calculation. Now, in the coordinate system, DE was equal to 2/3, same as AE. But how does that come from mass point? Alternatively, using mass point, we can compute the length AE and DE. But perhaps we need to use coordinate geometry or another method. Wait, since E is on AB such that AE:EB = 1:2, and D is the midpoint of AF. If we can show that DE is equal to AE, which is 1/3 of AB, then it's done. But maybe using vectors. Let me try vectors. Let’s assign vector coordinates. Let’s take A as the origin. Then vector A is 0. Let’s denote vector B as b and vector C as c. Since F is the midpoint of BC, vector F = (b + c)/2. D is the midpoint of AF, so vector D = (0 + (b + c)/2)/2 = (b + c)/4. Line CD connects point C (c) to D ((b + c)/4). The parametric equation for CD is c + t[( (b + c)/4 - c )] = c + t[(b - 3c)/4], where t ∈ [0,1].We need to find the intersection E of CD with AB. AB is the line from A (0) to B (b), so parametric equation is sb, where s ∈ [0,1].Set the two parametric equations equal:c + t(b - 3c)/4 = sbThis gives:c + (t/4)b - (3t/4)c = sbGrouping terms:(t/4 - s)b + (1 - 3t/4)c = 0Since b and c are linearly independent (as A, B, C are vertices of a triangle), their coefficients must be zero:t/4 - s = 01 - 3t/4 = 0Solving the second equation: 1 = 3t/4 ⇒ t = 4/3Then from the first equation: s = t/4 = (4/3)/4 = 1/3Therefore, point E is at s = 1/3 along AB, so vector E = (1/3)bSo, AE is the vector from A (0) to E (1/3b), which has length 1/3 |b|. DE is the vector from D to E. Vector D is (b + c)/4. Vector E is (1/3)b. So DE = E - D = (1/3b) - (b + c)/4 = (4/12b - 3/12b - 3/12c) = (1/12b - 3/12c) = (1b - 3c)/12But we need to compute the length of DE and show it's equal to AE. Wait, maybe instead of lengths, we can compute squared lengths. |AE|² = |(1/3)b|² = (1/3)² |b|² = (1/9)|b|²|DE|² = |(1b - 3c)/12|² = (1/144)(|b|² - 6b·c + 9|c|²)But we need to relate this to the given condition BD = BF. First, compute BD and BF in vectors. Point B is at b, point D is at (b + c)/4. So vector BD = D - B = (b + c)/4 - b = (-3b + c)/4|BD|² = (|-3b + c|²)/16 = (9|b|² - 6b·c + |c|²)/16Point F is at (b + c)/2. Vector BF = F - B = (b + c)/2 - b = (-b + c)/2|BF|² = (|-b + c|²)/4 = (|b|² - 2b·c + |c|²)/4Given BD = BF, so |BD|² = |BF|²:(9|b|² - 6b·c + |c|²)/16 = (|b|² - 2b·c + |c|²)/4Multiply both sides by 16:9|b|² - 6b·c + |c|² = 4|b|² - 8b·c + 4|c|²Bring all terms to left side:9|b|² -6b·c + |c|² -4|b|² +8b·c -4|c|² =0Simplify:(9|b|² -4|b|²) + (-6b·c +8b·c) + (|c|² -4|c|²) =05|b|² +2b·c -3|c|² =0So 5|b|² +2b·c -3|c|² =0. Let's note this equation. Now, go back to |DE|². From above:|DE|² = (1/144)(|b|² -6b·c +9|c|²)We need to show |DE|² = |AE|² = (1/9)|b|²So, set:(1/144)(|b|² -6b·c +9|c|²) = (1/9)|b|²Multiply both sides by 144:(|b|² -6b·c +9|c|²) = 16|b|²Bring all terms to left side:|b|² -6b·c +9|c|² -16|b|² =0Simplify:-15|b|² -6b·c +9|c|² =0Divide both sides by -3:5|b|² +2b·c -3|c|² =0Which is exactly the equation we obtained from BD = BF. Therefore, the condition BD = BF ensures that |DE|² = |AE|², hence DE = AE. Therefore, the proof is complete. Alternatively, using coordinate geometry as we did earlier by setting specific coordinates and verifying the condition also works. Another approach could be using Ceva's theorem. Let me explore that. Ceva's theorem states that for concurrent lines from the vertices of a triangle, the product of certain ratios equals 1. But in this case, lines CD and AF might be considered. Wait, E is the intersection of CD with AB. Let me think.If we consider point E on AB, D on AF, and C connected to D. Maybe Ceva's theorem can be applied to triangle ABF, but not sure. Alternatively, consider triangle ABC with cevians AF (median) and CD. Wait, Ceva’s Theorem states that for concurrent cevians AD, BE, CF in triangle ABC, (AF/FB)(BD/DC)(CE/EA) =1. But in our case, we have cevians AF (a median) and CD intersecting at E. But we need a third cevian to apply Ceva. Maybe not directly applicable. Alternatively, use Menelaus' theorem. Menelaus' theorem for triangle AFB with transversal CDE. Wait, Menelaus' theorem relates the ratios of a transversal cutting through the sides of a triangle. Alternatively, considering triangle AFB and the line CD passing through D and E. Let me attempt:In triangle AFB, the transversal CDE intersects AF at D, FB at some point, and AB at E. Wait, but CD starts at C, which is not part of triangle AFB. Maybe not straightforward. Alternatively, take triangle ADC. Line BE intersects AD at D and AC at C? Hmm, not sure. Alternatively, consider triangle ABD with cevian CD. But I think Ceva or Menelaus might complicate here. Given that we have already two approaches (coordinate geometry and vector) that worked, perhaps it's sufficient. Another approach is using similar triangles. Looking for similar triangles involving AE and DE. Since D is the midpoint of AF, and AF is a median, perhaps creating some parallels or similar triangles. Alternatively, construct a parallelogram or use midline theorem. Wait, since D is the midpoint of AF, and if we can show that DE is parallel to some line which might create similar triangles. Alternatively, let's consider triangles AED and some other triangle. Wait, let's look back at the coordinate geometry result. E divides AB in ratio 1:2 (AE:EB =1:2), and D is the midpoint of AF. In coordinates, DE turned out to be equal to AE. But maybe there's a way to see this through proportions. Since BD = BF, and F is the midpoint of BC, BF = FC. So BD = BF = FC. So BD = FC. Hmm, so BD = FC. Since F is the midpoint of BC, FC = BF, which is given equal to BD. So BD = FC. Maybe we can find some congruent triangles or parallelograms. Alternatively, consider triangle BFC and triangle BDC. Wait, BD = BF, FC is known. But not sure. Alternatively, look at triangle BDF. BD = BF, so it's an isosceles triangle. So angles at D and F are equal. But how does that help with AE and DE? Alternatively, since D is the midpoint of AF, and in the coordinate system, DE turned out to be equal to AE. So perhaps triangle AED is isosceles with AE = DE. To prove that, maybe show that angles at E are equal. Alternatively, use coordinate geometry to find slopes and show perpendicularity or something else. But in our coordinate example, the triangle wasn't necessarily right-angled. Alternatively, use the midpoint theorem. Wait, since D is the midpoint of AF, if we can connect D to E such that DE is part of some midline. Alternatively, construct midpoint of AB and see relations. But perhaps it's getting too vague. Given that both coordinate and vector methods worked, and mass point also indicated the ratios, perhaps the most straightforward is coordinate geometry with the given condition BD=BF leading to AE=DE. So, in conclusion, through coordinate geometry, we assigned coordinates to the triangle, applied the given condition BD=BF, derived the necessary relation between sides, computed the coordinates of E, and showed that AE=DE. Similarly, using vectors, we arrived at the same conclusion by leveraging the given condition to relate the vectors and show the equality of lengths. Hence, the problem is proven.

✅:To prove that (AE = DE) in triangle (ABC) with the given conditions, we can use coordinate geometry and vector analysis to systematically verify the required lengths.Step-by-Step Explanation:1. Assign Coordinates: - Place (A) at ((0, 0)), (B) at ((2b, 0)), and (C) at ((0, 2c)). - Midpoint (F) of (BC) is ((b, c)). - Midpoint (D) of (AF) is (left(frac{b}{2}, frac{c}{2}right)).2. Equation of Line (CD): - Parametric form of (CD): (x = frac{b}{2}t), (y = 2c - frac{3c}{2}t). - Intersection (E) with (AB) (where (y=0)): Solve (0 = 2c - frac{3c}{2}t) gives (t = frac{4}{3}). Thus, (E) is (left(frac{2b}{3}, 0right)).3. Verify (BD = BF): - Calculate (BD = frac{1}{2}sqrt{9b^2 + c^2}) and (BF = sqrt{b^2 + c^2}). - Set (BD = BF) leading to (5b^2 = 3c^2) or (c = bsqrt{frac{5}{3}}).4. Compute Lengths (AE) and (DE): - (AE = frac{2b}{3}) (distance from (A(0,0)) to (Eleft(frac{2b}{3}, 0right))). - (DE = sqrt{left(frac{2b}{3} - frac{b}{2}right)^2 + left(0 - frac{c}{2}right)^2}). - Substitute (c = bsqrt{frac{5}{3}}) into (DE): Simplifies to (frac{2b}{3}), hence (AE = DE).Conclusion:By methodically applying coordinate geometry and vector analysis under the condition (BD = BF), we demonstrated that (AE) and (DE) are equal in length. Thus, (AE = DE) is proven.(boxed{AE = DE})

🔑:1. Triangle and Median construction: Let us consider the triangle ( triangle ABC ) with median ( AF ) such that ( F ) is the midpoint of ( BC ).2. Midpoint Definition: The point ( D ) is the midpoint of segment ( AF ).3. Intersection Point: Let ( E ) be the point where the line ( CD ) intersects the side ( AB ).4. Equal Segments: It is given that ( BD = BF ).5. Angle Properties: - Since ( BD = BF ), ( triangle BDF ) is isosceles with ( BD ) and ( BF ) as its equal sides. - Consequently, ( angle BDF = angle BFD ) as base angles of an isosceles triangle are equal.6. Supplementary Angles: - Since ( angle BDF = angle BFD ), the supplementary pairs ( angle BDA ) and ( angle DFC ) must also be equal because they are the external angles of the isosceles triangle. 7. Congruent Triangles: - It is given ( AD = DF ) because ( D ) is the midpoint of ( AF ). - Also, ( BD = CF ) as given in the problem ( BD = BF ). - Thus, triangles ( triangle ABD ) and ( triangle DCF ) share two sides and the angle between them, proving the triangles are congruent by the SAS (Side-Angle-Side) postulate.8. Angle Correspondence: - Since ( triangle ABD cong triangle DCF ): - ( angle BAD = angle CDF ) because corresponding angles in congruent triangles are equal. 9. Vertical Angles: - The angles ( angle ADE = angle CDF ) are vertical angles and are therefore equal.10. Conclusion: - Since ( angle ADE = angle DAE ) (because both are equal to ( angle CDF )), triangle ( triangle ADE ) is isosceles. - Thus, ( AE = DE ).[boxed{AE = DE}]

🧠:Okay, so I need to prove that for every prime ( p > 100 ) and any integer ( r ), there exist integers ( a ) and ( b ) such that ( p ) divides ( a^2 + b^5 - r ). Hmm, let me think about how to approach this.First, the problem is about showing that the equation ( a^2 + b^5 equiv r mod p ) has solutions for any prime ( p > 100 ) and any integer ( r ). So, equivalently, we need to show that the set of values ( a^2 + b^5 ) modulo ( p ) covers all possible residues modulo ( p ). That means, for any given ( r ), there's some combination of ( a ) and ( b ) that hits that residue.I remember that in finite fields, like the field ( mathbb{F}_p ), certain equations have solutions depending on the properties of the field. Since ( p ) is a prime, ( mathbb{F}_p ) is a finite field with ( p ) elements. Maybe I can use some theorems from number theory related to finite fields.One approach might be to fix one variable and solve for the other. For example, fix ( b ) and then solve for ( a ), or fix ( a ) and solve for ( b ). Let me consider fixing ( b ) first. If I can show that for some ( b ), ( r - b^5 ) is a quadratic residue modulo ( p ), then there exists an ( a ) such that ( a^2 equiv r - b^5 mod p ), which would solve the problem.So, the problem reduces to showing that for some ( b ), ( r - b^5 ) is a quadratic residue modulo ( p ). Alternatively, we can rephrase this: the image of the map ( b mapsto b^5 ) modulo ( p ) must intersect with the set ( r - QR_p ), where ( QR_p ) is the set of quadratic residues modulo ( p ).Alternatively, maybe instead of fixing one variable, we can consider the sizes of the sets involved. The number of quadratic residues modulo ( p ) is ( frac{p + 1}{2} ), and the number of possible values of ( b^5 ) modulo ( p ) depends on the exponent 5. Since ( p ) is prime, the multiplicative group ( mathbb{F}_p^times ) is cyclic of order ( p - 1 ). The number of distinct fifth powers in ( mathbb{F}_p^times ) is ( frac{p - 1}{gcd(5, p - 1)} ). Therefore, if 5 divides ( p - 1 ), then the number of fifth powers is ( frac{p - 1}{5} ), otherwise, it's ( p - 1 ).Wait, if 5 does not divide ( p - 1 ), then the map ( x mapsto x^5 ) is a bijection on ( mathbb{F}_p^times ), right? Because in a cyclic group, if the exponent is coprime to the group order, then the map is a permutation. So, if ( gcd(5, p - 1) = 1 ), then every element is a fifth power. So in that case, the number of fifth powers is ( p - 1 ), but since we include 0, if ( b ) can be 0, then the total number of fifth powers modulo ( p ) is ( p ). Wait, no, if ( b ) is allowed to be 0, then ( b^5 = 0 ). So the set ( { b^5 mod p } ) is either size ( frac{p - 1}{gcd(5, p - 1)} + 1 ) (including 0) or something else?Wait, actually, ( b ) ranges over all integers modulo ( p ), so ( b^5 ) can be 0 when ( b = 0 ), and the rest are elements of ( mathbb{F}_p^times ). So if ( d = gcd(5, p - 1) ), then the number of distinct fifth powers in ( mathbb{F}_p^times ) is ( frac{p - 1}{d} ), and including 0, it's ( frac{p - 1}{d} + 1 ).Similarly, the number of quadratic residues modulo ( p ) is ( frac{p + 1}{2} ), since 0 is a quadratic residue (0^2 = 0), and there are ( frac{p - 1}{2} ) non-zero quadratic residues.So, perhaps using additive combinatorics or the pigeonhole principle, if the sumset of quadratic residues and fifth powers covers all residues modulo ( p ). Wait, but the problem is not exactly a sumset, because it's not the sum of a quadratic residue and a fifth power, but rather for each ( r ), there exists some quadratic residue ( a^2 ) and some fifth power ( b^5 ) such that ( a^2 + b^5 equiv r mod p ).Alternatively, think of it as for each ( r ), the equation ( a^2 equiv r - b^5 mod p ) must have a solution ( a ) for some ( b ). So for some ( b ), ( r - b^5 ) must be a quadratic residue. So, we need to show that the image of the map ( b mapsto r - b^5 ) intersects the set of quadratic residues non-trivially.Alternatively, if we fix ( r ), then we need that ( r - QR_p ) intersects the set of fifth powers. Since ( r ) is arbitrary, maybe shifting the set of quadratic residues by ( r ) will cover enough elements such that there's an overlap with fifth powers.But how can we ensure that? Maybe by estimating the sizes of the sets.Suppose that the number of quadratic residues is ( frac{p + 1}{2} ), and the number of fifth powers is ( frac{p - 1}{d} + 1 ), where ( d = gcd(5, p - 1) ). If the sum of these two sets' sizes exceeds ( p ), then by the pigeonhole principle, their sumset covers all residues. But wait, this is not exactly a sumset; it's more like for each ( r ), ( r ) is in the sumset QR + fifth powers. But the problem is different: for each ( r ), ( r ) is equal to some ( a^2 + b^5 ). So, the union over all ( b ) of ( { b^5 + a^2 } ) must cover all residues.Alternatively, if we fix ( r ), then we need that ( r - b^5 ) is in QR_p for some ( b ). So, if the set ( { r - b^5 mod p } ) as ( b ) varies must intersect QR_p. The size of ( { r - b^5 mod p } ) is equal to the number of distinct fifth powers. If the number of fifth powers is large enough, and QR_p is also a large set, then their intersection must be non-empty.Using the Cauchy-Davenport theorem or similar additive combinatorics results? Wait, but in the multiplicative group, maybe not directly applicable. Alternatively, use the fact that if two sets are large enough, their product (in some sense) must cover the whole space.But let's think in terms of probabilities. The probability that a random element is a quadratic residue is about 1/2. The number of fifth powers is at least ( frac{p - 1}{5} + 1 approx p/5 ). So, if we have roughly p/5 elements in the fifth powers set, and we shift them by r, then the shifted set has the same size. The intersection with QR_p would be roughly (p/5)*(1/2) = p/10 elements. But this is heuristic.Wait, but maybe using the fact that the number of pairs (a, b) is p * p = p^2, and the number of possible equations is p, so by pigeonhole, each residue is hit multiple times. But that's not precise. The problem is to show that each residue is hit at least once.Alternatively, use the Weil conjectures or bounds on character sums. For example, consider the character sum over additive characters, but this might be complicated.Wait, here's a thought. Let me consider the equation ( a^2 + b^5 = r mod p ). Let's fix ( b ), then ( a^2 = r - b^5 mod p ). For each ( b ), there's either 0, 1, or 2 solutions for ( a ). The total number of solutions (a, b) is roughly the number of ( b ) such that ( r - b^5 ) is a quadratic residue, multiplied by 1 or 2.So, the total number of solutions is approximately ( 2 times ) (number of ( b ) such that ( r - b^5 ) is QR). If we can show that this number is at least 1 for each ( r ), then we are done.But how do we estimate the number of ( b ) such that ( r - b^5 ) is QR?This seems related to character sums. Let me recall that the number of solutions to an equation can be expressed using multiplicative characters. The number of ( b ) such that ( r - b^5 ) is a quadratic residue is equal to:( sum_{b} frac{1 + chi(r - b^5)}{2} ),where ( chi ) is the quadratic character modulo ( p ). So, the number is:( frac{p}{2} + frac{1}{2} sum_{b} chi(r - b^5) ).Therefore, if we can show that the character sum ( sum_{b} chi(r - b^5) ) is small in absolute value, then the number of such ( b ) is roughly ( p/2 ), which would certainly be greater than 1, hence ensuring a solution.So, using the Hölder inequality or Weil's bound for character sums. The Weil bound states that for a non-trivial additive character ( psi ) and a polynomial ( f ) of degree ( n ), ( | sum_{x in mathbb{F}_p} psi(f(x)) | leq (n - 1)sqrt{p} ). But here we have a multiplicative character sum. The relevant bound would be the Weil bound for multiplicative characters.Weil's bound for multiplicative character sums: if ( f(x) ) is a polynomial that is not a perfect power (i.e., cannot be written as ( g(x)^m ) for some polynomial ( g ) and ( m > 1 )), then:( left| sum_{x in mathbb{F}_p} chi(f(x)) right| leq (d - 1)sqrt{p} ),where ( d ) is the degree of ( f ).In our case, ( f(b) = r - b^5 ). So, the degree is 5. Assuming that ( f(b) ) is not a perfect square, cube, etc. But since ( f(b) = r - b^5 ), it's a linear term in ( r ), but degree 5 in ( b ). Unless ( 5 ) is a multiple of the character order. Wait, ( chi ) is a quadratic character, so order 2. Since 5 and 2 are coprime, ( f(b) ) is not a perfect square (unless ( r - b^5 ) is a square, which is exactly what we are counting). So, the polynomial ( f(b) = r - b^5 ) is not a perfect square as a polynomial, so it's square-free (assuming ( r ) is a constant). Therefore, the Weil bound applies, and we have:( left| sum_{b in mathbb{F}_p} chi(r - b^5) right| leq (5 - 1)sqrt{p} = 4sqrt{p} ).Therefore, the character sum is bounded by ( 4sqrt{p} ). Then, the number of ( b ) such that ( r - b^5 ) is a quadratic residue is at least:( frac{p}{2} - frac{1}{2} times 4sqrt{p} = frac{p}{2} - 2sqrt{p} ).Since ( p > 100 ), ( sqrt{p} > 10 ), so ( 2sqrt{p} < p/10 ) for ( p > 100 ). Wait, let's check for ( p = 101 ), ( 2sqrt{101} approx 20.1 ), while ( p/2 approx 50.5 ), so ( 50.5 - 20.1 approx 30.4 ), which is still positive. For larger ( p ), this lower bound ( frac{p}{2} - 2sqrt{p} ) increases. Therefore, the number of ( b ) such that ( r - b^5 ) is a quadratic residue is at least ( frac{p}{2} - 2sqrt{p} ), which is certainly positive for ( p > 100 ). Therefore, there exists at least one ( b ) such that ( r - b^5 ) is a quadratic residue, hence there exists an ( a ) with ( a^2 equiv r - b^5 mod p ), so ( a^2 + b^5 equiv r mod p ).Therefore, the proof is complete by using the Weil bound to show that the character sum is small, hence the number of valid ( b ) is positive.Wait, but let me check if all the conditions for applying Weil's bound are satisfied. The polynomial ( f(b) = r - b^5 ) must not be a perfect power in the multiplicative sense. Since ( f(b) ) is a degree 5 polynomial, and 5 is prime, it can't be written as a perfect square, cube, etc., unless it's a constant polynomial. But since ( f(b) = r - b^5 ) has degree 5, which is not a multiple of the character's order (which is 2). Therefore, the multiplicative character ( chi ) is applied to a polynomial that's not a perfect square, so Weil's bound applies. Therefore, the bound holds, and the rest follows.Hence, for any prime ( p > 100 ) and any integer ( r ), there exist integers ( a ) and ( b ) such that ( p ) divides ( a^2 + b^5 - r ).Final Answerboxed{text{For every prime } p > 100 text{ and integer } r, text{ such integers } a text{ and } b text{ exist.}}

✅:To prove that for every prime ( p > 100 ) and every integer ( r ), there exist integers ( a ) and ( b ) such that ( p ) divides ( a^2 + b^5 - r ), we need to show that the equation ( a^2 + b^5 equiv r mod p ) has solutions for any prime ( p > 100 ) and any integer ( r ).1. Quadratic Residues and Fifth Powers: - The set of quadratic residues modulo ( p ) has size ( frac{p + 1}{2} ). - The set of fifth powers modulo ( p ) depends on ( gcd(5, p - 1) ). If ( gcd(5, p - 1) = d ), the number of distinct fifth powers in ( mathbb{F}_p^times ) is ( frac{p - 1}{d} ), including 0, the total is ( frac{p - 1}{d} + 1 ).2. Character Sum Approach: - We need to show that for some ( b ), ( r - b^5 ) is a quadratic residue modulo ( p ). - Using the quadratic character ( chi ), the number of such ( b ) is given by: [ sum_{b} frac{1 + chi(r - b^5)}{2} = frac{p}{2} + frac{1}{2} sum_{b} chi(r - b^5) ]3. Weil Bound Application: - The Weil bound for multiplicative character sums states that for a polynomial ( f(x) ) of degree ( d ) that is not a perfect power, the character sum is bounded by ( (d - 1)sqrt{p} ). - For the polynomial ( f(b) = r - b^5 ), the degree is 5, and the Weil bound gives: [ left| sum_{b in mathbb{F}_p} chi(r - b^5) right| leq 4sqrt{p} ]4. Lower Bound on Solutions: - The number of ( b ) such that ( r - b^5 ) is a quadratic residue is at least: [ frac{p}{2} - 2sqrt{p} ] - For ( p > 100 ), this lower bound is positive, ensuring there exists at least one ( b ) such that ( r - b^5 ) is a quadratic residue.5. Conclusion: - For such ( b ), there exists an ( a ) with ( a^2 equiv r - b^5 mod p ), hence ( a^2 + b^5 equiv r mod p ).Thus, for every prime ( p > 100 ) and every integer ( r ), there exist integers ( a ) and ( b ) such that ( p ) divides ( a^2 + b^5 - r ).[boxed{text{For every prime } p > 100 text{ and integer } r, text{ such integers } a text{ and } b text{ exist.}}]

🔑:Fix a prime ( p > 100 ) and consider an integer ( r ). The goal is to show that there exist integers ( a ) and ( b ) such that ( p ) divides ( a^{2} + b^{5} - r ).# Step 1: Introduction of Residue ClassesDefine the set ( mathcal{P} = {0, 1, ldots, p-1} ), which represents the residue classes modulo ( p ). For every integer ( r in mathcal{P} ), let ( S_{r} = {(a, b) in mathcal{P} times mathcal{P} : a^2 + b^5 equiv r pmod{p}} ) be the set of pairs ( (a, b) ) satisfying the congruence. Let ( s_{r} = |S_{r}| ) denote the size of this set. We aim to prove that ( s_{r} > 0 ) for all ( r in mathcal{P} ).# Step 2: Counting Solutions with LemmaWe will use the following lemma: Lemma: Let ( N ) be the number of quadruples ( (a, b, c, d) in mathcal{P}^{4} ) such that ( a^2 + b^5 equiv c^2 + d^5 pmod{p} ). Then,[ N = sum_{r in mathcal{P}} s_{r}^{2} ]and [ N leq p left( p^{2} + 4p - 4 right). ]Proof of the Lemma:1. Calculation of ( sum_{r in mathcal{P}} s_{r}^{2} ): For each residue class ( r ), there are exactly ( s_{r} ) pairs ( (a, b) ) such that ( a^2 + b^5 equiv r pmod{p} ), and thereby ( s_{r} ) pairs ( (c, d) ) such that ( c^2 + d^5 equiv r pmod{p} ). Therefore, the number of quadruples ( (a, b, c, d) ) such that ( a^2 + b^5 equiv c^2 + d^5 equiv r ) is ( s_{r}^{2} ). Summing over all ( r in mathcal{P} ), [ N = sum_{r in mathcal{P}} s_{r}^{2}. ]2. Bounding ( N ): - For ( b^{5} equiv d^{5} pmod{p} ), if ( b equiv d equiv 0 ), there is one solution. If ( b not equiv 0 ), there can be at most 5 solutions (since ( x^5 equiv r ) has at most 5 solutions in modular arithmetic for prime modulus). Hence, there are at most ( 5p - 4 ) such pairs. - For ( a^2 equiv c^2 pmod{p} ), which gives ( a equiv pm c ), there are ( 2p - 1 ) pairs. - If ( b^5 notequiv d^5 pmod{p} ), ( (a, c) ) are distinct and satisfy ( (a - c)(a + c) = d^5 - b^5 ). Since ( a - c ) determines ( a + c ), there are ( p - 1 ) suitable pairs ( (a, c) ). Therefore, [ N leq (5p - 4)(2p - 1) + (p^{2} - (5p - 4))(p - 1). ] Simplifying, [ N leq p left( p^{2} + 4p - 4 right). ]# Step 3: Proof by ContradictionSuppose ( S_{r} = emptyset ) for some ( r in mathcal{P} ). This would mean no pairs ( (a, b) ) satisfy ( a^2 + b^5 equiv r pmod{p} ).Define the set ( T = { x^{10} mid x in mathcal{P} setminus {0} } ) of nonzero tenth powers modulo ( p ). Each residue class is a tenth power of at most 10 elements. Hence, ( |T| geq frac{p - 1}{10} geq 4 ).For every ( t in T ), ( S_{tr} = emptyset ), because if ( (a, b) in S_{tr} ) with ( t equiv z^{10} ), then ( (z^{-5}a, z^{-2}b) in S_{r} ), contradicting ( S_{r} = emptyset ).Thus, there are at least ( frac{p - 1}{10} geq 4 ) empty sets ( S_{1}, S_{2}, ldots, S_{p-1} ), leaving at most ( p - 4 ) nonempty sets.By the AM-QM Inequality,[ sum_{r in mathcal{P} setminus r T} s_{r}^2 geq frac{1}{p-4} left( sum_{r in mathcal{P} setminus r T} s_{r} right)^2 = frac{p^4}{p-4} > p (p^2 + 4p - 4),]which is a contradiction to the lemma.Hence, every ( r in mathcal{P} ) must be expressible as ( a^2 + b^5 ) for some integers ( a, b ). # Conclusion[boxed{text{For every prime } p>100 text{ and every integer } r, text{ there exist integers } a text{ and } b text{ such that } p text{ divides } a^2 + b^5 - r.}]

🧠:Okay, so I need to prove that DEGH is a parallelogram. Let me start by understanding the problem setup.We have triangle ABC with three external triangles: ANC, CLB, and BKA. Each of these triangles is erected outside ABC, and they have some angle conditions. Specifically, the angles at NAC, KBA, and LCB are equal, and the angles at NCA, KAB, and LBC are equal. Then D, E, G, H are midpoints of AB, LK, CA, and NA respectively. We need to show that the quadrilateral formed by these midpoints, DEGH, is a parallelogram.First, I should recall that a quadrilateral is a parallelogram if both pairs of opposite sides are parallel and equal in length, or if one pair of opposite sides are both parallel and equal, or if the diagonals bisect each other. So, my goal is to show one of these conditions for DEGH.Since D, E, G, H are midpoints, maybe the Midline Theorem (also known as the Midsegment Theorem) could be useful here. The Midline Theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half as long. So, if I can find triangles where these midpoints are connected, maybe I can apply this theorem.But first, let me try to visualize the figure. ABC is the main triangle. Triangles ANC, CLB, and BKA are outside. Let me try to sketch this mentally. Point N is outside AC such that triangle ANC is erected with angles at A and C equal to those in the other triangles. Similarly, L is outside BC, and K is outside AB. Wait, the angles at NAC, KBA, and LCB are equal. Let's denote this common angle as α. Similarly, the angles at NCA, KAB, and LBC are equal; let's denote this common angle as β. So, each of these external triangles has two angles equal to α and β. Since the sum of angles in a triangle is 180 degrees, α + β + angle at the third vertex = 180. But the third vertex in each case is part of the external triangle. For example, in triangle ANC, the angles at N, A, and C are α, β, and the remaining angle at N. Wait, no, the angles at A and C in triangle ANC are given as α and β. So angle NAC = α, angle NCA = β. Then angle at N would be 180 - α - β. Similarly for the other triangles.This seems important. So all three external triangles have two angles α and β, and the third angle 180 - α - β. Therefore, these triangles are similar to each other? If they have the same angles, they are similar. So triangles ANC, CLB, and BKA are similar. That might be useful.Given that, perhaps the sides of these triangles are proportional. But I need to relate this to the midpoints.Let me note the midpoints:- D is the midpoint of AB.- E is the midpoint of LK.- G is the midpoint of CA.- H is the midpoint of NA.So, H is the midpoint of NA, which is part of triangle ANC. G is the midpoint of CA. D is the midpoint of AB. E is the midpoint of LK, which is part of triangle CLB? Wait, LK is part of triangle CLB? Wait, triangle CLB has vertices C, L, B. So LK is not a side of that triangle. Wait, maybe LK is a side of another triangle? Wait, L is a vertex of triangle CLB, and K is a vertex of triangle BKA. So, LK is a segment connecting L and K. Hmm. Maybe I need to figure out the positions of points N, L, K.Alternatively, perhaps using vectors or coordinate geometry could help here. Since the problem involves midpoints and parallelograms, coordinate geometry might be a good approach. Let me try assigning coordinates to the points.Let me place triangle ABC in the coordinate plane. Let’s assign coordinates:Let’s let point A be at (0, 0), point B at (2b, 0), and point C at (2c, 2d). Choosing coordinates with midpoints in mind, since midpoints will be easier to compute if coordinates are even numbers. So, midpoint D of AB would then be at (b, 0). Midpoint G of CA would be at (c, d). H is the midpoint of NA; but I need to find coordinates for N first.But to find N, L, K, I need to use the angle conditions. The triangles ANC, CLB, and BKA are erected outside ABC with angles α and β at specific vertices.Given that angles at NAC and KBA and LCB are equal to α, and angles at NCA, KAB, and LBC are equal to β. Since each of these triangles has two angles α and β, they are similar. So, triangle ANC ~ triangle CLB ~ triangle BKA.Therefore, their corresponding sides are proportional. Let’s denote the ratios of similarity. Let me suppose that triangle ANC, CLB, BKA are similar with scaling factors. But without loss of generality, maybe the ratios can be determined based on the sides of ABC? Hmm, this is getting a bit abstract.Alternatively, maybe using complex numbers would help. Representing points as complex numbers and then using rotation and scaling to construct points N, L, K.Since the triangles are similar and externally erected, perhaps each is constructed by rotating a side of ABC by angle α or β and scaling. Wait, but the angles α and β are given at specific vertices.Let me think about triangle ANC. It's erected on AC, outside ABC. The angles at A and C are α and β. So, point N is such that in triangle ANC, angle at A is α, angle at C is β. So, the direction from A to N makes an angle α with AC, and the direction from C to N makes an angle β with CA.Alternatively, perhaps constructing point N by rotating AC around A by angle α and around C by angle β? Wait, maybe using rotation transformations.Suppose we rotate point C around A by angle α to get a direction, and rotate point A around C by angle β to get another direction; the intersection of these two rotated lines would be point N.Similarly for points L and K. Since the angles are the same, each of these points is constructed by rotating the respective sides by α and β.Since all three triangles are similar, the rotations would be similar, leading to proportional sides.But perhaps this is too vague. Let me try to formalize it.Let’s consider triangle ANC. To construct N, we can consider that from point A, we have a direction making angle α with AC, and from point C, a direction making angle β with CA. The intersection of these two directions is point N.Similarly, for triangle CLB: from point C, a direction making angle α with CB, and from point B, a direction making angle β with BC. The intersection is point L.For triangle BKA: from point B, a direction making angle α with BA, and from point A, a direction making angle β with AB. The intersection is point K.If these angles α and β are the same for each construction, then the triangles ANC, CLB, BKA are similar.Given that, maybe the key is to show that the midpoints D, E, G, H form a parallelogram by showing that the vectors between them are equal or that their coordinates satisfy the parallelogram conditions.Let me try coordinates. Let’s assign coordinates to ABC first. Let me set coordinate system with A at (0,0), B at (2,0), and C at (0,2). Then midpoints would be easier: D is midpoint of AB at (1,0), G is midpoint of CA at (0,1). But we need to find coordinates of N, L, K.Wait, but given the angles α and β, constructing N, L, K might be complex. Alternatively, maybe there is symmetry or transformation that can be applied.Alternatively, perhaps using vectors. Let’s denote vectors for points A, B, C as vectors a, b, c. Then points N, L, K can be expressed in terms of a, b, c, and rotations.But to handle the angles α and β, perhaps using rotation matrices. For example, in triangle ANC, angle at A is α, angle at C is β. So, vector n - a is a rotation of vector c - a by angle α, scaled by some factor. Similarly, vector n - c is a rotation of vector a - c by angle β, scaled by another factor.Wait, perhaps. Let me formalize this. Let’s denote that in triangle ANC, angle at A is α, angle at C is β. Then, the direction from A to N is a rotation of the direction from A to C by α. Similarly, the direction from C to N is a rotation of the direction from C to A by β.But the rotations could be clockwise or counterclockwise depending on the orientation. Since the triangles are erected outside ABC, the rotations are likely in a specific direction. Let me assume they are rotated outward, so the direction of rotation would be consistent.Suppose from point A, rotating AC by angle α to get AN. Similarly, from point C, rotating CA by angle β to get CN. The intersection of these two lines is point N.Similarly for the other points.If I can express points N, L, K in terms of A, B, C with these rotations, then I can find coordinates for N, L, K, and subsequently find midpoints E, H, etc.But this seems involved. Let me attempt to define rotation matrices.Let’s denote a rotation by angle θ as R(θ). If we are rotating a vector v by θ, then the rotated vector is R(θ)v.Assuming that the rotation is counterclockwise unless specified otherwise. But since the triangles are erected externally, the direction of rotation might be different. Let me consider that.For triangle ANC, erected outside ABC. So, if we are at point A, the direction from A to N is obtained by rotating AC by α outward. Similarly, at point C, the direction from C to N is obtained by rotating CA by β outward.Depending on the orientation of triangle ABC, outward could mean different directions. This is getting a bit confusing. Maybe using complex numbers would help here, as rotations can be easily represented.Let me represent points as complex numbers. Let’s assign complex coordinates:Let A = 0, B = 2b, C = 2c + 2di (using complex plane where real and imaginary parts are coordinates). Then midpoints D = (0 + 2b)/2 = b, G = (0 + 2c + 2di)/2 = c + di.Now, to find N, L, K.Starting with point N in triangle ANC. At point A (0), angle NAC = α, angle NCA = β. So, in complex plane terms, the line AN is obtained by rotating AC by α degrees. Similarly, line CN is obtained by rotating CA by β degrees.Assuming AC is the vector from A to C, which is 2c + 2di. Rotating this vector by α (outward, so direction depends on the orientation). Wait, but in which direction? If ABC is oriented counterclockwise, then the external triangles would be rotated in the clockwise direction. Hmm.Alternatively, perhaps the rotation is such that the triangle ANC is outside ABC. So, if ABC is counterclockwise, then to construct N outside, we might rotate AC around A clockwise by α. Similarly, rotate CA around C clockwise by β.Assuming that, let's model the rotations.The direction from A to N is AC rotated by -α (clockwise) around A. Similarly, the direction from C to N is CA rotated by -β (clockwise) around C.In complex numbers, rotating a vector v by angle θ around point a is given by a + (v - a) * e^{-iθ}. Wait, no. If we have a point p, rotating it around a by angle θ is a + ( p - a ) * e^{iθ} if counterclockwise, or * e^{-iθ} if clockwise.So, to construct point N:From A: Take vector AC (which is C - A = 2c + 2di) and rotate it clockwise by α around A (which is 0). So the direction vector becomes (2c + 2di) * e^{-iα}.But we need to scale this appropriately to get point N. Wait, but in triangle ANC, the angles at A and C are α and β. So the ratio of sides AN / AC = sin β / sin (α + β), by the Law of Sines.Wait, Law of Sines in triangle ANC: AN / sin β = CN / sin α = AC / sin (180 - α - β) = AC / sin (α + β).Therefore, AN = AC * sin β / sin (α + β)Similarly, CN = AC * sin α / sin (α + β)Therefore, if we can express vectors AN and CN accordingly.Let’s denote AC = |C - A| = |2c + 2di| = 2√(c² + d²). Then AN = 2√(c² + d²) * sin β / sin (α + β), and CN = 2√(c² + d²) * sin α / sin (α + β).Therefore, vector AN is in the direction of AC rotated by -α (clockwise), scaled by (sin β / sin (α + β)).Similarly, vector CN is in the direction of CA rotated by -β (clockwise), scaled by (sin α / sin (α + β)).Therefore, point N can be expressed as:N = A + (C - A) * e^{-iα} * (sin β / sin (α + β))Similarly, point N can also be expressed from point C:N = C + (A - C) * e^{-iβ} * (sin α / sin (α + β))Since both expressions must equal N, we can set them equal and solve for consistency. But perhaps this is overcomplicating. Let's proceed with the first expression.Thus, N = (2c + 2di) * e^{-iα} * (sin β / sin (α + β))Similarly, points L and K can be constructed by analogous rotations on their respective sides.For point L in triangle CLB:Using the same logic, angle at C is α, angle at B is β. So, Law of Sines in triangle CLB: CL / sin β = LB / sin α = CB / sin (α + β)Vector CL is CB rotated by -α (clockwise) around C, scaled by sin β / sin (α + β)Similarly, vector LB is LC rotated by -β around B.But again, constructing L:L = C + (B - C) * e^{-iα} * (sin β / sin (α + β))Similarly, point K in triangle BKA:K = B + (A - B) * e^{-iα} * (sin β / sin (α + β))Wait, but all three triangles have angles α and β, so the scaling factors would be similar.But note that the sides being rotated are CB, BA, and AC for triangles CLB, BKA, and ANC respectively.This seems quite involved, but perhaps if we proceed with these expressions, we can find coordinates for N, L, K, then find midpoints E, H, and then compute vectors for DEGH to check if it's a parallelogram.Alternatively, maybe there's a symmetry or transformation that can relate these points.Wait, the problem states that all three external triangles have the same angles α and β, so they are similar. Therefore, the points N, L, K are constructed in a similar fashion relative to sides AC, CB, BA. Therefore, there might be rotational symmetry or a spiral similarity connecting them.Alternatively, since D, E, G, H are midpoints, perhaps using vector midpoints and showing that the vectors between them are equal.Let’s denote vectors for all points.Let me assign vectors to points:Let’s set point A as the origin, so a = 0. Let b and c be the vectors for points B and C.Then, midpoints:D is midpoint of AB: d = (a + b)/2 = b/2G is midpoint of CA: g = (c + a)/2 = c/2H is midpoint of NA: h = (n + a)/2 = n/2E is midpoint of LK: e = (l + k)/2So, to find e, we need l and k.Now, need expressions for n, l, k.From earlier, using the Law of Sines and rotations:n = a + ( c - a ) * e^{-iα} * ( sin β / sin (α + β ) )But since a = 0, simplifies to n = c * e^{-iα} * ( sin β / sin (α + β ) )Similarly, l = c + ( b - c ) * e^{-iα} * ( sin β / sin (α + β ) )k = b + ( a - b ) * e^{-iα} * ( sin β / sin (α + β ) )Simplify k:k = b + ( -b ) * e^{-iα} * ( sin β / sin (α + β ) )= b [ 1 - e^{-iα} * ( sin β / sin (α + β ) ) ]Similarly, l:l = c + ( b - c ) * e^{-iα} * ( sin β / sin (α + β ) )= c [ 1 - e^{-iα} * ( sin β / sin (α + β ) ) ] + b * e^{-iα} * ( sin β / sin (α + β ) )And n:n = c * e^{-iα} * ( sin β / sin (α + β ) )Therefore, h = n / 2 = ( c / 2 ) * e^{-iα} * ( sin β / sin (α + β ) )Now, e = ( l + k ) / 2Let’s compute l + k:From above,l + k = [ c (1 - e^{-iα} * s ) + b e^{-iα} * s ] + [ b (1 - e^{-iα} * s ) ]where s = sin β / sin (α + β )Expanding,= c (1 - e^{-iα} s ) + b e^{-iα} s + b (1 - e^{-iα} s )= c (1 - e^{-iα} s ) + b [ e^{-iα} s + 1 - e^{-iα} s ]= c (1 - e^{-iα} s ) + b (1 )Therefore,e = ( l + k ) / 2 = [ b + c (1 - e^{-iα} s ) ] / 2Now, we have d = b / 2, g = c / 2, h = ( c e^{-iα} s ) / 2, and e as above.We need to compute vectors DE, EG, GH, HD and check if opposite sides are equal and parallel.First, compute DE: e - d = [ ( b + c (1 - e^{-iα} s ) ) / 2 ] - b / 2 = [ c (1 - e^{-iα} s ) ] / 2Next, compute GH: h - g = ( c e^{-iα} s / 2 ) - c / 2 = c / 2 ( e^{-iα} s - 1 )We need to see if DE = GH. That is, whether [ c (1 - e^{-iα} s ) ] / 2 = c / 2 ( e^{-iα} s - 1 )But (1 - e^{-iα} s ) = - ( e^{-iα} s - 1 ), so DE = - GH. Therefore, DE and GH are negatives of each other, meaning they are equal in magnitude and opposite in direction, so DE is equal and parallel to HG (since GH is from G to H, HG would be -GH, so DE = HG). Therefore, DE is equal and parallel to HG, which is one pair of sides.Now, check the other pair: EG and DH.Compute EG: g - e = c / 2 - [ ( b + c (1 - e^{-iα} s ) ) / 2 ] = [ c / 2 - b / 2 - c (1 - e^{-iα} s ) / 2 ] = [ - b / 2 + c ( 1 - (1 - e^{-iα} s ) ) / 2 ] = [ - b / 2 + c e^{-iα} s / 2 ]Compute DH: h - d = h - b / 2 = ( c e^{-iα} s / 2 ) - b / 2 = [ - b + c e^{-iα} s ] / 2So EG = ( - b + c e^{-iα} s ) / 2, and DH = ( - b + c e^{-iα} s ) / 2. Therefore, EG = DH. Therefore, EG is equal and parallel to DH.Therefore, since both pairs of opposite sides are equal and parallel, DEGH is a parallelogram.Wait, but in the computation above, DE = - GH implies DE is equal in length and opposite in direction to GH, which would mean DE is equal and parallel to HG. Similarly, EG = DH. So yes, both pairs of opposite sides are equal and parallel, hence DEGH is a parallelogram.Therefore, the proof is complete.Final Answerboxed{DEGH text{ is a parallelogram}}

✅:To prove that (DEGH) is a parallelogram, we start by noting that (D), (E), (G), and (H) are midpoints of (AB), (LK), (CA), and (NA) respectively. We use vector analysis and the properties of midpoints and similar triangles.1. Assigning Coordinates and Midpoints: - Let (A), (B), and (C) be points in the complex plane with vectors (mathbf{a}), (mathbf{b}), and (mathbf{c}). - Midpoints: (D = frac{mathbf{a} + mathbf{b}}{2}), (G = frac{mathbf{c} + mathbf{a}}{2}), (H = frac{mathbf{n} + mathbf{a}}{2}), and (E = frac{mathbf{l} + mathbf{k}}{2}).2. Constructing Points (N), (L), and (K): - Using rotations and the Law of Sines, we express (mathbf{n}), (mathbf{l}), and (mathbf{k}) in terms of (mathbf{a}), (mathbf{b}), (mathbf{c}), and angles (alpha) and (beta). - For example, (mathbf{n} = mathbf{c} cdot e^{-ialpha} cdot frac{sin beta}{sin(alpha + beta)}).3. Calculating Vectors: - Compute vectors for (D), (E), (G), and (H): - (mathbf{d} = frac{mathbf{b}}{2}) - (mathbf{g} = frac{mathbf{c}}{2}) - (mathbf{h} = frac{mathbf{n}}{2}) - (mathbf{e} = frac{mathbf{l} + mathbf{k}}{2})4. Checking Parallelogram Conditions: - Compute vectors (DE), (GH), (EG), and (DH): - (DE = frac{mathbf{c}(1 - e^{-ialpha} cdot frac{sin beta}{sin(alpha + beta)})}{2}) - (GH = frac{mathbf{c}(e^{-ialpha} cdot frac{sin beta}{sin(alpha + beta)} - 1)}{2}) - (EG = frac{-mathbf{b} + mathbf{c} cdot e^{-ialpha} cdot frac{sin beta}{sin(alpha + beta)}}{2}) - (DH = frac{-mathbf{b} + mathbf{c} cdot e^{-ialpha} cdot frac{sin beta}{sin(alpha + beta)}}{2})5. Verifying Equality and Parallelism: - (DE = -GH), showing (DE parallel GH) and (DE = GH). - (EG = DH), showing (EG parallel DH) and (EG = DH).Since both pairs of opposite sides (DE parallel GH) and (EG parallel DH) are equal and parallel, (DEGH) is a parallelogram.[boxed{DEGH text{ is a parallelogram}}]

🔑:We are given that triangles triangle ANC , triangle CLB , and triangle BKA are erected outside of triangle triangle ABC , with the specific angle conditions:1. ( angle NAC = angle KBA = angle LCB )2. ( angle NCA = angle KAB = angle LBC )This implies that the triangles are similar: ( triangle ANC sim triangle CLB sim triangle BKA ).We aim to show that quadrilateral DEGH is a parallelogram, where D , E , G , and H are midpoints of sides AB , LK , CA , and NA respectively. Step-by-Step1. Midpoints and Parallel lines: - Let F be the midpoint of BK . By properties of midpoints: [ D = text{midpoint of } AB, E = text{midpoint of } LK, G = text{midpoint of } CA, H = text{midpoint of } NA ]2. Parallel lines and triangle similarity: - Since F is midpoint of BK : [ DF parallel AK quad text{and} quad EF parallel LB ]3. Angle considerations: - Considering angles in the constructions: [ angle DFE = 180^circ - left( angle DFK + angle EFB right) = 180^circ - left( angle KAB + angle ABK right) - angle LBP ] - Since ( angle KAB ) and ( angle LBP ) are given equal in construction: [ angle DFE = angle ABC ]4. Triangle similarity: - Given ratio of triangle pairs: [ frac{DF}{EF} = frac{AK}{LB} = frac{AB}{CB} ] - Thus, triangles triangle DFE and triangle ABC are similar by Angle-Angle-Angle similarity criteria. [ triangle DFE sim triangle ABC ]5. Parallelism: - With triangle similarity: [ DE parallel NC quad text{and} quad GH parallel DE ]6. Congruence and geometric properties: - We also need to show equivalence considering the specific points: [ angle EDR = angle FDR - angle FDE = angle KAB + angle BAC - angle FDE = angle KAB ] - Midpoint properties and ratios: [ DE = frac{NC}{2} = GH ] Conclusion:By showing that ( DE parallel GH ) and ( DE = GH ), we conclude that ( DEGH ) is a parallelogram.Thus, ( boxed{DEGH text{ is a parallelogram}} ).