🧠:Okay, so I need to prove that for any number a, raising it to the power of n can be done with no more than 2 log₂(n) multiplications. Hmm, let's start by thinking about how exponentiation works. The straightforward way is multiplying a by itself n-1 times, but that's obviously inefficient, especially for large n. So there must be a smarter method.Wait, I remember something about exponentiation by squaring. That's where you break down the exponent into powers of two, right? Let me recall. For example, if you want to compute a^8, instead of multiplying a seven times, you square a to get a², then square that to get a⁴, then square again to get a⁸. That's three multiplications instead of seven. Since 8 is 2³, log₂(8) is 3, and indeed it took 3 multiplications. So maybe the number of multiplications is related to log₂(n).But the problem states no more than 2 log₂(n) multiplications. Why the factor of 2? Let me think. Maybe when the exponent isn't a power of two, you need some extra steps. Let's take an example where n is not a power of two. Let's say n = 5. How would you compute a⁵ using exponentiation by squaring?Starting with a, square it to get a² (1 multiplication), square that to get a⁴ (2 multiplications), then multiply by a to get a⁵ (3 multiplications). So here, log₂(5) is approximately 2.32, so 2 log₂(5) is about 4.64, and we used 3 multiplications, which is less. Hmm, so maybe the factor of 2 accounts for cases where you have to do both squaring and multiplying by the base when the exponent is odd?Let's take another example, say n = 6. How would that work? a^6 can be done as a² (1), a³ by multiplying a² * a (2), then square a³ to get a^6 (3). So three multiplications. log₂(6) ≈ 2.58, 2 log₂(6) ≈ 5.16, so 3 is less than that. Wait, but maybe another approach: a^2 (1), a^3 (2), a^6 (3). Alternatively, a^2 (1), a^4 (2), a^6 by multiplying a^4 * a^2 (3). Either way, three multiplications. Still under 2 log₂(n).Wait, let me check n=7. To compute a^7. Let's see: a² (1), a³ (2), a^6 (3), a^7 (4). So four multiplications. log₂(7) ≈ 2.8, 2 log₂(7) ≈ 5.6. So four is still less than 5.6. So maybe even with worst case, the number is bounded by 2 log₂(n). Hmm.So the idea is to express the exponent n in binary, perhaps? Let me recall that exponentiation by squaring uses the binary representation of n to determine when to multiply and when to square. For each bit in the binary representation, except the leading 1, you square the current result, and if the bit is 1, you also multiply by a. So the number of multiplications would be the number of bits minus 1 (for the initial squaring steps) plus the number of 1s in the binary representation minus 1 (since the leading 1 doesn't require a multiply). Wait, maybe that's a bit fuzzy. Let me clarify.Suppose n is written in binary as b_k b_{k-1} ... b_1 b_0, where b_k is 1 (the leading bit). Then starting from the leading bit, you initialize the result as a. Then for each subsequent bit, you square the result, and if the bit is 1, you multiply by a. So for each bit after the first, you do a squaring (which is one multiplication), and possibly another multiplication if the bit is 1. Therefore, the total number of multiplications is (number of bits - 1) + (number of 1s - 1). Because the first bit doesn't require a squaring or multiplication. Wait, let me check with n=5, which is 101 in binary. Number of bits is 3, number of 1s is 2. So according to that formula: (3 -1) + (2 -1) = 2 +1=3. Which matches the earlier example where a^5 took 3 multiplications. Similarly, n=7 is 111, so (3-1)+(3-1)=2+2=4, which matches.So the total number of multiplications is (k) + (t -1), where k is the number of bits minus 1 (since we start from the leading bit which is position k, and go down to 0), and t is the number of 1s. Wait, maybe not exactly. Let me re-examine.Alternatively, each bit after the first contributes at least one multiplication (the squaring), and possibly an extra one if the bit is 1. So for each bit after the first, that's one squaring (so one multiplication) and if the bit is 1, an additional multiplication. Therefore, total multiplications would be (number of bits -1) + (number of 1s -1). Because the leading bit is 1, so the number of 1s minus 1 gives the extra multiplications. So total is (k) + (t -1), where k = number of bits -1. Since the number of bits is floor(log₂n) +1. Therefore, k = floor(log₂n). So the total multiplications would be floor(log₂n) + (t -1). Since t is the number of 1s in the binary representation, which is at most the number of bits. Therefore, the worst case is when all bits are 1, which would be k + (k +1 -1) = 2k. But wait, floor(log₂n) is k. For example, n=7 is 111, floor(log₂7)=2 (since log₂7≈2.8). Then t=3. So total multiplications would be 2 + (3 -1)=4, which is correct. But 2k would be 4, which is the same. Wait, so in the worst case, the number of multiplications is 2 floor(log₂n). But since floor(log₂n) ≤ log₂n, then 2 floor(log₂n) ≤ 2 log₂n. Therefore, the number of multiplications is bounded by 2 log₂n.Ah, so that's where the factor of 2 comes from. Because in the worst case, each bit after the first contributes two multiplications: one for squaring and one for multiplying by a. However, if the number is a power of two, like n=8 (1000 in binary), then t=1, so the total multiplications would be floor(log₂8)=3, plus (1-1)=0, so 3 multiplications, which is exactly log₂8=3. So in that case, it's better, but the upper bound is 2 log₂n.Therefore, regardless of the binary representation of n, the number of multiplications required using exponentiation by squaring is at most 2 log₂n. That makes sense. So the key is that the number of multiplications is related to the number of bits in the binary expansion of n, and each bit (after the first) can contribute at most two multiplications.Alternatively, another way to look at it is through the recursive formula: a^n = (a^(n/2))^2 if n is even, and a^n = a*(a^((n-1)/2))^2 if n is odd. Each time you're reducing the exponent by half, and each step requires at most two multiplications: one squaring and possibly one multiplication by a. Therefore, each recursive step adds at most two multiplications, and since you divide the exponent by 2 each time, the number of steps is log₂n, leading to 2 log₂n multiplications.Wait, but actually in each step, whether n is even or odd, you do one squaring and potentially one multiplication. So for each bit in the binary representation, you might have two operations. But let me check with n=5.Starting with n=5 (odd):1. a^5 = a * (a^2)^2. To compute a^2 (1 multiplication), square it to get a^4 (2), multiply by a to get a^5 (3). So three multiplications. log₂5 ≈2.32, so 2*2.32≈4.64, so three is less.But maybe the recursive approach counts each squaring and multiplication as separate steps. Let's model this as a recurrence relation. Let T(n) be the number of multiplications needed to compute a^n.If n=1, T(n)=0 (since it's just a).If n is even, T(n) = T(n/2) +1 (since we square a^(n/2)).If n is odd, T(n) = T((n-1)/2) +2 (since we square a^((n-1)/2) and multiply by a).So the recurrence is T(n) ≤ T(n/2) +2, since for odd n, it's T((n-1)/2) +2 ≤ T(n/2) +2 because (n-1)/2 < n/2. Therefore, solving this recurrence, we can use the master theorem or iterate it. Let's see:T(n) ≤ T(n/2) +2T(n) ≤ T(n/2) +2 ≤ T(n/4) +2 +2 = T(n/4) +4Continuing this, after k steps, T(n) ≤ T(n/2^k) +2kWhen n/2^k =1, then k=log₂n, so T(n) ≤ T(1) +2 log₂n =0 +2 log₂n.Therefore, T(n) ≤2 log₂n. Hence, the number of multiplications is at most 2 log₂n.Therefore, this recurrence approach gives the upper bound of 2 log₂n.But wait, in reality, for n=1, T(1)=0, n=2: T(2)=T(1)+1=1, which is log₂2=1. n=3: T(3)=T(1)+2=2. Which is 2 log₂3≈3.16, so 2 is less. Similarly, n=4: T(4)=T(2)+1=2. Which is log₂4=2. So the bound of 2 log₂n is an upper bound that's not tight for powers of two, but it's valid for all n.Therefore, this recurrence shows that the number of multiplications required is at most 2 log₂n.Hence, the proof would involve showing that using the exponentiation by squaring method, whether through binary expansion or recursive halving, the number of multiplications is bounded by 2 log₂n.Another way to think about it is through the binary representation of n. Suppose n has k+1 bits in its binary form. Then, each bit after the leading one requires a squaring (1 multiplication), and if the bit is a 1, an additional multiplication. Therefore, for each of the k bits after the first, we have at most 2 multiplications, leading to a total of 2k. Since k is floor(log₂n), the total number of multiplications is at most 2 floor(log₂n), which is less than 2 log₂n.Therefore, combining these perspectives, we can formalize the proof by induction or using the recurrence relation, demonstrating that the exponentiation by squaring method ensures that the number of multiplications does not exceed 2 log₂n.In summary, the key steps are:1. Decompose n into its binary representation.2. For each bit in the binary representation (after the leading 1), perform a squaring operation.3. If the bit is 1, perform an additional multiplication by the base a.4. Since each bit contributes at most two multiplications and the number of bits is log₂n +1, the total number of multiplications is bounded by 2 log₂n.Therefore, this method ensures that computing aⁿ requires no more than 2 log₂n multiplications.

✅:To prove that (a^n) can be computed with no more than (2 log_{2} n) multiplications, we use exponentiation by squaring, which leverages the binary representation of (n) and recursively halves the exponent. Here's the step-by-step justification:--- Key Steps and Proof1. Binary Representation of (n): - Express (n) in binary: (n = b_k b_{k-1} dots b_1 b_0), where (b_k = 1) (leading bit). - The number of bits is (k + 1 = lfloor log_2 n rfloor + 1).2. Exponentiation by Squaring: - Initialize result as (a) (corresponding to the leading bit (b_k = 1)). - For each subsequent bit (b_i) ((i < k)): - Square the current result (1 multiplication). - If (b_i = 1), multiply the result by (a) (1 additional multiplication). - Total multiplications: At most (2k) (since each of the (k) bits after the first contributes at most 2 multiplications).3. Upper Bound on Multiplications: - Since (k = lfloor log_2 n rfloor), the total multiplications are bounded by: [ 2k leq 2 log_2 n. ]4. Recurrence Relation: - Let (T(n)) denote the number of multiplications to compute (a^n). - Recurrence: [ T(n) leq begin{cases} 0 & text{if } n = 1, T(n/2) + 1 & text{if } n text{ is even}, T((n-1)/2) + 2 & text{if } n text{ is odd}. end{cases} ] - Solving this gives (T(n) leq 2 log_2 n) (see *detailed derivation below*).--- Example Validation- For (n = 7) (binary (111)): - Compute (a^2) (1), (a^3) (2), (a^6) (3), (a^7) (4). - Total: (4) multiplications. Since (2 log_2 7 approx 5.6), (4 leq 5.6).--- ConclusionBy decomposing (n) into its binary form and applying exponentiation by squaring, the number of required multiplications is at most (2 log_2 n). This method ensures efficiency by halving the exponent at each step and using at most two multiplications per bit in the binary representation.Final Answer: (a^n) can be computed with (boxed{2 log_{2} n}) or fewer multiplications using exponentiation by squaring.

🔑:1. Binary Representation of ( n ): Consider the binary representation of the integer ( n ). It can be expressed as: [ n = a_0 + a_1 cdot 2 + a_2 cdot 2^2 + ldots + a_m cdot 2^m ] where ( a_i in {0, 1} ) for ( i = 0, 1, 2, ldots, m ) and ( a_m neq 0 ).2. Bounds on ( m ): The integer ( m ) represents the position of the highest bit (from the right) in the binary representation of ( n ). By definition of the binary system, we have: [ 2^m leq n < 2^{m+1} ] Taking the base-2 logarithm on both sides, we get: [ m leq log_2 n < m + 1 ] Hence: [ m = lfloor log_2 n rfloor ]3. Computing Powers of ( a ): To compute ( a^n ), first compute the powers of ( a ) that are necessary using repeated squaring: [ a, a^2, a^4, a^8, ldots, a^{2^m} ] Each subsequent power can be computed by squaring the previous power, which requires ( m ) multiplications.4. Forming ( a^n ): The next step is to multiply the appropriate subset of the previously computed powers to obtain ( a^n ). Namely, we multiply the terms corresponding to the binary representation of ( n ). For example, if the binary representation of ( n ) includes terms ( 2^i ) where ( a_i = 1 ): [ a^n = a^{a_0 cdot 2^0 + a_1 cdot 2 + a_2 cdot 2^2 + ldots + a_m cdot 2^m} = prod_{i=0}^{m} (a^{2^i})^{a_i} ] In this product, there are at most ( m+1 ) terms being multiplied (since ( a_m neq 0 )).5. Total Multiplications: - Step (a): Calculating ( a, a^2, a^4, ldots, a^{2^m} ) requires exactly ( m ) multiplications. - Step (b): Multiplying at most ( m+1 ) terms to form ( a^n ) requires at most ( m ) multiplications. Therefore, the total number of multiplications required is: [ m + m = 2m ] Given that ( m leq log_2 n < m + 1 ), we have: [ 2m leq 2 log_2 n ]6. Conclusion: We can compute ( a^n ) using no more than ( 2 log_2 n ) multiplications. [ boxed{2 log_2 n text{ multiplications}} ]

🧠:Okay, let's try to tackle this problem. So, we have a set A of 1000 positive integers, ordered as a₁ < a₂ < ... < a₁₀₀₀, and each element is at most 2017. The condition is that for any indices i and j (both between 1 and 1000), if the sum i + j is in A (meaning there's some k where a_{i+j} = a_i + a_j), then actually a_i + a_j must be an element of A. So, if i + j is an index in A (i.e., i + j ≤ 1000), then the sum of the corresponding elements a_i + a_j must also be in A. But the way it's phrased is a bit tricky. Let me parse it again.The problem states: "for any 1 ≤ i, j ≤ 1000, i + j ∈ A implies a_i + a_j ∈ A." Wait, hold on. Wait, the wording here might be tripping me up. It says if i + j is in A, then a_i + a_j is in A. But i + j is an integer, not an element of A. Wait, that can't be right. Wait, the set A contains elements a₁, a₂, ..., a₁₀₀₀. So, when they say "i + j ∈ A", does that mean that the integer i + j is an element of the set A? But A is a set of numbers from 1 up to 2017, with 1000 elements. So, if i + j is one of those elements, then a_i + a_j must also be in A.Wait, that's the interpretation? So, if the sum of the indices i and j (which are integers from 1 to 1000) is an element of the set A, then the sum of the corresponding elements a_i + a_j must also be in A. But that seems a bit odd because i + j can range from 2 to 2000 (since i and j are up to 1000 each), and A is a subset of numbers up to 2017. Hmm. So, for example, if i + j is equal to some a_k in A, then a_i + a_j must also be in A. Wait, but i and j are indices, not elements. So, for example, if we take two indices i and j, compute their sum (i + j), and if that sum is actually an element of the set A (i.e., there exists some k where a_k = i + j), then the sum of the elements a_i + a_j must also be in A.Wait, that makes more sense. So the condition is: whenever the sum of two indices i and j is an element of A (i.e., i + j ∈ A), then the sum of the corresponding elements a_i + a_j must be in A. So, for example, if 5 is in A (so 5 is some a_k), then for any i, j where i + j = 5, we must have that a_i + a_j is in A. But since A has 1000 elements, and all elements are ≤ 2017, but indices go up to 1000. Hmm. This is a bit confusing.Alternatively, maybe the problem is misphrased, and it's supposed to say "if i + j is an index in A", i.e., if i + j ≤ 1000, then a_{i+j} = a_i + a_j. But that's not what it says. Wait, let me check the original problem again.The problem says: "for any 1 ≤ i, j ≤ 1000, i + j ∈ A implies a_i + a_j ∈ A". So, if i + j is an element of A, then a_i + a_j must also be an element of A. So, the indices i and j are just used to pick elements from the set A, but the sum of the indices (i + j) is checked if it's in A, and if so, the sum of the elements must be in A.This is a bit of a strange condition. Let's think of an example. Suppose that in set A, the number 3 is present. Then, for any i and j such that i + j = 3 (i.e., (i,j) = (1,2) or (2,1)), then a₁ + a₂ must be in A. Similarly, if 4 is in A, then all pairs (1,3), (2,2), (3,1) would require that a₁ + a₃, a₂ + a₂, etc., are in A.But since A is a set of 1000 elements up to 2017, and the indices go up to 1000, this seems like a recursive condition. The problem is asking how many such sets A exist.First, maybe we can rephrase the condition. Let me try to understand it again.Given the set A = {a₁, a₂, ..., a₁₀₀₀}, with a₁ < a₂ < ... < a₁₀₀₀ ≤ 2017. The condition is: for any i, j between 1 and 1000, if i + j is an element of A, then a_i + a_j is an element of A.Wait, but i and j are indices, which go from 1 to 1000. So, i + j can be from 2 to 2000. But the elements of A are numbers from a₁ up to at most 2017, and A has 1000 elements. So, for some i, j, i + j might be in A (as an element), and in that case, a_i + a_j must be in A.But since A is a set of 1000 elements, with a₁ < a₂ < ... < a₁₀₀₀ ≤ 2017, this condition is linking the structure of A (its elements) with the sums of its elements. It's a kind of closure property: if the sum of two indices is in A, then the sum of the corresponding elements is in A.Alternatively, maybe A is required to be closed under some operation. But not exactly closed under addition, unless the sum of indices is in A. It's a bit tangled.Alternatively, perhaps the key is that A has to be an additive basis or something, but with specific constraints.Wait, maybe we can think of A as a set where if a certain number (i + j) is present in A, then the sum a_i + a_j must also be present. So, for each element x in A, if x can be written as i + j (for some indices i, j), then the sum of the elements a_i + a_j must be in A.But since x is an element of A, and the indices i and j are such that i + j = x. Wait, but i and j are indices, not the elements. So, for example, if x is in A, and x can be written as i + j where i and j are indices (from 1 to 1000), then a_i + a_j must be in A. But x is an element of A, so x = a_k for some k. Therefore, for each a_k in A, if there exist indices i and j such that i + j = a_k, then a_i + a_j must be in A.Hmm, this is quite a complex condition. Let's try to unpack it step by step.First, note that A has 1000 elements, all ≤ 2017. So, the elements are 1000 distinct integers starting from some a₁ ≥ 1, up to a₁₀₀₀ ≤ 2017. The ordering is a₁ < a₂ < ... < a₁₀₀₀.The condition is: For any i, j (indices from 1 to 1000), if i + j is in A (i.e., i + j is one of the elements a₁, a₂, ..., a₁₀₀₀), then the sum a_i + a_j must be in A.Wait, but i + j is an integer between 2 and 2000. So, for each pair of indices i, j, check if their sum (as integers) is an element of A. If yes, then the sum of the corresponding elements a_i + a_j must also be in A.So, for example, if 2 is in A, then for i = 1, j = 1, since 1 + 1 = 2, then a₁ + a₁ = 2a₁ must be in A. Similarly, if 3 is in A, then for i = 1, j = 2 (since 1 + 2 = 3), then a₁ + a₂ must be in A. Also, for i = 2, j = 1, same thing. So, for every element x in A, if x can be expressed as the sum of two indices i and j, then the sum of the elements a_i and a_j must be in A.But since the elements of A are numbers, and the indices are from 1 to 1000, the key is that when an element x is in A, any decomposition of x into a sum of two indices (i + j) forces the sum of a_i and a_j to be in A.This seems like a recursive condition. For example, if 2 is in A, then 2a₁ must be in A. Then, if 2a₁ is in A, and 2a₁ can be written as i + j for some indices i, j, then a_i + a_j must be in A. But 2a₁ is an element of A, so 2a₁ = a_k for some k. So, if 2a₁ can be written as i + j (as an integer), then a_i + a_j must be in A. Wait, but 2a₁ is an element of A, but writing it as i + j would require that i + j = a_k, which is another element of A. So, if a_k can be written as i + j, then a_i + a_j must be in A. This is getting convoluted.Alternatively, maybe the set A has to be such that for any x in A, if x can be written as the sum of two indices i and j (where i and j are between 1 and 1000), then the sum a_i + a_j is also in A. But x itself is an element of A, so x = a_m for some m. So, if a_m = i + j, then a_i + a_j must be in A. Wait, but i and j are indices, so they range from 1 to 1000. Thus, x = a_m = i + j. But since a_m is an element of A, and A's elements are at least a₁ ≥ 1, but x = i + j, which is at least 2. So, for example, if a_m is 2, then i + j = 2 implies i = 1, j = 1, so a₁ + a₁ = 2a₁ must be in A.But then 2a₁ must be in A. Let's denote that as a_p = 2a₁. Then, if a_p can be written as i + j (for some indices i, j), then a_i + a_j must be in A. For example, if p is even, say p = 2q, then i = q, j = q would give a_q + a_q = 2a_q, which would have to be in A. But this seems like a chain reaction, requiring that various multiples of a₁ are in A.Alternatively, maybe A has to contain all numbers of the form a_i + a_j whenever i + j is in A. But how does this constrain the structure of A?Alternatively, maybe the set A has to be an additive semigroup, but restricted by the indices. Wait, but additive semigroups are closed under addition, but here the condition is more specific.Wait, let's think of a simple case. Suppose that A is an arithmetic progression. For example, A = {k, k + d, k + 2d, ..., k + 999d} where k + 999d ≤ 2017. Then, if we have such a progression, maybe the condition is satisfied?But let's check. Suppose A is an arithmetic progression starting at a₁ with difference d. Then, for any i, j, a_i + a_j = 2a₁ + (i + j - 2)d. If i + j is in A, which would mean that i + j = a_m = a₁ + (m - 1)d. Then, a_i + a_j = 2a₁ + (i + j - 2)d = 2a₁ + (a₁ + (m - 1)d - 2)d. Wait, this seems messy. Maybe not.Alternatively, if A is the set {1, 2, 3, ..., 1000}, but here A is supposed to have 1000 elements up to 2017, so the minimal case would be {1, 2, ..., 1000}, and the maximal case would be {1018, 1019, ..., 2017}. But the problem states that a₁ < a₂ < ... < a₁₀₀₀ ≤ 2017.Wait, but the problem is asking for the number of sets A satisfying the condition. So, perhaps only certain sets satisfy this condition. Maybe all the sets A must be arithmetic progressions? Or maybe they have to be additive subsemigroups of the natural numbers?Alternatively, perhaps the only possible sets are intervals, i.e., sets of consecutive integers. For example, if A is {k, k+1, ..., k + 999}, then maybe the condition is satisfied. Let's test this.Suppose A is {1000, 1001, ..., 1999} (but wait, 1999 - 1000 + 1 = 1000, so that's 1000 elements). If we take any i and j, then i + j would range from 2 to 2000. But A is from 1000 to 1999. So, if i + j is in A (i.e., between 1000 and 1999), then a_i + a_j = (1000 + i - 1) + (1000 + j - 1) = 2000 + i + j - 2. Since i + j is between 1000 and 1999, then a_i + a_j is between 2000 + 1000 - 2 = 2998 and 2000 + 1999 - 2 = 3997. But A only goes up to 1999, so a_i + a_j is way larger than the maximum element of A, which is 1999. Hence, the condition would fail because a_i + a_j is not in A. Therefore, this set does not satisfy the condition.Alternatively, if A is an interval starting at 1: {1, 2, ..., 1000}. Then, for any i, j, i + j ranges from 2 to 2000. If i + j is in A (i.e., i + j ≤ 1000), then a_i + a_j = i + j, which is in A. So in this case, the condition holds because a_i + a_j = i + j, which is in A. So, this set A = {1, 2, ..., 1000} satisfies the condition.Wait, that's a good point. If A is the set {1, 2, ..., 1000}, then for any i, j, if i + j is in A (i.e., i + j ≤ 1000), then a_i + a_j = i + j, which is exactly the element a_{i+j} in A. Therefore, this set satisfies the condition. Similarly, if A is any interval {k, k+1, ..., k + 999} where k + 999 ≤ 2017, then let's check.Suppose A is {k, k+1, ..., k + 999}. Then, a_i = k + i - 1. For any i, j, if i + j is in A, which would mean that k ≤ i + j ≤ k + 999. But since i and j are indices from 1 to 1000, i + j ranges from 2 to 2000. So, for i + j to be in A, we must have k ≤ i + j ≤ k + 999. Then, a_i + a_j = (k + i - 1) + (k + j - 1) = 2k + i + j - 2. For this to be in A, we need 2k + i + j - 2 ≤ k + 999, so k + i + j - 2 ≤ 999, so i + j ≤ 1001 - k. But since i + j ≥ k (as i + j is in A, which starts at k), we need a complicated relation here. This seems restrictive. For example, take k = 1. Then, A = {1, 2, ..., 1000}, as before. Then, a_i + a_j = i + j. If i + j is in A (i.e., i + j ≤ 1000), then a_i + a_j = i + j is in A. So it works. For k = 2, A = {2, 3, ..., 1001}. Then, a_i + a_j = 2 + i - 1 + 2 + j - 1 = i + j + 2. If i + j is in A, i.e., 2 ≤ i + j ≤ 1001. Then, a_i + a_j = i + j + 2 must be in A, which requires that i + j + 2 ≤ 1001, so i + j ≤ 999. But i + j can be up to 1001, so if i + j is 1000 or 1001, then a_i + a_j = 1002 or 1003, which are not in A (since A only goes up to 1001). Therefore, the condition is not satisfied for k = 2. Therefore, only when k = 1 does this interval set satisfy the condition.Therefore, the interval {1, 2, ..., 1000} works. What about other sets?Suppose A is an arithmetic progression with difference 1 starting at 1, which is exactly {1, 2, ..., 1000}. So this is one possible set. But maybe there are others. Let's see.Suppose A is the set {1, 3, 5, ..., 1999} (the first 1000 odd numbers). But wait, 1999 is the 1000th odd number. Then, check the condition. For example, take i = 1, j = 1. Then, i + j = 2. Is 2 in A? No, because A contains only odd numbers. So, the implication is vacuously true here. If i + j is not in A, then there's no requirement. But for some i, j, i + j might be in A. For example, take i = 1, j = 2. Then, i + j = 3, which is in A. Then, a_i + a_j = 1 + 3 = 4, which is not in A (since A has only odd numbers). Therefore, this set does not satisfy the condition. So, the set of odd numbers doesn't work.Another idea: maybe A has to be a additive subsemigroup of the natural numbers. An additive subsemigroup is a set closed under addition. For example, the set of all numbers ≥ k for some k. But in our case, A has only 1000 elements, so it can't be all numbers ≥ k. But if we take a shifted version. However, the minimal such semigroup is the set {1, 2, 3, ...}, but again, A has finite size.Alternatively, perhaps A has to be a progression where a_i = i. Then, as in the first example, it works. But are there other progressions?Wait, if A is an arithmetic progression with a difference d, starting at 1. So, a_i = 1 + (i - 1)d. Then, a_i + a_j = 2 + (i + j - 2)d. Now, if i + j is in A, which would mean that 1 + ( (i + j) - 1 )d = 1 + (i + j - 1)d is an element of A. But a_i + a_j = 2 + (i + j - 2)d. For this to be in A, there must exist some k such that 1 + (k - 1)d = 2 + (i + j - 2)d. Solving for k:1 + (k - 1)d = 2 + (i + j - 2)d ⇒ (k - 1)d = 1 + (i + j - 2)d ⇒ k - 1 = (1/d) + (i + j - 2). If d divides 1, then d = 1. Therefore, only when d = 1 does this equation have an integer solution k. Then, k = 1 + (i + j - 2) + 1/d ⇒ since d=1, k = i + j. Therefore, a_i + a_j = 2 + (i + j - 2)(1) = i + j, which is exactly a_{i + j} when d=1. Therefore, only arithmetic progression with d=1 (the trivial case) satisfies the condition that a_i + a_j is in A when i + j is in A. Therefore, arithmetic progressions with d > 1 do not satisfy the condition.So, that suggests that the only arithmetic progression that works is the one with difference 1, starting at 1. But are there non-arithmetic progression sets that satisfy the condition?Suppose we take the set A = {1, 2, ..., 1000}. Then, adding some larger numbers but keeping the condition. Wait, but if we add a larger number, say replace 1000 with 1001. Then, the set becomes {1, 2, ..., 999, 1001}. Now, check the condition. For example, take i = 999 and j = 2. Then, i + j = 1001, which is in A. Then, a_i + a_j = 999 + 2 = 1001, which is in A. That works. But what about i = 999 and j = 1. Then, i + j = 1000, which is not in A anymore. So, the implication is vacuous here. But if we have another pair where i + j is in A. Let's say we have i = 500 and j = 501. Then, i + j = 1001, which is in A. Then, a_i + a_j = 500 + 501 = 1001, which is in A. So, that works. But what if we add a number that's not at the end? For example, replace 500 with 1001. So the set is {1, 2, ..., 499, 1001, 501, ..., 1000}. Wait, but the set must be ordered, so inserting 1001 in the middle would mess up the ordering. Therefore, we can't do that. So, the only way to have a larger number is to have it at the end. But replacing 1000 with 1001 seems possible. But then, when i + j = 1001, which is in A, then a_i + a_j must be 1001. For example, take i = 1 and j = 1000 (since 1 + 1000 = 1001). Then, a_i + a_j = 1 + 1000 = 1001, which is in A. Similarly, i = 2 and j = 999, sum is 1001, a_i + a_j = 2 + 999 = 1001, which is in A. Wait, this seems to work. But does this modified set {1, 2, ..., 999, 1001} satisfy the condition for all i, j?Wait, let's check another pair. Suppose i = 500 and j = 500. Then, i + j = 1000, which is not in A (since we removed 1000 and added 1001). So, the condition doesn't require anything here. But if i + j is in A, which is now up to 1001. For example, i = 1 and j = 1000: 1 + 1000 = 1001 ∈ A, and a_i + a_j = 1 + 1001? Wait, no: in this modified set, a_j where j = 1000 is 1001. So, a_i + a_j = 1 + 1001 = 1002. But 1002 is not in A. Therefore, this modified set does not satisfy the condition. Therefore, replacing 1000 with 1001 breaks the condition.Wait, that's a problem. So, even though i + j = 1001 is in A, the sum a_i + a_j = 1 + 1001 = 1002 needs to be in A, but it's not. Therefore, this modified set does not satisfy the condition. Therefore, we can't just replace the last element with a larger one.Alternatively, perhaps we can only add numbers in a way that the sum a_i + a_j is still in A. For example, if we have a set where a_i = i for all i except the last few, but adjusted such that any required sums are still present. But this seems difficult.Alternatively, maybe the only set that satisfies the condition is the interval {1, 2, ..., 1000}. Let's check if there are others.Suppose we take A = {2, 3, ..., 1001}. Then, check the condition. For example, take i = 1 and j = 1. Then, i + j = 2, which is in A. Then, a_i + a_j = 2 + 2 = 4. Is 4 in A? Yes, because A is {2, 3, ..., 1001}. So, 4 is in A. Similarly, take i = 1, j = 2. Then, i + j = 3, which is in A. a_i + a_j = 2 + 3 = 5 ∈ A. Similarly, take i = 2, j = 2. i + j = 4 ∈ A. a_i + a_j = 3 + 3 = 6 ∈ A. Seems okay. Wait, maybe this works?Wait, let's formalize this. If A is {k, k+1, ..., k + 999}, then a_i = k + i - 1. Then, for any i, j, if i + j ∈ A, which means k ≤ i + j ≤ k + 999. Then, a_i + a_j = (k + i - 1) + (k + j - 1) = 2k + i + j - 2. For this to be in A, we need:k ≤ 2k + i + j - 2 ≤ k + 999Simplify the left inequality:k ≤ 2k + i + j - 2 ⇒ 0 ≤ k + i + j - 2 ⇒ which is always true since k ≥ 1, i, j ≥ 1, so k + i + j - 2 ≥ 1 + 1 + 1 - 2 = 1 ≥ 0.Right inequality:2k + i + j - 2 ≤ k + 999 ⇒ k + i + j ≤ 1001 ⇒ i + j ≤ 1001 - k.But since i + j can be as large as 2000 (if i and j are both 1000), but in our case, i and j are indices from 1 to 1000, so i + j can be up to 2000. However, for the condition to apply, i + j must be in A, i.e., i + j ∈ {k, k+1, ..., k + 999}. Therefore, the maximum possible i + j that triggers the condition is k + 999. Therefore, we need that for all i, j such that k ≤ i + j ≤ k + 999, we have 2k + i + j - 2 ≤ k + 999 ⇒ k + i + j - 2 ≤ 999 ⇒ i + j ≤ 1001 - k.But since i + j can be up to k + 999 (as per the condition), combining these gives:k + 999 ≤ 1001 - k ⇒ 2k ≤ 2 ⇒ k ≤ 1.Therefore, the only possible k where the right inequality holds for all required i + j is when k ≤ 1. But k must be at least 1 (since it's a positive integer). Therefore, k = 1 is the only possibility. Therefore, the only interval set that satisfies the condition is A = {1, 2, ..., 1000}. Any other interval starting at k > 1 will not satisfy the condition because when i + j is near the upper end of A (k + 999), the sum a_i + a_j would exceed the maximum element of A.Therefore, this suggests that the only interval set satisfying the condition is {1, 2, ..., 1000}. But the problem says "determine the number of sets A", so maybe there are more sets than just this interval.Wait, let's think differently. Suppose that A is such that a_i = i for all i. Then, as we saw earlier, this set satisfies the condition. But what if we permute some elements? For example, swap a₁ and a₂. Then, A = {2, 1, 3, 4, ..., 1000}, but wait, no, the set must be ordered increasingly. So swapping is not allowed; the elements must be in increasing order. Therefore, the only set where a_i = i is {1, 2, ..., 1000}.Alternatively, suppose we take A = {1, 2, 4, 8, ..., 2^999}. But this is a set of powers of 2. Let's see if this satisfies the condition. Take i = 1, j = 1: i + j = 2 ∈ A. Then, a_i + a_j = 1 + 1 = 2 ∈ A. Good. i = 1, j = 2: i + j = 3. Is 3 in A? No, so no requirement. i = 2, j = 2: i + j = 4 ∈ A. Then, a_i + a_j = 2 + 2 = 4 ∈ A. Good. i = 1, j = 3: i + j = 4 ∈ A. a_i + a_j = 1 + 4 = 5. Is 5 in A? No. Therefore, the condition fails. So, this set doesn't work.Alternatively, maybe A is a set where each element is double the previous one: a_i = 2^{i-1}. Then, similar issues as above.Alternatively, perhaps there's a different structure. Suppose that A is the union of multiple intervals. But ensuring the condition holds would be difficult.Wait, another approach: the problem resembles the problem of additive bases. A is a basis of order 2 if every number can be expressed as a sum of two elements. But here, the condition is different: if i + j is in A, then a_i + a_j must be in A.Alternatively, think of A as a set where the operation i + j ↦ a_i + a_j is closed within A. That is, the image of the addition of indices (when it lands in A) must result in an element of A.This seems similar to a homomorphism. If we consider the mapping from indices to elements, φ: i ↦ a_i, then the condition is: if i + j ∈ A (as an element), then φ(i) + φ(j) ∈ A.But φ is strictly increasing since A is ordered. So, φ is a strictly increasing function from {1, 2, ..., 1000} to {1, 2, ..., 2017}.If we suppose that φ is linear, φ(i) = i. Then, as before, this works. Are there other linear functions? Suppose φ(i) = c * i. Then, the condition would require that if i + j ∈ A (i.e., c(i + j) ∈ A?), no, wait. Wait, φ(i) = c * i, so A = {c, 2c, ..., 1000c}. Then, the elements are c, 2c, ..., 1000c. For the condition: if i + j is in A, which would mean that i + j = k * c for some k. Then, a_i + a_j = c * i + c * j = c(i + j) = c * (k * c) = k c^2. For this to be in A, we need k c^2 = m * c for some m, i.e., k c = m. Since k and m are integers, this requires that c divides m. But since m must be ≤ 1000, and k is such that k c = i + j (which is ≤ 2000), so c divides m and m = k c. But m must be ≤ 1000. Therefore, k c ≤ 1000, but i + j = k c ≤ 2000, so k c ≤ 2000. Therefore, unless c = 1, it's impossible to satisfy this for all required k. For example, take c = 2. Then, A = {2, 4, ..., 2000}, but since 2000 ≤ 2017, okay. Then, if i + j is in A, i + j = 2k. Then, a_i + a_j = 2i + 2j = 2(i + j) = 4k. For this to be in A, 4k must be in A, i.e., 4k ≤ 2000 ⇒ k ≤ 500. But i + j can be up to 2000 (if i and j are both 1000), which would mean k = 1000. Then, 4k = 4000, which is way beyond A's maximum of 2000. Therefore, this fails. Therefore, only c = 1 works, giving A = {1, 2, ..., 1000}.Therefore, linear functions other than the identity don't work.Another idea: maybe A is an affine function, a_i = c + i. Then, similar to the interval case. Let's check with c=0: a_i = i. Works. c=1: a_i = 1 + i. Then, A = {2, 3, ..., 1001}. As checked before, this doesn't work because, for example, i + j = 2 implies a_i + a_j = 2 + 2 = 4 must be in A, which it is, but for larger i + j, the sum a_i + a_j = (1 + i) + (1 + j) = 2 + i + j. If i + j is in A (which is 2 to 1001), then 2 + i + j needs to be in A. But i + j can be up to 1001, so 2 + i + j can be up to 1003, which is beyond A's maximum of 1001. Therefore, this fails.Therefore, only the case where a_i = i works.Wait, but earlier when we considered A = {1, 2, ..., 1000}, we saw that it satisfies the condition because whenever i + j is in A (i.e., i + j ≤ 1000), then a_i + a_j = i + j, which is in A. So, that works. What if we shift all elements by a constant? For example, A = {k, k+1, ..., k + 999}. Then, a_i = k + i - 1. Then, the condition would require that if i + j is in A, then a_i + a_j = (k + i - 1) + (k + j - 1) = 2k + i + j - 2 is in A. But since A is {k, k+1, ..., k + 999}, the sum 2k + i + j - 2 must be in that interval. So, 2k + i + j - 2 ≤ k + 999 ⇒ k + i + j - 2 ≤ 999 ⇒ i + j ≤ 1001 - k. But since i + j can be as large as 2000, but in the condition, i + j is in A, which is between k and k + 999. Therefore, the maximum i + j that triggers the condition is k + 999. Therefore, substituting, k + 999 ≤ 1001 - k ⇒ 2k ≤ 2 ⇒ k ≤ 1. So, k must be 1. Hence, only A = {1, 2, ..., 1000} satisfies the condition among all interval sets.So, this suggests that the only interval set that works is {1, 2, ..., 1000}. But maybe there are non-interval sets that also satisfy the condition.Suppose we take A = {1, 2, ..., 999, 1001}. So, replacing the last element with 1001. Let's check the condition. For i + j in A. For most i, j, i + j will be in {2, ..., 1999}, but A contains {1, ..., 999, 1001}. So, for i + j = 1001, which is in A, we need a_i + a_j ∈ A. a_i + a_j is 1001 only if i + j = 1001. Wait, but a_i = i for i from 1 to 999, and a₁₀₀₀ = 1001. So, if i + j = 1001, then one of i or j must be 1000, because 1001 = 1 + 1000. So, take i = 1, j = 1000: a_i + a_j = 1 + 1001 = 1002, which needs to be in A. But 1002 is not in A (since A only goes up to 1001). Therefore, this set doesn't satisfy the condition. Therefore, changing the last element breaks the condition.Alternatively, suppose we add an extra element somewhere else. For example, A = {1, 2, ..., 500, 502, 503, ..., 1001}. So, we removed 501 and added 1001. Let's see. Take i = 500 and j = 1. Then, i + j = 501, which is not in A (since we removed 501). So, no requirement. But take i = 500 and j = 2: i + j = 502, which is in A. Then, a_i + a_j = 500 + 502 = 1002, which is not in A. Hence, the condition fails.Alternatively, if we remove a middle element and add a larger one, but then some sums would require the larger element to be included, but it might not be.Alternatively, consider adding a duplicate. But no, the set must have distinct elements.Alternatively, think of a set where a_i = i for i from 1 to 999, and a₁₀₀₀ = 1001. Then, as before, when i + j = 1001, which is in A, a_i + a_j = 1001 + 1 = 1002, not in A.It's challenging to see how to modify the set {1, ..., 1000} without breaking the condition. Every time you change an element, you might create a situation where a required sum is not present.Another approach: suppose A is an additive basis of order 2, but such that all sums are within A. But since A has only 1000 elements, and the number of possible sums is much larger, this is not feasible unless A is structured in a specific way.Alternatively, think recursively. Suppose that the first element a₁ must be 1. Otherwise, if a₁ > 1, then take i = j = 1: i + j = 2. If 2 is in A, then a₁ + a₁ must be in A. But since a₁ > 1, a₁ + a₁ ≥ 2a₁ ≥ 2*2 = 4. If 2 is in A, then 4 must be in A. Similarly, if 3 is in A, then for i = 1, j = 2, a₁ + a₂ must be in A. But if a₁ ≥ 2, then a₁ + a₂ ≥ 2 + 3 = 5. This might create a chain where higher elements are required, but since A has to stay within 2017 and have only 1000 elements, this could be impossible. Therefore, perhaps a₁ must be 1.Let’s test this hypothesis. Suppose a₁ = 1. Then, take i = j = 1: i + j = 2. If 2 is in A, then a₁ + a₁ = 2 must be in A. So, 2 must be in A. Similarly, i = 1, j = 2: i + j = 3. If 3 is in A, then a₁ + a₂ = 1 + a₂ must be in A. But a₂ is the second smallest element in A, which is at least 2. So, 1 + a₂ must be in A. Since A is ordered, a₃ ≥ a₂ + 1. If a₂ = 2, then 1 + a₂ = 3, so 3 must be in A. Then, a₃ = 3. Similarly, i = 1, j = 3: i + j = 4. If 4 is in A, then 1 + 3 = 4 must be in A. Hence, a₄ = 4. Continuing this pattern, we get a_i = i. Therefore, if a₁ = 1 and we require that all numbers are present to satisfy the conditions when i + j is in A, we end up with A = {1, 2, ..., 1000}.Alternatively, suppose that a₁ = 1, but we skip some elements. For example, A = {1, 3, 4, ..., 1000, 1001}. Then, check the condition. Take i = j = 1: i + j = 2. If 2 is in A? No. So, no requirement. Then, take i = 1, j = 2: i + j = 3, which is in A. Then, a₁ + a₂ = 1 + 3 = 4 must be in A. 4 is in A. Then, i = 1, j = 3: i + j = 4, which is in A. a₁ + a_j = 1 + 4 = 5, must be in A. 5 is not in A (since we skipped 2 and 5 is not skipped yet). Wait, no, in this example, we only skipped 2. So, A is {1, 3, 4, 5, ..., 1001}. Then, 5 is in A. So, a₁ + a₃ = 1 + 4 = 5 is in A. Okay. Next, i = 2, j = 2: i + j = 4 ∈ A. Then, a₂ + a₂ = 3 + 3 = 6 must be in A. 6 is in A. Similarly, i = 2, j = 3: i + j = 5 ∈ A. a₂ + a₃ = 3 + 4 = 7 must be in A. 7 is in A. Wait, but in this case, even though we skipped 2, the set might still satisfy the condition because the required sums are present. But wait, initially, we have to make sure that all required sums are included. For example, when we skip 2, then i + j = 2 is not in A, so we don't have to include 1 + 1 = 2. However, when i + j is in A, like 3, 4, etc., we need the sums to be in A. But if we skip 2, but include all numbers from 3 onwards, but then the sum a_i + a_j where i + j corresponds to elements in A must also be in A.Wait, in this example, A = {1, 3, 4, 5, ..., 1001}. Then, a₁ = 1, a₂ = 3, a₃ = 4, a₄ = 5, and so on. Now, take i = 1, j = 1: i + j = 2 not in A, so no problem. i = 1, j = 2: i + j = 3 ∈ A. Then, a₁ + a₂ = 1 + 3 = 4 ∈ A. Yes, 4 is in A. i = 1, j = 3: i + j = 4 ∈ A. Then, a₁ + a₃ = 1 + 4 = 5 ∈ A. Yes. i = 2, j = 2: i + j = 4 ∈ A. Then, a₂ + a₂ = 3 + 3 = 6 ∈ A. Yes. i = 2, j = 3: i + j = 5 ∈ A. Then, a₂ + a₃ = 3 + 4 = 7 ∈ A. Yes. i = 3, j = 3: i + j = 6 ∈ A. Then, a₃ + a₃ = 4 + 4 = 8 ∈ A. Yes. Continuing this, it seems like as long as we include all numbers from 3 onwards, but we skipped 2, then the condition is satisfied. But wait, but when i = 4, j = 4: i + j = 8 ∈ A. Then, a₄ + a₄ = 5 + 5 = 10 ∈ A. Yes, 10 is in A. Similarly, this seems to work. However, notice that a_i in this case is not equal to i. For example, a₂ = 3, a₃ = 4, etc. So, the indices are shifted.But wait, how is a_i defined? The set A is {1, 3, 4, 5, ..., 1001}, so a₁ = 1, a₂ = 3, a₃ = 4, a₄ = 5, ..., a_999 = 1001 - 1 = 1000, a_1000 = 1001. Wait, but in this case, a_i = i + 1 for i ≥ 2. But let's check if this works for all i, j.Take i = 1, j = 1000: i + j = 1001, which is in A. Then, a₁ + a_1000 = 1 + 1001 = 1002. Is 1002 in A? No, because A goes up to 1001. Therefore, this set does not satisfy the condition. Therefore, even though some sums work, the higher sums fail.Therefore, this approach of skipping elements may not work because when you get to the higher indices, the required sums fall outside the set.Alternatively, suppose we construct a set where a_i = i for i from 1 to 999, and a_1000 = 2017. Then, check the condition. For example, take i = 999 and j = 1: i + j = 1000. If 1000 is in A, then a_999 + a_1 = 999 + 1 = 1000 must be in A. But in this set, a_1000 = 2017, so 1000 is not in A. Therefore, the implication is vacuous. But other pairs: i = 1000, j = 1: i + j = 1001. If 1001 is in A, which it's not, so no problem. But if i + j is in A, which now is {1, 2, ..., 999, 2017}. So, the only element that's large is 2017. For example, i = 1000, j = 1000: i + j = 2000. If 2000 is in A, which it's not (since A contains up to 2017, but only 1, ..., 999, 2017), then no requirement. However, if there exists some i, j such that i + j = 2017, then a_i + a_j must be in A. For example, i = 1 and j = 2016, but j is only up to 1000. Therefore, i and j can only go up to 1000, so i + j ≤ 2000. Hence, 2017 cannot be written as i + j for i, j ≤ 1000. Therefore, the condition is vacuous for 2017. Therefore, this set might satisfy the condition. But let's check another pair. Take i = 500, j = 500: i + j = 1000. Since 1000 is not in A, no requirement. Take i = 999, j = 1: i + j = 1000 not in A. Take i = 999, j = 2: i + j = 1001 not in A. Therefore, all pairs i, j where i + j is in A are those where i + j ≤ 999. Therefore, since a_i = i for i ≤ 999, then a_i + a_j = i + j, which is in A (since i + j ≤ 999). Therefore, this modified set where a_1000 = 2017 seems to satisfy the condition.Wait, is that true? Let's check. For example, take i = 500, j = 499: i + j = 999, which is in A. Then, a_i + a_j = 500 + 499 = 999, which is in A. If i = 1, j = 998: i + j = 999, a_i + a_j = 1 + 998 = 999 ∈ A. If i = 2, j = 997: 2 + 997 = 999, a_i + a_j = 2 + 997 = 999 ∈ A. All these work. For i + j in A (which is up to 999), the sums a_i + a_j = i + j are in A. For i + j = 1000 or more, since they are not in A, there's no requirement. Therefore, this set A = {1, 2, ..., 999, 2017} satisfies the condition.Therefore, this suggests that there are more sets beyond the initial interval {1, ..., 1000} that satisfy the condition. Specifically, we can replace the last element with any number up to 2017, as long as we don't affect the closure condition for the lower elements.Wait, but wait. If we set a_1000 to 2017, does that affect any condition? For i + j being in A (which is up to 999), we need a_i + a_j in A. But since a_i = i for i ≤ 999, and a_1000 = 2017, then for i, j ≤ 999, a_i + a_j = i + j ∈ A (since i + j ≤ 999 + 999 = 1998, but A only contains up to 999 and 2017). Wait, hold on. If i and j are both up to 999, then i + j can be up to 1998. But in our modified set A = {1, 2, ..., 999, 2017}, the elements are from 1 to 999 and 2017. So, if i + j is in A, which is between 2 and 999 or 2017. But since i and j are at most 1000 (but in this case, up to 999 for i, j ≤ 999). Wait, but the indices are still up to 1000. So, if i = 1000 and j = 1, then i + j = 1001, which is not in A, so no problem. If i = 999 and j = 1000, i + j = 1999, which is not in A. But in the modified set, a_1000 = 2017. So, if there exists i, j such that i + j is in A (which is up to 999 or 2017), but i and j are indices from 1 to 1000.So, if i + j = 2017, which is in A, then a_i + a_j must be in A. However, i and j are indices from 1 to 1000, so i + j = 2017 implies that one of the indices is at least 2017 - 1000 = 1017. But since the indices only go up to 1000, it's impossible. Therefore, i + j can never be 2017. Therefore, the only elements in A that are attainable as i + j are from 2 to 2000, but A only contains up to 999 and 2017. Therefore, the condition only requires that for i + j in {2, 3, ..., 999}, which are all in A, then a_i + a_j must be in A. Since a_i + a_j = i + j for i, j ≤ 999, which are in A. For indices involving 1000, since a_1000 = 2017, and i + j would have to be in A (which would require i + j ≤ 999 or i + j = 2017). But i + j ≤ 1000 + 1000 = 2000, so i + j = 2017 is impossible. Therefore, for all i, j where i + j is in A, it's only when i + j ≤ 999. In those cases, since i and j are both ≤ 999, their sum is i + j ≤ 1998, but in A, the elements up to 999 are included. Wait, hold on. If i and j are both ≤ 999, then i + j can be up to 1998. But in A, the elements are only up to 999 and 2017. Therefore, if i + j is between 1000 and 1998, it's not in A, so no requirement. Only if i + j is between 2 and 999, which is in A, then a_i + a_j must be in A. But in this case, for i, j ≤ 999, i + j ≤ 1998, but A only contains up to 999 and 2017. So, if i + j is in A (i.e., between 2 and 999), then a_i + a_j = i + j is in A. Therefore, the condition is satisfied. The element 2017 in A doesn't affect anything because there's no i, j with i + j = 2017. Therefore, this set A = {1, 2, ..., 999, 2017} satisfies the condition.Therefore, this suggests that we can take the initial interval {1, 2, ..., n} and replace elements from n+1 to 1000 with larger numbers, as long as the replaced numbers are large enough that i + j can't reach them. Specifically, if we replace a_k with some number greater than 2000 - k + 1, then i + j can't reach it. Wait, no. Let's think carefully.Actually, the only requirement is that for any i, j, if i + j is in A, then a_i + a_j is in A. So, if we have a set where the lower part is an interval {1, 2, ..., m}, and the upper part consists of numbers larger than m, chosen such that no i + j equals those upper numbers. Then, the condition is satisfied because for the lower part, the sums are closed, and for the upper part, since they can't be expressed as i + j (indices), there's no requirement.But how can we ensure that the upper elements can't be expressed as i + j? Since i and j are indices from 1 to 1000, their sum i + j can go up to 2000. Therefore, any element in A greater than 2000 cannot be expressed as i + j, so for those elements, the condition is vacuously true. Therefore, if we take A to be {1, 2, ..., m} union S, where S is a set of numbers greater than 2000, and the total number of elements is 1000. Then, since elements in S are greater than 2000, they can't be expressed as i + j for any i, j ≤ 1000 (since i + j ≤ 2000). Therefore, for elements in S, there's no i, j such that i + j is equal to them, so the condition is automatically satisfied. For the lower part {1, 2, ..., m}, we need that for any i, j, if i + j is in {1, 2, ..., m}, then a_i + a_j is in A. But since a_i = i for i ≤ m, then a_i + a_j = i + j, which is in {1, 2, ..., 2m - 1}. Therefore, to have closure, we need 2m - 1 ≤ m, which implies m ≤ 1. But if m = 1, then A is {1} union S, which is only possible if S has 999 elements. But S must consist of numbers from 2001 to 2017, but 2017 - 2001 + 1 = 17 numbers. So, we can't have 999 elements there. Therefore, this approach doesn't work.Wait, perhaps the previous example was a special case. If we take m = 999, then A = {1, 2, ..., 999} union {2017}. Here, the lower part {1, ..., 999} is closed under addition in the sense that if i + j is in A (which is up to 999), then a_i + a_j = i + j is in A. The element 2017 is too large to be expressed as i + j, so no requirements on it. Therefore, this set works. Similarly, we could replace the last element with any number from 1000 to 2017, but wait, if we replace it with a number less than or equal to 2000, then i + j could potentially reach it.For example, if we take A = {1, 2, ..., 999, 1000}. Then, this is the original interval set which works. If we replace 1000 with 1001, as before, but earlier example showed that it doesn't work because i + j = 1001 would require a_i + a_j = 1002 to be in A, but it's not. However, in the case where we replace 1000 with a number greater than 2000, say 2001, then i + j can't reach 2001 (since i + j ≤ 2000). Therefore, A = {1, 2, ..., 999, 2001} satisfies the condition. Similarly, we can replace multiple elements.Wait, but the problem states that the set A has exactly 1000 elements, all ≤ 2017. So, if we take the first m elements as {1, 2, ..., m}, and the remaining 1000 - m elements as numbers from 2001 - (1000 - m) + 1 to 2017. Wait, but we need to choose numbers greater than 2000. However, 2017 - 2000 = 17, so there are only 17 numbers from 2001 to 2017. Therefore, the maximum number of elements we can include in S is 17. Therefore, if we take m = 1000 - 17 = 983, then A can be {1, 2, ..., 983} union {2001, 2002, ..., 2017}, totaling 983 + 17 = 1000 elements. In this case, the lower 983 elements are closed under addition (i + j in A implies a_i + a_j = i + j in A), and the upper 17 elements are too large to be expressed as i + j, so no requirements. Therefore, this set would satisfy the condition.But wait, but in this case, when i + j is in A, which could be up to 983. For example, take i = 983, j = 1: i + j = 984. If 984 is in A, then a_i + a_j = 983 + 1 = 984 must be in A. But 984 is not in A, because A only includes up to 983 and then jumps to 2001. Therefore, this set does not satisfy the condition.Ah, right. Therefore, in order for the lower part to be closed, m must be such that for all i + j ≤ m, the sum i + j is in A. But if A is {1, 2, ..., m} union S, then for i + j ≤ m, the sum is in A, but for i + j > m, if it's in A, then we need to check. However, if S consists of numbers greater than m, but potentially some numbers between m + 1 and 2000 could be in S. If any number between m + 1 and 2000 is in S, then we have to ensure that for any i, j with i + j equal to that number, a_i + a_j is in A. But a_i + a_j would be i + j if i and j are in the lower part, or something else if they're in S. But if S contains numbers between m + 1 and 2000, then it's possible that i + j equals to one of those numbers, requiring a_i + a_j to be in A. However, a_i and a_j could be in the lower part or in S. This complicates things.Therefore, to safely replace elements without violating the condition, we need to replace elements with numbers that cannot be expressed as i + j for any i, j ≤ 1000. Since i + j can be at most 2000, any number greater than 2000 cannot be expressed as such. Therefore, replacing elements with numbers greater than 2000 is safe. Numbers from 2001 to 2017 are safe. Therefore, if we take A = {1, 2, ..., n} union {2001, 2002, ..., 2017}, where n + 17 = 1000 ⇒ n = 983. But then, as before, the lower part {1, ..., 983} needs to be closed under addition whenever i + j is in A. However, in this case, A contains {1, ..., 983} union {2001, ..., 2017}. Therefore, for i + j in {1, ..., 983}, the sum a_i + a_j = i + j must be in A, which is true. For i + j in {2001, ..., 2017}, but since i + j can be at most 2000, there's no such case. Therefore, this set satisfies the condition.Wait, but in this case, if we take A = {1, 2, ..., 983} union {2001, ..., 2017}, then for any i, j where i + j is in A (which can only be from 2 to 983), the sum a_i + a_j = i + j is in A. Therefore, this set works. Additionally, there's no i, j such that i + j is in {2001, ..., 2017}, so those elements don't impose any conditions. Therefore, this set satisfies the requirements.Similarly, we can replace any number of elements from the initial interval {1, ..., 1000} with numbers from 2001 to 2017, as long as the total number of elements is 1000. However, there are only 17 numbers from 2001 to 2017. Therefore, we can replace up to 17 elements from the initial interval with these larger numbers.For example, we can replace 1 element: A = {1, 2, ..., 999, 2001}. This set has 1000 elements. It satisfies the condition because the lower 999 elements are closed under addition (when i + j is in A, which is up to 999, the sum is in A), and the last element 2001 is too large to be expressed as i + j.Similarly, we can replace 2 elements: A = {1, 2, ..., 998, 2001, 2002}, and so on, up to replacing 17 elements: A = {1, 2, ..., 983, 2001, ..., 2017}.Therefore, the number of such sets A is equal to the number of ways to choose how many elements we replace from the initial interval {1, ..., 1000} with the 17 available large numbers (2001 to 2017). Since we can replace any number of elements from 0 to 17 (but we have to maintain the set size as 1000), each time replacing k elements from the end of the interval with k elements from 2001 to 2017.However, note that the numbers from 2001 to 2017 are 17 distinct numbers, so replacing k elements requires choosing k elements from these 17. Additionally, we need to replace the last k elements of the initial interval. Since the set must be ordered increasingly, we can't replace arbitrary elements; we have to replace the largest elements. Therefore, the number of possible sets is the number of subsets of {2001, ..., 2017} of size k, for k from 0 to 17, combined with the remaining elements {1, 2, ..., 1000 - k}.However, since we need the total number of elements to be 1000, and replacing k elements means that we remove k elements from {1, ..., 1000} and add k elements from {2001, ..., 2017}. The number of ways to do this is the number of ways to choose k elements from the 17 available, and replace the last k elements of the initial interval with them. Since the order in the set must be increasing, the added elements must be larger than the remaining elements from the initial interval. Therefore, for each k from 0 to 17, the number of sets is C(17, k). Therefore, the total number of sets is the sum_{k=0}^{17} C(17, k) = 2^17. However, this is only if we can replace any number of elements. But wait, replacing k elements requires that we remove the last k elements from {1, ..., 1000} and add k elements from {2001, ..., 2017}. Since the elements must remain in order, we can only replace the last k elements with the k new elements, arranged in order. Therefore, for each k from 0 to 17, the number of sets is C(17, k), since we choose k elements from 17 and place them at the end. Therefore, the total number of sets is the sum_{k=0}^{17} C(17, k) = 2^17.But wait, 2^17 is 131072. However, the problem states that the elements must be in increasing order and all elements are ≤ 2017. Therefore, replacing k elements from the end with k elements from 2001-2017, which are all larger than 1000, maintains the increasing order. For each k, the number of possible sets is C(17, k), because we choose k elements from the 17 available and add them in increasing order. Therefore, the total number is sum_{k=0}^{17} C(17, k) = 2^17. However, when k = 0, we have the original set {1, ..., 1000}, and when k = 17, we have {1, ..., 983} union {2001, ..., 2017}.But wait, another thing to consider: when replacing k elements, the new elements must be added in increasing order, and they must be greater than the remaining elements from the original set. Since we're replacing the last k elements, the remaining elements are {1, ..., 1000 - k}, and the new elements are {2001, ..., 2017} chosen k elements. Since 1000 - k < 2001 for all k ≤ 17 (since 1000 - 17 = 983 < 2001), the new elements are indeed larger, so the entire set remains increasing. Therefore, the total number of sets is indeed 2^17, as each element from 2001 to 2017 can be either included or not included, but since we can only replace up to 17 elements, it's actually the sum_{k=0}^{17} C(17, k) = 2^17.But wait, no. If we have 17 elements available (2001-2017), and we want to choose any subset of them to add to the set A, replacing the corresponding number of elements from the end of {1, ..., 1000}, then for each of the 17 elements, we have a binary choice: include it or not. However, the number of elements we include must be such that we remove the same number from the original set. Therefore, if we include k elements from 2001-2017, we must remove k elements from {1, ..., 1000}, specifically the last k. Therefore, the number of sets is the number of subsets of {2001-2017}, which is 2^17, but each subset corresponds to a unique set A where the elements {2001-2017} that are in the subset are added, and the corresponding number of elements from the end of {1-1000} are removed. However, the size of the set must be exactly 1000. Therefore, if we choose a subset S of {2001-2017} of size k, then we must remove k elements from {1-1000}, so the total size is 1000 - k + k = 1000. Therefore, the number of such sets is indeed the number of subsets of {2001-2017}, which is 2^17. But wait, {2001-2017} has 17 elements, so the number of subsets is 2^17. However, each subset corresponds to a unique set A: for each subset S, A is ({1, 2, ..., 1000 - |S|} union S). Since S is a subset of {2001-2017}, and |S| can be from 0 to 17.Therefore, the total number of sets A satisfying the condition is 2^17 = 131072.But let's verify this with an example. For k = 1, there are C(17, 1) = 17 sets: each replaces the last element (1000) with one of the 17 numbers from 2001 to 2017. For k = 2, C(17, 2) sets: replace the last two elements (999, 1000) with two numbers from 2001 to 2017. And so on, up to k = 17, which is C(17, 17) = 1 set: replace the last 17 elements (984 to 1000) with all 17 numbers from 2001 to 2017. The total number is indeed 2^17.Therefore, the answer should be 2^17, which is 131072. However, the problem states that the elements of A are a₁ < a₂ < ... < a₁₀₀₀ ≤ 2017. Since the numbers from 2001 to 2017 are 17 numbers, and 2017 is included, we can indeed use all 17 of them. Therefore, the total number of sets is 2^17.But let's confirm once more. Each subset of the 17 large numbers corresponds to a unique set A. For each such subset S, we create A by taking the numbers from 1 up to 1000 - |S|, and then adding the numbers in S. Since the numbers in S are all larger than 1000 - |S| (because 1000 - |S| ≥ 1000 - 17 = 983, and S contains numbers ≥ 2001), the set A is increasing. Moreover, A has exactly 1000 elements. For each such set A, the condition is satisfied because:1. For any i, j where i + j is in the lower part (1 to 1000 - |S|), the sum a_i + a_j = i + j is in A.2. For any i, j where i + j is in the upper part (S), but since i + j ≤ 2000 and S contains numbers ≥ 2001, there are no such i, j. Therefore, the condition is vacuous for the upper part.Therefore, all such sets satisfy the condition, and the number of such sets is 2^17 = 131072.But wait, another thought: what if we replace elements not from the end? For example, can we replace some middle elements with larger ones? Suppose we take A = {1, 2, ..., 500, 2001, 2002, ..., 20017} but this is not possible since 2017 is the upper limit. Wait, no, the upper numbers are only up to 2017. But if we replace middle elements, we might break the increasing order. For instance, if we remove 500 and add 2001, the set would need to be {1, ..., 499, 501, ..., 1000, 2001}, but this skips 500 and adds 2001 at the end. However, this set would have 1000 elements, but the element 500 is missing, which might affect the closure condition. For example, if i + j = 500, which is not in A, so no requirement. But other sums: take i = 499 and j = 1. Then, i + j = 500, which is not in A. But if i + j is in A, say 501, which is in A. Then, a_i + a_j must be in A. However, a_i for i = 501 is now 502 (since we removed 500), so a_i + a_j might not correspond. Wait, this is getting complicated. But in our previous reasoning, we only considered replacing the last k elements with larger ones, which doesn't affect the closure of the lower part. If we replace middle elements, the indices shift, and the closure property might break. Therefore, to ensure the closure property, we can only replace the last k elements, as replacing others would break the sequence and cause the sums not to align.Therefore, the only valid sets are those where the lower part is an unbroken sequence from 1 to m, and the upper part is a subset of {2001, ..., 2017}. This requires that the lower part is exactly {1, ..., m}, and the upper part is a subset of size 1000 - m from {2001, ..., 2017}. Since {2001, ..., 2017} has 17 elements, 1000 - m ≤ 17 ⇒ m ≥ 1000 - 17 = 983. Therefore, m can be from 983 to 1000, inclusive. For each m, the number of sets is C(17, 1000 - m). Therefore, summing over m from 983 to 1000, we get sum_{k=0}^{17} C(17, k) = 2^17.Hence, the number of sets A is 2^17 = 131072.Final AnswerThe number of sets ( A ) is (boxed{131072}).

✅:Given the set of positive integers ( A = {a_1, a_2, ldots, a_{1000}} ) where ( a_1 < a_2 < cdots < a_{1000} leq 2017 ), we need to determine the number of sets ( A ) that satisfy the condition: for any ( 1 leq i, j leq 1000 ), if ( i + j in A ), then ( a_i + a_j in A ).Key steps leading to the solution:1. Understanding the Condition: The condition requires that if the sum of two indices ( i ) and ( j ) is an element of ( A ), then the sum of the corresponding elements ( a_i ) and ( a_j ) must also be an element of ( A ).2. Initial Interval Set: The set ( A = {1, 2, ldots, 1000} ) satisfies the condition because ( a_i = i ), and if ( i + j in A ), then ( a_i + a_j = i + j in A ).3. Replacing Elements with Larger Numbers: We can replace elements from the end of the interval ( {1, 2, ldots, 1000} ) with numbers from 2001 to 2017. These larger numbers cannot be expressed as the sum of two indices ( i ) and ( j ) (since ( i + j leq 2000 )), thus not affecting the closure condition.4. Counting Valid Sets: The numbers from 2001 to 2017 are 17 distinct numbers. For each subset of these 17 numbers, we can replace the corresponding number of elements from the end of the interval ( {1, 2, ldots, 1000} ). The number of such subsets is ( 2^{17} ).5. Conclusion: The total number of valid sets ( A ) is ( 2^{17} = 131072 ).Thus, the number of sets ( A ) that satisfy the given conditions is (boxed{131072}).

🔑:1. Let ( k ) be the largest integer such that ( a_k leq 2000 ). Therefore, [ a_{1000} > a_{999} > cdots > a_{k+1} geq 2001. ]2. If ( a_k > 1000 ), there exists an integer ( i ) such that [ a_k = 1000 + i in A. ]3. Hence, it follows that [ a_{1000} + a_i in A. ] But since ( a_{1000} + a_i > a_{1000} ), we have a contradiction.4. Therefore, ( a_i leq 1000 ) for all ( i = 1, 2, cdots, k ).5. Observe that ( a_i geq i ) for all ( i = 1, 2, cdots, 1000 ). Hence, [ a_k geq k. ]6. If ( a_k > k ), then there exists an integer ( i ) such that ( i = a_k - k ), and thus ( i + k = a_k in A ). Consequently, there exists an integer ( m ), such that [ a_m = a_i + a_k in A. ]7. Also, ( i + k = a_k leq 1000 ).8. Therefore, the elements of ( A ) which are not less than 2001 are at most 17. Hence, [ k geq 1000 - 17 = 983. ]9. We also have ( i leq a_k - k leq 1000 - k leq 17 < k ).10. Hence, ( a_m = a_i + a_k < a_k + a_k leq 2000 ).11. Noting that ( 2000 > a_m = a_i + a_k > a_k ), which contradicts the maximality of ( a_k ).12. Therefore, ( a_k = k ). Thus ( A ) is structured as ( {1, 2, cdots, k, a_{k+1}, a_{k+2}, cdots, a_{1000}} ), where ( a_{k+1}, a_{k+2}, cdots, a_{1000} ) are all greater than 2000.13. When ( k = 983, 984, cdots, 1000 ), the number of such ( A ) is given by: [ binom{17}{17} + binom{17}{16} + cdots + binom{17}{1} + binom{17}{0} = 2^{17}. ] Conclusion:Therefore, the number of subsets ( A ) that satisfy the given conditions is [boxed{2^{17}}.]

🧠:Okay, so I need to figure out if point D inside the acute angle BAC can be such that the angle CAD is twice the angle BAD, and also D is twice as far from AC as it is from AB. Hmm, let me try to visualize this. First, let me sketch the angle BAC. Since it's acute, both BA and CA are forming an angle less than 90 degrees. Point D is somewhere inside this angle. The problem states that angle CAD is twice angle BAD. So, if I let angle BAD be θ, then angle CAD is 2θ. Therefore, the total angle BAC is θ + 2θ = 3θ. That makes sense, so the angle BAC must be three times the angle BAD. Now, the other condition is that the distance from D to AC is twice the distance from D to AB. Let me recall that the distance from a point to a line can be calculated using the formula for the perpendicular distance. So, if I denote the distance from D to AB as h, then the distance from D to AC is 2h. Maybe I can model this with coordinate geometry. Let me place point A at the origin. Let’s set AB along the x-axis and AC in the plane such that angle BAC is 3θ. Wait, but θ is the angle BAD, which is part of the problem. Hmm, maybe I need a different approach. Alternatively, I can use coordinate axes. Let's place point A at (0,0), AB along the positive x-axis, and AC in the first quadrant making an angle of 3θ with AB. Then, point D lies inside this angle. Since angle BAD is θ, and angle CAD is 2θ, then AD is the angle bisector that splits BAC into θ and 2θ. Wait, no, it's not exactly an angle bisector because the division is in the ratio 1:2. Maybe using vectors or coordinates would help. Let me try coordinates. Let’s set AB along the x-axis, and AC in such a way that angle BAC is 3θ. Let me assign coordinates to points B and C. Suppose AB is the x-axis, so point B is at (b, 0) for some b > 0. Similarly, point C is somewhere in the first quadrant. The exact coordinates might depend on the angle. Alternatively, maybe using trigonometry. If I consider AD making an angle θ with AB, then the direction of AD is θ from AB. Similarly, since angle CAD is 2θ, then AD is 2θ from AC. Wait, but how does that relate to the position of D? Let me think. The distance from D to AB is h, and to AC is 2h. Let's denote the coordinates of D as (x, y). Then, the distance from D to AB (which is the x-axis) is just y. The distance from D to AC can be calculated using the formula for distance from a point to a line. First, I need the equation of line AC. Since angle BAC is 3θ, and AB is along the x-axis, the slope of AC is tan(3θ). Therefore, the equation of AC is y = tan(3θ) x. The distance from point D (x, y) to line AC is given by |Ax + By + C| / sqrt(A² + B²), where Ax + By + C = 0 is the equation of the line. For line AC, the equation can be rewritten as tan(3θ) x - y = 0. Therefore, the distance from D to AC is |tan(3θ) x - y| / sqrt(tan²(3θ) + 1). Simplify the denominator: sqrt(tan²(3θ) + 1) = sec(3θ). So, the distance becomes |tan(3θ) x - y| / sec(3θ) = |sin(3θ) x - cos(3θ) y|, since tan(3θ) = sin(3θ)/cos(3θ) and multiplying numerator and denominator by cos(3θ). Wait, let me check that again. If the distance is |tan(3θ) x - y| / sec(3θ), then:tan(3θ) = sin(3θ)/cos(3θ), so substitute:| (sin(3θ)/cos(3θ)) x - y | / (1/cos(3θ)) )Multiply numerator and denominator by cos(3θ):| sin(3θ) x - cos(3θ) y | / 1So yes, the distance is | sin(3θ) x - cos(3θ) y |.But since D is inside the angle BAC, the distance should be positive. So, maybe we can drop the absolute value if we consider the sign. But perhaps it's safer to keep the absolute value for now. Given that the distance from D to AC is twice the distance from D to AB. The distance to AB is y, so we have:| sin(3θ) x - cos(3θ) y | = 2y.Also, since D lies on the angle bisector? Wait, no. Wait, D is such that angle BAD = θ and angle CAD = 2θ. So, the direction of AD is such that it splits angle BAC into θ and 2θ. Therefore, AD is not the bisector, but a line that makes angle θ with AB and 2θ with AC. Therefore, the coordinates of D can be represented in terms of θ. Let me parametrize point D along the direction determined by angle θ from AB. Let’s assume that AD has length r. Then, coordinates of D would be (r cosθ, r sinθ). Wait, that might be a good approach. If we consider AD making angle θ with AB, then coordinates of D are (r cosθ, r sinθ). Now, we need to ensure that this point D also satisfies the distance condition: distance from D to AC is twice the distance from D to AB. The distance from D to AB is y-coordinate, which is r sinθ. The distance from D to AC, as we found earlier, is | sin(3θ) x - cos(3θ) y |. Substituting x = r cosθ and y = r sinθ:Distance to AC = | sin(3θ) * r cosθ - cos(3θ) * r sinθ | = r | sin(3θ) cosθ - cos(3θ) sinθ |.Simplify the expression inside the absolute value:sin(3θ) cosθ - cos(3θ) sinθ = sin(3θ - θ) = sin(2θ). Because sin(A - B) = sinA cosB - cosA sinB. Wait, actually, that's sin(A - B) = sinA cosB - cosA sinB. So, in our case, sin(3θ) cosθ - cos(3θ) sinθ = sin(3θ - θ) = sin(2θ). Therefore, the distance from D to AC is r | sin(2θ) |. Since θ is part of an acute angle, 3θ must also be acute, so θ < 30 degrees. Therefore, sin(2θ) is positive. So, distance to AC is r sin(2θ). Given the condition that distance to AC is twice the distance to AB:r sin(2θ) = 2 * r sinθ.We can divide both sides by r (assuming r ≠ 0, which it isn't since D is a point inside the angle):sin(2θ) = 2 sinθ.But wait, sin(2θ) is equal to 2 sinθ cosθ. So:2 sinθ cosθ = 2 sinθ.Divide both sides by 2 sinθ (assuming sinθ ≠ 0, which it isn't since θ is an angle between 0 and 30 degrees):cosθ = 1.Therefore, cosθ = 1 implies θ = 0 degrees. But θ is part of the angle BAD, which can't be zero if D is a point inside the angle. Therefore, this leads to a contradiction. Hmm, that suggests that our assumption that such a point D exists leads to a contradiction, meaning that the answer is no. But wait, maybe I made a mistake somewhere. Let me check the steps again. First, parametrizing D as (r cosθ, r sinθ) because angle BAD is θ. Then, calculating the distance to AC as | sin(3θ) x - cos(3θ) y |, which becomes r | sin(2θ) |. Then, equating to 2y = 2 r sinθ. So, sin(2θ) = 2 sinθ, leading to 2 sinθ cosθ = 2 sinθ, then cosθ = 1. Yes, that seems correct. Therefore, the only solution is θ = 0, which is impossible. Therefore, there is no such point D that satisfies both conditions. But wait, the problem says "Could the point D end up being twice as far from AC as from AB?" So, according to this, the answer is no. However, maybe there's another approach where this is possible. Let me think again. Alternatively, maybe I should not parametrize D along the angle θ, but instead use another method. Let's try using coordinate geometry with angle BAC = 3θ, AB along x-axis, and coordinates accordingly. Let’s set point A at (0,0), AB along x-axis, and AC at angle 3θ. Let’s assume AB has length 1 for simplicity, so point B is at (1,0). Point C can be at some coordinate (cos(3θ), sin(3θ)) to make AC length 1, but maybe that complicates things. Alternatively, let’s just set AC as the line y = tan(3θ) x. Point D is somewhere inside the angle. Let's denote the coordinates of D as (x,y). The condition angle BAD = θ and angle CAD = 2θ. The angle between AD and AB is θ, and the angle between AD and AC is 2θ. So, the direction of AD is such that it makes angle θ with AB and 2θ with AC. Alternatively, using the angle conditions, perhaps we can relate the slopes. The slope of AD is y/x. The angle between AD and AB (x-axis) is θ, so tanθ = y/x. Therefore, y = x tanθ. Additionally, the angle between AD and AC is 2θ. The slope of AC is tan(3θ). The angle between two lines with slopes m1 and m2 is given by:tanφ = |(m2 - m1)/(1 + m1 m2)|.So, the angle between AD (slope m1 = tanθ) and AC (slope m2 = tan3θ) is 2θ. Therefore:tan(2θ) = |(tan3θ - tanθ)/(1 + tan3θ tanθ)|.Let’s compute the right-hand side:(tan3θ - tanθ)/(1 + tan3θ tanθ) = [sin3θ/cos3θ - sinθ/cosθ] / [1 + (sin3θ/cos3θ)(sinθ/cosθ)]Simplify numerator:= [ (sin3θ cosθ - sinθ cos3θ ) / (cos3θ cosθ) ] Denominator:= [ (cos3θ cosθ + sin3θ sinθ ) / (cos3θ cosθ) ]So, overall:[ (sin3θ cosθ - sinθ cos3θ ) / (cos3θ cosθ) ] / [ (cos3θ cosθ + sin3θ sinθ ) / (cos3θ cosθ) ] = [sin3θ cosθ - sinθ cos3θ] / [cos3θ cosθ + sin3θ sinθ]The numerator is sin(3θ - θ) = sin2θ. The denominator is cos(3θ - θ) = cos2θ. Therefore, the whole expression is sin2θ / cos2θ = tan2θ. Therefore, tan(2θ) = |tan2θ|. Since angles are positive and less than 90 degrees, tan2θ is positive, so absolute value can be removed. Therefore, tan(2θ) = tan2θ, which is an identity. Therefore, this condition is automatically satisfied. So, this tells us that if we have a point D lying on a line making angle θ with AB, then the angle between AD and AC is indeed 2θ. Therefore, such a point D exists for any θ, but we also have the distance condition. Earlier, we found that if D is on the line making angle θ with AB, then its coordinates are (x, y) = (r cosθ, r sinθ). Then, the distance to AC is r sin2θ, and distance to AB is r sinθ. The condition is r sin2θ = 2 r sinθ, leading to sin2θ = 2 sinθ, which simplifies to 2 sinθ cosθ = 2 sinθ, hence cosθ = 1, which implies θ = 0. But θ = 0 would mean D is on AB, which contradicts D being inside the angle. Therefore, the only solution is trivial, which isn't allowed. Therefore, such a point D cannot exist. But wait, this seems too straightforward. Maybe there's an error in assuming that D must lie along that specific line. Let me think again. Alternatively, perhaps D is not on the line making angle θ with AB, but somewhere else. Wait, but angle BAD is θ, so AD must make an angle θ with AB. Therefore, D has to lie on that line. So, if angle BAD is θ, then direction of AD is fixed, hence D must lie along that direction. Therefore, parametrizing D along that direction is correct, and then the distance condition leads to a contradiction. Hence, no such point exists. Alternatively, maybe using area considerations. Suppose we construct D such that distances are in the ratio 2:1. The ratio of areas of triangles ABD and ACD might relate to these distances and angles. But perhaps another approach is using the angle bisector theorem or some trigonometric identities. Wait, angle BAD = θ, angle CAD = 2θ. Let’s denote the distances: distance from D to AB is h, distance to AC is 2h. Then, in triangle ABD, the area is (1/2)*AB*h. In triangle ACD, the area is (1/2)*AC*2h. But AB and AC are sides of the angle, but their lengths are not given. Alternatively, consider the ratio of areas. But maybe this isn't helpful. Wait, perhaps using the law of sines. In triangle ABD and triangle ACD. Wait, let's consider triangle ABD: angle at A is θ, the distance from D to AB is h, which is the height. Similarly, in triangle ACD, angle at A is 2θ, distance from D to AC is 2h. But how do these relate? Alternatively, in triangle ABD, if h is the height from D to AB, then h = AD sinθ. Similarly, in triangle ACD, the height from D to AC is 2h = AD sin2θ. Therefore, from triangle ABD: h = AD sinθ.From triangle ACD: 2h = AD sin2θ.Substitute h from the first equation into the second:2(AD sinθ) = AD sin2θ.Divide both sides by AD (assuming AD ≠ 0):2 sinθ = sin2θ.But sin2θ = 2 sinθ cosθ, so:2 sinθ = 2 sinθ cosθ.Divide both sides by 2 sinθ (again, sinθ ≠ 0):1 = cosθ.Which implies θ = 0, again leading to a contradiction. Therefore, this confirms the previous result. Hence, such a point D cannot exist because it would require θ = 0, which is not possible inside the angle. Therefore, the answer is no, it's not possible for point D to be twice as far from AC as from AB under the given angle conditions.

✅:Step-by-Step Explanation and Answer:To determine if point ( D ) inside the acute angle ( angle BAC ) can satisfy both the angular condition ( angle CAD = 2angle BAD ) and the distance condition (twice as far from ( AC ) as from ( AB )), we proceed as follows:1. Angular Relationships: - Let ( angle BAD = theta ), hence ( angle CAD = 2theta ). - The total angle ( angle BAC = theta + 2theta = 3theta ).2. Coordinate Setup: - Place ( A ) at the origin ((0,0)), ( AB ) along the x-axis, and ( AC ) forming an angle ( 3theta ) with ( AB ).3. Parametrizing Point ( D ): - Since ( D ) lies such that ( angle BAD = theta ), parametrize ( D ) along a line making angle ( theta ) with ( AB ). Let ( AD = r ). - Coordinates of ( D ): ( (rcostheta, rsintheta) ).4. Distance Conditions: - Distance from ( D ) to ( AB ) (x-axis) is ( y = rsintheta ). - Distance from ( D ) to ( AC ): The line ( AC ) has equation ( y = tan(3theta)x ). Using the distance formula: [ text{Distance} = frac{|tan(3theta)x - y|}{sqrt{tan^2(3theta) + 1}} = |sin(3theta)x - cos(3theta)y| ] - Substituting ( x = rcostheta ), ( y = rsintheta ): [ text{Distance} = |rcosthetasin(3theta) - rsinthetacos(3theta)| = r|sin(2theta)| ]5. Equating Distances: - Given the distance to ( AC ) is twice that to ( AB ): [ rsin(2theta) = 2(rsintheta) ] - Simplifying: ( sin(2theta) = 2sintheta ). - Using ( sin(2theta) = 2sinthetacostheta ): [ 2sinthetacostheta = 2sintheta implies costheta = 1 ] - This implies ( theta = 0 ), which is impossible for a valid point ( D ) inside the angle.Conclusion:No such point ( D ) exists within the acute angle ( angle BAC ) that satisfies both conditions. The requirement leads to a contradiction, necessitating ( theta = 0 ), which places ( D ) on ( AB ), outside the interior of the angle.Answer: (boxed{text{No}})

🔑:We are given a point (D) inside an acute angle ( angle text{BAC} ) such that ( angle text{CAD} ) is twice the size of ( angle text{BAD} ). The problem asks if point (D) could be twice as far from the line ( AC ) than it is from the line ( AB ).To find the answer to this question, let us go through the following steps:1. Introduce Perpendiculars: Drop a perpendicular from point (D) to the line ( AB ), denoted as ( DE ). On the ray ( AC ), mark a segment ( AF ) such that ( AF = AD ).2. Form a Triangle: Construct the isosceles triangle ( ADF ), where ( AD = AF ). 3. Median and Angle Bisector Properties: Draw the median ( AG ) in the triangle ( ADF ). Since ( AG ) is a median in the isosceles triangle, it is also the angle bisector and the altitude. Thus, the angles ( angle DAG ) and ( angle DAE ) are equal, and right triangles ( triangle AED ) and ( triangle AGD ) are congruent by the hypotenuse-leg (HL) congruence theorem.4. Equality of Distances: Because ( triangle AED ) and ( triangle AGD ) are congruent, we have: [ DG = GE ] Also, note: [ DF = 2DG = 2DE ] However, the segment ( DF ) is not perpendicular to ( AC ), and it is therefore longer than the perpendicular distance from ( D ) to ( AC ), denoted ( DH ).5. Consequence: Since ( DF = 2DE ) but ( DH < DF ), we cannot have: [ DH = 2DE ] Conclusion:Thus, point ( D ) cannot be twice as far from the line ( AC ) as it is from the line ( AB ). [boxed{text{No}}]

🧠:Okay, let's try to tackle this tennis tournament problem. So, we have n women and 2n men. Each player plays exactly once against every other player. No ties, so every match results in a win for either a woman or a man. The ratio of matches won by women to those won by men is 7 to 5. We need to find n, and the options are given as 2, 4, 6, 7, or none of these. First, let me understand the setup. There are n women and 2n men. So, total players are n + 2n = 3n. Each player plays every other player exactly once. That means the total number of matches is the combination of 3n players taken 2 at a time. The formula for combinations is C(3n, 2) which is (3n)(3n - 1)/2. Since each match is between two players, and there are no ties, the total number of matches is also equal to the total number of wins, right? Each match has one winner, so total wins are equal to total matches. So, the total number of wins by women plus the total number of wins by men equals (3n)(3n - 1)/2.The problem states that the ratio of women's wins to men's wins is 7:5. So, if I let the number of women's wins be 7k and men's wins be 5k, then 7k + 5k = 12k = (3n)(3n - 1)/2. Therefore, 12k = (3n)(3n - 1)/2. So, k = (3n)(3n - 1)/(24). But I don't know if that helps yet. Maybe there's another way to approach this.Alternatively, the ratio of women's wins to men's wins is 7/5. So, if W is the number of women's wins and M is men's wins, then W/M = 7/5, which means W = (7/5)M. Also, W + M = total matches = (3n)(3n - 1)/2. So substituting W, we have (7/5)M + M = (12/5)M = (3n)(3n - 1)/2. Therefore, M = (5/12)*(3n)(3n - 1)/2 = (5*(3n)(3n - 1))/24. Similarly, W = (7/5)M = (7/5)*(5*(3n)(3n - 1))/24) = (7*(3n)(3n - 1))/24. But maybe this is not directly helpful. Let's think differently.But perhaps instead of looking at total wins, we need to consider how many matches are between women only, men only, and between women and men. Because in each of these categories, the wins can be assigned differently. So, first, let's compute the number of matches between women: since there are n women, the number of such matches is C(n, 2) = n(n - 1)/2. All these matches are women vs women, so the winner is a woman. So, all these matches contribute to women's wins. Similarly, the number of matches between men: there are 2n men, so C(2n, 2) = 2n(2n - 1)/2 = n(2n - 1). All these matches are men vs men, so the winner is a man. Then, the remaining matches are between women and men. The number of such inter-gender matches is n*2n = 2n². Each of these matches is between a woman and a man, so the winner could be either a woman or a man. Let's denote the number of these inter-gender matches won by women as x. Then, the number won by men is 2n² - x.Therefore, total women's wins = wins in women's matches + wins in inter-gender matches = C(n, 2) + x = [n(n - 1)/2] + x.Total men's wins = wins in men's matches + wins in inter-gender matches = C(2n, 2) + (2n² - x) = [2n(2n - 1)/2] + 2n² - x = [n(2n - 1)] + 2n² - x.So, the ratio of women's wins to men's wins is [n(n - 1)/2 + x] / [n(2n - 1) + 2n² - x] = 7/5.So, we can set up the equation:[n(n - 1)/2 + x] / [n(2n - 1) + 2n² - x] = 7/5.Let me write that equation out:( [n(n - 1)/2 + x] ) / ( [n(2n - 1) + 2n² - x] ) = 7/5.Let me compute the denominator:n(2n - 1) + 2n² - x = 2n² - n + 2n² - x = 4n² - n - x.Numerator:n(n - 1)/2 + x = (n² - n)/2 + x.So, equation becomes:( (n² - n)/2 + x ) / (4n² - n - x) = 7/5.Cross-multiplying:5*( (n² - n)/2 + x ) = 7*(4n² - n - x).Let me multiply out both sides:Left side: 5*( (n² - n)/2 + x ) = (5/2)(n² - n) + 5x.Right side: 7*(4n² - n - x) = 28n² -7n -7x.So, bringing all terms to one side:(5/2)(n² - n) + 5x -28n² +7n +7x = 0.Combine like terms:First, terms with n²: (5/2)n² -28n² = (5/2 -28)n² = (-51/2)n².Terms with n: -5n/2 +7n = (-5/2 +7)n = (9/2)n.Terms with x:5x +7x =12x.So overall:(-51/2)n² + (9/2)n +12x =0.Hmm, but we need another equation to solve for x and n. Wait, perhaps we can find x in terms of n? Since x is the number of inter-gender matches won by women. But x can be any integer between 0 and 2n². However, the problem doesn't give us more information. Wait, but maybe there's a way to relate x through another equation. Alternatively, perhaps there's an assumption that the ratio of women's wins to men's wins is 7:5, so if we can express the total women's wins and total men's wins in terms of n, then we can set up the ratio. Wait, but in our previous approach, x is involved, which is an unknown. But maybe x is not necessary? Wait, perhaps not. Let me check.Wait, another thought: in the inter-gender matches (2n² matches), each match is between a woman and a man. So, for each such match, either a woman wins or a man wins. Let’s suppose that each woman has a certain number of wins against men, and each man has a certain number of wins against women. But perhaps we can think in terms of total wins. Let’s denote x as the total number of inter-gender matches won by women, so men have 2n² - x wins in those matches. Then, as we had before, total women's wins are C(n, 2) + x and men's are C(2n, 2) + (2n² - x). Then, the ratio is 7:5.Therefore, the equation is [C(n, 2) + x] / [C(2n, 2) + 2n² - x] = 7/5.But since x is a variable here, we need to solve for x and n. But we have two variables. Wait, but maybe x can be expressed in terms of n. Let me see.Alternatively, maybe we can use the fact that the total number of matches is fixed, so women's wins plus men's wins equals total matches. So:C(n, 2) + x + C(2n, 2) + (2n² - x) = C(3n, 2).But let me compute that:Left side: C(n,2) + C(2n,2) + x + (2n² - x) = C(n,2) + C(2n,2) + 2n².Which is equal to [n(n -1)/2] + [2n(2n -1)/2] + 2n².Simplify:First term: n(n -1)/2.Second term: [2n(2n -1)]/2 = n(2n -1).Third term: 2n².So total left side: n(n -1)/2 + n(2n -1) + 2n².Let me compute this:First term: (n² -n)/2.Second term: 2n² -n.Third term: 2n².Adding them up:(n² -n)/2 + 2n² -n + 2n² = (n² -n)/2 + 4n² -n.Convert to common denominator:(n² -n)/2 + (8n² - 2n)/2 = (n² -n +8n² -2n)/2 = (9n² -3n)/2.But the total number of matches is C(3n, 2) = 3n*(3n -1)/2 = (9n² -3n)/2. So that checks out. Therefore, the equation is valid. So, since x cancels out, that equation doesn't give us new information. So we need another equation. But the problem gives us the ratio of women's wins to men's wins as 7/5. So, the only equation we have is:[ C(n,2) + x ] / [ C(2n,2) + (2n² - x) ] = 7/5.Therefore, substituting in the values:[ n(n -1)/2 + x ] / [ 2n(2n -1)/2 + 2n² - x ] = 7/5.Simplify denominator:2n(2n -1)/2 = n(2n -1). So denominator becomes n(2n -1) + 2n² -x = 2n² -n + 2n² -x = 4n² -n -x.Numerator: n(n -1)/2 +x.So, (n(n -1)/2 +x)/(4n² -n -x) =7/5.Cross multiply:5(n(n -1)/2 +x) =7(4n² -n -x).Multiply both sides by 2 to eliminate the fraction:5(n(n -1) + 2x) =14(4n² -n -x).Expand left side:5n(n -1) +10x =5n² -5n +10x.Right side:14(4n² -n -x)=56n² -14n -14x.So the equation becomes:5n² -5n +10x =56n² -14n -14x.Bring all terms to left side:5n² -5n +10x -56n² +14n +14x =0.Combine like terms:(5n² -56n²) + (-5n +14n) + (10x +14x)= (-51n²)+(9n)+(24x)=0.So:-51n² +9n +24x=0.Simplify by dividing by 3:-17n² +3n +8x=0.So,8x=17n² -3n.Therefore,x=(17n² -3n)/8.But x is the number of inter-gender matches won by women. Since x must be an integer (as you can't have a fraction of a match), the right-hand side must be integer. Therefore, (17n² -3n) must be divisible by 8. Let's check the options given (n=2,4,6,7) to see which ones satisfy this.First, n=2:17*(2)^2 -3*(2)=17*4 -6=68-6=62. 62 divided by 8 is 7.75. Not integer. So x would be 7.75, which is invalid. So n=2 is out.n=4:17*(4)^2 -3*4=17*16 -12=272-12=260. 260/8=32.5. Still not integer. So n=4 is invalid.n=6:17*(6)^2 -3*6=17*36 -18=612-18=594. 594/8=74.25. Not integer. So n=6 is out.n=7:17*(7)^2 -3*7=17*49 -21=833-21=812. 812 divided by 8 is 101.5. Not integer. Hmm. So none of the options A to D give an integer x. Therefore, the answer is E) none of these.But wait, the problem is from an exam where options are given, including E. But maybe I made a mistake somewhere. Let me check my reasoning again.First, total women's wins: C(n,2) + x = [n(n-1)/2] + x.Total men's wins: C(2n,2) + (2n² - x) = [2n(2n -1)/2] +2n² -x = [n(2n -1)] +2n² -x =2n² -n +2n² -x=4n² -n -x.Then, the ratio of women's wins to men's wins is 7/5. Therefore, [n(n-1)/2 + x]/[4n² -n -x] =7/5. Cross multiplying, 5(n(n-1)/2 +x)=7(4n² -n -x). Multiplying both sides by 2:5(n(n -1) +2x)=14(4n² -n -x). Then expanding:5n² -5n +10x=56n² -14n -14x. Bringing everything to left: -51n² +9n +24x=0, which simplifies to 24x=51n² -9n, then x=(51n² -9n)/24. Simplify by dividing numerator and denominator by 3: x=(17n² -3n)/8. So yes, that is correct.So x must be an integer. Therefore, (17n² -3n) must be divisible by 8. Let's check the answer choices again.n=2: 17*(4) -6=68-6=62. 62/8=7.75. Not integer.n=4:17*16 -12=272-12=260. 260/8=32.5. Not integer.n=6:17*36 -18=612-18=594. 594/8=74.25. Not integer.n=7:17*49 -21=833-21=812. 812/8=101.5. Not integer.So indeed, none of the options A to D give an integer x. Therefore, the answer must be E) none of these.But wait, let me check if there's an error in the problem setup. Let me verify the total number of matches. Total players:3n. Total matches: C(3n,2)= (3n)(3n -1)/2. Correct.Number of women's matches: C(n,2)=n(n-1)/2. Men's matches: C(2n,2)=2n(2n -1)/2= n(2n -1). Inter-gender matches: n*2n=2n². Total matches: n(n-1)/2 +n(2n -1) +2n². Let's compute that:n(n-1)/2 +n(2n -1) +2n².First term: (n² -n)/2.Second term:2n² -n.Third term:2n².Adding all together:(n² -n)/2 +2n² -n +2n² = (n² -n)/2 +4n² -n.Convert to halves:(n² -n)/2 + (8n² -2n)/2 = (9n² -3n)/2. Which is equal to C(3n,2)=3n(3n -1)/2=(9n² -3n)/2. Correct. So that's okay.Therefore, the setup is correct. Then, proceeding to x=(17n² -3n)/8. So, x must be an integer. Since none of the given options for n result in integer x, then answer E is correct. So, even though E is an option, let's just confirm once more.Alternatively, maybe there's a different approach. Let's compute total women's wins and total men's wins.Total women's wins: W = C(n,2) + x = [n(n -1)/2] + x.Total men's wins: M = C(2n,2) + (2n² -x) = [2n(2n -1)/2] +2n² -x = [2n² -n] +2n² -x =4n² -n -x.Given that W/M =7/5. So:[n(n -1)/2 +x] / [4n² -n -x] =7/5.Cross-multiplying:5[n(n -1)/2 +x] =7[4n² -n -x].Multiply both sides by 2:5[n(n -1) +2x] =14[4n² -n -x].So,5n(n -1) +10x =56n² -14n -14x.Bring all terms to left:5n² -5n +10x -56n² +14n +14x=0.Combine terms:-51n² +9n +24x=0.Which leads to 24x=51n² -9n.Then x=(51n² -9n)/24= (17n² -3n)/8. So same as before.Thus, since x must be integer, (17n² -3n) must be divisible by 8. Let's check n=3, for example:17*9 -9=153-9=144. 144/8=18. So x=18. Then, if n=3, which is not among the options, but x=18. So n=3 is a possible solution. But since it's not given, then E is correct.But the options given are 2,4,6,7. None of these n=3. Therefore, the answer is E. So yes, this seems correct. So I think the answer is E.

✅:To solve the problem, we analyze the total matches and wins in the tournament. There are ( n ) women and ( 2n ) men, leading to ( 3n ) total players. The total number of matches is ( binom{3n}{2} = frac{3n(3n - 1)}{2} ). These matches are categorized into three types:1. Women vs. Women: ( binom{n}{2} = frac{n(n - 1)}{2} ) matches, all won by women.2. Men vs. Men: ( binom{2n}{2} = n(2n - 1) ) matches, all won by men.3. Women vs. Men: ( n times 2n = 2n^2 ) matches, with ( x ) wins by women and ( 2n^2 - x ) wins by men.Total wins by women: ( W = frac{n(n - 1)}{2} + x ). Total wins by men: ( M = n(2n - 1) + 2n^2 - x ). Given the ratio ( frac{W}{M} = frac{7}{5} ), we substitute ( W ) and ( M ): [frac{frac{n(n - 1)}{2} + x}{4n^2 - n - x} = frac{7}{5}]Cross-multiplying and simplifying: [5left(frac{n(n - 1)}{2} + xright) = 7(4n^2 - n - x)][5(n(n - 1) + 2x) = 14(4n^2 - n - x)][5n^2 - 5n + 10x = 56n^2 - 14n - 14x][-51n^2 + 9n + 24x = 0 implies 24x = 51n^2 - 9n implies x = frac{17n^2 - 3n}{8}]For ( x ) to be an integer, ( 17n^2 - 3n ) must be divisible by 8. Testing the given options:- ( n = 2 ): ( 17(4) - 6 = 62 ), ( 62 div 8 = 7.75 ) (not integer).- ( n = 4 ): ( 17(16) - 12 = 260 ), ( 260 div 8 = 32.5 ) (not integer).- ( n = 6 ): ( 17(36) - 18 = 594 ), ( 594 div 8 = 74.25 ) (not integer).- ( n = 7 ): ( 17(49) - 21 = 812 ), ( 812 div 8 = 101.5 ) (not integer).None of the options yield an integer ( x ), so the correct answer is E) none of these.boxed{E}

🔑:1. Let k represent the number of matches won by the women against the men. Let W and M represent the total number of matches won by the women and men, respectively.2. The total number of participants is 3n (since there are n women and 2n men).3. We need to determine the number of matches: * Total number of matches among all participants: [ frac{3n(3n-1)}{2} ] * Matches among men: [ frac{2n(2n-1)}{2} = n(2n-1) ] * Matches among women: [ frac{n(n-1)}{2} ] * Matches between men and women: [ n cdot 2n = 2n^2 ]4. The total number of matches is: [ frac{3n(3n-1)}{2} = frac{9n^2 - 3n}{2} ]5. Given that the ratio of matches won by women to men is 7:5, we have: [ frac{W}{M} = frac{7}{5} Rightarrow W = frac{7M}{5} ]6. The total number of matches: [ W + M = frac{9n^2 - 3n}{2} ]7. Since W = frac{7M}{5}, we can rewrite the equation as: [ frac{7M}{5} + M = frac{9n^2 - 3n}{2} ]8. Combine like terms: [ frac{12M}{5} = frac{9n^2 - 3n}{2} ]9. Solving for M: [ 12M cdot 2 = 5(9n^2 - 3n) ] [ 24M = 45n^2 - 15n ] [ M = frac{45n^2 - 15n}{24} ]10. With (W = frac{7M}{5}): [ W = frac{7(45n^2 - 15n)}{24 cdot 5} ] [ W = frac{7(45n^2 - 15n)}{120} ] [ W = frac{315n^2 - 105n}{120} ] Simplifying: [ W = frac{21n^2 - 7n}{8} ]11. Since k leq 2n^2: [ k = frac{45n^2 - 9n}{8} text{ should be an integer} ] Trying integer values for n leq 3: [ n = 3 rightarrow k = frac{45(3)^2 - 9(3)}{8} = frac{45 cdot 9 - 27}{8} = frac{405 - 27}{8} = frac{378}{8} = 47.25 , text{(not integer)} ]12. Given, by verifying allowable integers: [ k leq 2n^2 rightarrow n = 3 text{ falsify} ]13. Therefore the correct option is: [ boxed{E} ]

🧠:Alright, so I need to prove these four set theory statements. Let me take them one by one. I remember that for subset proofs, I usually need to show that if an element is in the left-hand set, it must also be in the right-hand set. For equalities, I have to show both subsets. Let me start with the first one.First Statement: A - D ⊆ (A - B) ∪ (B - C) ∪ (C - D)Okay, A minus D is all elements in A that are not in D. The right side is the union of three differences: A-B, B-C, and C-D. So I need to show that any element in A but not in D must be in at least one of these three sets.Let me think. Take an arbitrary element x in A - D. That means x is in A and x is not in D. Now, where could x be? If x is not in B, then x would be in A - B, so that's part of the union. If x is in B, then maybe we have to look further. If x is in B but not in C, then x would be in B - C. If x is in C, then since x is not in D, x would be in C - D. So either way, if x is in A - D, then x has to be in one of those three sets. That seems to cover all possibilities.Wait, let's structure this more formally. Let x ∈ A - D. Then x ∈ A and x ∉ D. Now, consider two cases:1. x ∉ B: Then x ∈ A - B, so x ∈ (A - B) ∪ (B - C) ∪ (C - D).2. x ∈ B: Now, check if x ∈ C or not. a. x ∉ C: Then x ∈ B - C, so again in the union. b. x ∈ C: Since x ∈ C and x ∉ D, x ∈ C - D, which is part of the union.Therefore, in all cases, x is in the union. Thus, A - D is a subset of the union. That seems to work. Maybe I should verify with an example. Suppose A = {1,2,3}, D = {3}, B = {2}, C = {2}. Then A - D = {1,2}. The right side: A - B = {1,3}, B - C = ∅, C - D = ∅. So union is {1,3}, but A - D is {1,2}. Wait, this doesn't match. Hmm, that's a problem. Wait, in my example, x = 2 is in A - D (since 2 is in A and not in D). But according to the right-hand side, A - B is {1,3}, B - C is empty, C - D is empty. So union is {1,3}, which does not include 2. But 2 is in A - D. That contradicts the inclusion. Therefore, my proof must be wrong.Wait, that's confusing. Let me check the example again. A = {1,2,3}, B = {2}, C = {2}, D = {3}. Then A - D is {1,2}. Now, (A - B) is {1,3}, (B - C) is empty, (C - D) is {2} - {3} = {2}. Wait, C is {2}, so C - D is {2} - {3} = {2}. So union is {1,3} ∪ ∅ ∪ {2} = {1,2,3}. But A - D is {1,2}, which is a subset of {1,2,3}. So in this case, it does hold. Because 2 is in C - D? Wait, C is {2}, D is {3}, so 2 is in C and not in D, so yes, 2 ∈ C - D. So the union includes 2. Therefore, in the example, A - D is {1,2}, and the union is {1,2,3}, so it is a subset. So maybe my initial confusion was misplaced. Wait, but in the first case, when x is in A - D, and if x is in B and in C, then x is in C - D. So in the example, x=2 is in A, B, C, and not in D, so it's in C - D, hence in the union. So my original reasoning was correct. Maybe I made a miscalculation earlier. So the example actually works. So the proof is valid. Therefore, first statement is proven.Second Statement: A △ C ⊆ (A △ B) ∪ (B △ C)A △ C is the symmetric difference, which is (A - C) ∪ (C - A). Similarly, (A △ B) ∪ (B △ C) is the union of (A - B) ∪ (B - A) and (B - C) ∪ (C - B). Need to show that any element in A △ C is in at least one of those sets.Take x ∈ A △ C. So x is in A - C or x is in C - A.Case 1: x ∈ A - C. Then x ∈ A and x ∉ C. Now, check if x is in B.Subcase 1a: x ∉ B. Then x ∈ A - B, which is part of A △ B, so x ∈ (A △ B) ∪ (B △ C).Subcase 1b: x ∈ B. Then since x ∈ B and x ∉ C, x ∈ B - C, which is part of B △ C, so x ∈ the union.Case 2: x ∈ C - A. Similarly, x ∈ C and x ∉ A. Check if x is in B.Subcase 2a: x ∉ B. Then x ∈ C - B, which is part of B △ C, so x is in the union.Subcase 2b: x ∈ B. Then since x ∈ B and x ∉ A, x ∈ B - A, which is part of A △ B, so x is in the union.Therefore, in all cases, x is in the union. Hence, A △ C is a subset of (A △ B) ∪ (B △ C). That seems to work.Alternatively, I remember that symmetric difference is associative? Not exactly, but maybe the inclusion can be thought of in terms of transitivity. If x is in A and not in C, then either it's not in B (so in A - B) or it is in B (so in B - C). Similarly for the other direction. So yes, the proof seems solid.Third Statement: (A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A) = (A ∩ B) ∪ (B ∩ C) ∪ (C ∩ A)This is an equality, so I need to show both inclusions.First, show that left side is a subset of the right side. Take x in the left side. Then x is in all three unions: A ∪ B, B ∪ C, and C ∪ A. So x is in at least two of the sets A, B, C. Wait, because to be in all three unions, x must be in A or B, and in B or C, and in C or A. Let me see.If x is in (A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A), then:1. x ∈ A ∪ B2. x ∈ B ∪ C3. x ∈ C ∪ ASo, for x to be in all three, it must be in at least two of the sets. For example, if x is in A, then from 2, x ∈ B ∪ C. So if x is in A but not in B, then x must be in C. Similarly, if x is in B but not in C, then from 3, x must be in A. If x is in C but not in A, then from 1, x must be in B. If x is in all three, then it's in all intersections. So, x is either in all three, or in exactly two of them. In any case, x is in A ∩ B, or B ∩ C, or C ∩ A. Therefore, x is in the right side.Conversely, take x in (A ∩ B) ∪ (B ∩ C) ∪ (C ∩ A). Then x is in at least one of the intersections. Suppose x ∈ A ∩ B. Then x is in A and B, so x ∈ A ∪ B, x ∈ B ∪ C (since x ∈ B), and x ∈ C ∪ A (since x ∈ A). Therefore, x is in the left side. Similarly, if x ∈ B ∩ C or C ∩ A, same reasoning. Therefore, right side is a subset of left side. Hence equality holds.Alternatively, I can use distributive laws. Let's expand the left side:First, (A ∪ B) ∩ (B ∪ C) = B ∪ (A ∩ C) by distributive law: (A ∪ B) ∩ (B ∪ C) = B ∪ (A ∩ C). Then intersect with (C ∪ A):(B ∪ (A ∩ C)) ∩ (C ∪ A) = (B ∩ (C ∪ A)) ∪ ((A ∩ C) ∩ (C ∪ A)) [distributive]= (B ∩ (C ∪ A)) ∪ (A ∩ C) [since (A ∩ C) ∩ (C ∪ A) = A ∩ C]Now, B ∩ (C ∪ A) = (B ∩ C) ∪ (B ∩ A). So overall:Left side becomes (B ∩ C) ∪ (B ∩ A) ∪ (A ∩ C) = (A ∩ B) ∪ (B ∩ C) ∪ (C ∩ A), which is the right side. Therefore, equality holds. That's another way to see it.Fourth Statement: (A - B) △ B = A ∪ BFirst, recall that symmetric difference is defined as (X - Y) ∪ (Y - X). So (A - B) △ B = [(A - B) - B] ∪ [B - (A - B)].Compute each part:[(A - B) - B] = A - B - B = A - (B ∪ B) = A - B.[B - (A - B)] = B - (A ∩ B^c) = B ∩ (A ∩ B^c)^c = B ∩ (A^c ∪ B) = (B ∩ A^c) ∪ (B ∩ B) = (B - A) ∪ B = B.Therefore, (A - B) △ B = (A - B) ∪ B = A ∪ B (since A - B and B are disjoint, their union is A ∪ B).Wait, let me check that again. Wait, (A - B) ∪ B = A ∪ B. Yes, because A - B is all elements in A not in B, and B is all elements in B, so together they are A ∪ B.Alternatively, using the definition:(A - B) △ B = (A - B) ∪ B - (A - B) ∩ B. But since A - B and B are disjoint, their intersection is empty. So (A - B) △ B = (A - B) ∪ B = A ∪ B. Hence, equality holds.Let me verify with an example. Let A = {1,2}, B = {2,3}. Then A - B = {1}, (A - B) △ B = {1} △ {2,3} = {1,2,3}. A ∪ B = {1,2,3}, so they are equal. Another example: A = {1}, B = {2}. Then A - B = {1}, △ B = {1} △ {2} = {1,2} = A ∪ B. If A ⊆ B, say A = {1}, B = {1,2}. Then A - B = ∅, ∅ △ B = B, which is A ∪ B = {1,2}. Correct. So the equality holds.Therefore, all four statements are proven.

✅:Proofs:1. ( A - D subseteq (A - B) cup (B - C) cup (C - D) )*Proof:*Let ( x in A - D ). Then ( x in A ) and ( x notin D ). - If ( x notin B ), then ( x in A - B ). - If ( x in B ): - If ( x notin C ), then ( x in B - C ). - If ( x in C ), then ( x in C - D ) (since ( x notin D )). Thus, ( x ) belongs to ( (A - B) cup (B - C) cup (C - D) ). Hence, ( A - D subseteq (A - B) cup (B - C) cup (C - D) ).2. ( A triangle C subseteq (A triangle B) cup (B triangle C) )*Proof:*Let ( x in A triangle C ). Then ( x in A - C ) or ( x in C - A ). - Case 1: ( x in A - C ): - If ( x notin B ), then ( x in A - B subseteq A triangle B ). - If ( x in B ), then ( x in B - C subseteq B triangle C ). - Case 2: ( x in C - A ): - If ( x notin B ), then ( x in C - B subseteq B triangle C ). - If ( x in B ), then ( x in B - A subseteq A triangle B ). Thus, ( x in (A triangle B) cup (B triangle C) ). Hence, ( A triangle C subseteq (A triangle B) cup (B triangle C) ).3. ( (A cup B) cap (B cup C) cap (C cup A) = (A cap B) cup (B cap C) cup (C cap A) )*Proof:*- Left to Right: Let ( x ) be in the left-hand side. Then ( x in A cup B ), ( x in B cup C ), and ( x in C cup A ). This implies ( x ) must be in at least two of ( A, B, C ), so ( x in (A cap B) cup (B cap C) cup (C cap A) ).- Right to Left: Let ( x in (A cap B) cup (B cap C) cup (C cap A) ). If ( x in A cap B ), then ( x in A cup B ), ( x in B cup C ), and ( x in C cup A ). Similarly for the other intersections. Thus, ( x ) is in the left-hand side. Hence, equality holds by mutual inclusion.4. ( (A - B) triangle B = A cup B )*Proof:*Using the definition of symmetric difference: [(A - B) triangle B = [(A - B) - B] cup [B - (A - B)]]Simplify each component: - ( (A - B) - B = A - B ) (since ( B subseteq B )), - ( B - (A - B) = B ) (since ( A - B ) and ( B ) are disjoint). Thus, [(A - B) triangle B = (A - B) cup B = A cup B](because ( A - B ) and ( B ) are disjoint and their union covers all elements of ( A cup B )). Hence, ( (A - B) triangle B = A cup B ).

🔑:Let's prove each assertion individually in order.# 1. Proving ( A - D subseteq (A - B) cup (B - C) cup (C - D) ):To prove this inclusion, consider any element ( x in A - D ). By definition:- ( x in A ) and- ( x notin D ).Now, we will examine several cases based on the membership of ( x ) in ( B ) and ( C ):- Case 1: ( x notin B ). If ( x notin B ), then necessarily ( x in A - B ). Therefore, ( x in (A - B) ).- Case 2: ( x in B ) and ( x notin C ). If ( x in B ) and ( x notin C ), then ( x in B - C ). Therefore, ( x in (B - C) ).- Case 3: ( x in C ). If ( x in C ), given ( x notin D ), then ( x in C - D ). Therefore, ( x in (C - D) ).In all possible cases, ( x ) belongs to at least one of the sets in the union ((A - B) cup (B - C) cup (C - D) ). Hence,[x in A - D Rightarrow x in (A - B) cup (B - C) cup (C - D).]Thus, we have shown ( A - D subseteq (A - B) cup (B - C) cup (C - D) ).# 2. Proving ( A triangle C subseteq (A triangle B) cup (B triangle C) ):Recall the definition of the symmetric difference ( A triangle C ):[A triangle C = (A - C) cup (C - A).]Consider an element ( x in A triangle C ). Therefore, either:- ( x in A ) and ( x notin C ), or- ( x in C ) and ( x notin A ).Let's consider these two cases:- Case 1: ( x in A ) and ( x notin C ). - Subcase 1: ( x notin B ). If ( x notin B ), then ( x in A triangle B ). - Subcase 2: ( x in B ). If ( x in B ), then ( x notin C ) implies ( x in B triangle C ). Therefore, ( x in (A triangle B) cup (B triangle C) ).- Case 2: ( x in C ) and ( x notin A ). - Subcase 1: ( x in B ). If ( x in B ), then ( x in B triangle C ). - Subcase 2: ( x notin B ). If ( x notin B ), then ( x in A triangle B ). Therefore, ( x in (A triangle B) cup (B triangle C) ).Thus, we have shown that ( x in A triangle C Rightarrow x in (A triangle B) cup (B triangle C) ), and therefore[A triangle C subseteq (A triangle B) cup (B triangle C).]# 3. Proving ( (A cup B) cap (B cup C) cap (C cup A) = (A cap B) cup (B cap C) cup (C cap A) ):Consider any element ( x in (A cup B) cap (B cup C) cap (C cup A) ). This means:[x in (A cup B), quad x in (B cup C), quad text{and} quad x in (C cup A).]For ( x ) to be in each of these sets, ( x ) must be in at least two of the sets ( A, B, ) and ( C ). So, ( x ) is in:- Either ( A cap B ),- Or ( B cap C ),- Or ( C cap A ).In other words,[x in (A cap B) cup (B cap C) cup (C cap A).]Similarly, consider any element ( x in (A cap B) cup (B cap C) cup (C cap A) ). This means ( x ) is in one of the intersections:- If ( x in A cap B ), then ( x in (A cup B)), ( x in (B cup C)), and ( x in (C cup A) ),- If ( x in B cap C ), then ( x in (B cup C)), ( x in (C cup A)), and ( x in (A cup B) ),- If ( x in C cap A ), then ( x in (C cup A)), ( x in (A cup B)), and ( x in (B cup C) ).Thus,[x in (A cup B) cap (B cup C) cap (C cup A).]Combining both directions, we conclude that [(A cup B) cap (B cup C) cap (C cup A) = (A cap B) cup (B cap C) cup (C cap A).]# 4. Proving ( (A - B) triangle B = A cup B ):First, recall the definition of symmetric difference:[(A - B) triangle B = (A - B - B) cup (B - (A - B)) = (A - B) cup (B - A + B).]Notice that ( A - B ) and ( B ) are disjoint by definition since ( x in A - B implies x notin B ). Therefore,[(A - B) cup (B - A + B) = A - B cup B = (A - B) cup B = A cup B.]Thus, we have shown that [(A - B) triangle B = A cup B.]Hence,[boxed{(A - B) triangle B = A cup B.}]

🧠:Okay, let me try to figure out this problem. So, we have two quadratic inequalities, right? The proposition P says that these two inequalities have the same solution set. Then proposition Q states that the ratios of the coefficients a1/a2, b1/b2, and c1/c2 are all equal. The question is whether Q is necessary, sufficient, both, or neither for P.Hmm. Let me start by recalling what necessary and sufficient conditions mean. A necessary condition must be true for the original statement to be true, but it might not be enough on its own. A sufficient condition means that if it's true, then the original statement must be true, but there might be other ways the original statement could be true without it. So, if Q is a necessary and sufficient condition for P, then P and Q imply each other. If it's only sufficient, then Q implies P, but P doesn't necessarily imply Q. If it's necessary, then P implies Q, but Q doesn't necessarily imply P. If neither, then there's no implication either way.Alright, let's consider quadratic inequalities. The general form is ax² + bx + c > 0. The solution set depends on the quadratic's direction (the coefficient a) and its discriminant (b² - 4ac). Also, where the quadratic is positive depends on whether it opens upwards or downwards and its roots.First, if two quadratics have the same solution set for the inequality > 0, they must open in the same direction. Otherwise, one might be positive outside the roots and the other inside, but if they open in the same direction, their solution sets could be the same. But maybe even more specific.Let me think. Suppose both quadratics open upwards (a1 > 0 and a2 > 0). Then the inequality ax² + bx + c > 0 holds where the quadratic is above the x-axis, which is either between the roots or outside, depending on the direction. Wait, if a quadratic opens upwards and has two real roots, then it's positive outside the interval between the roots. If it doesn't have real roots (discriminant negative), then it's always positive. Similarly, if a quadratic opens downward (a < 0), then it's positive between the roots (if real roots exist) or always negative if no real roots.So, for two quadratics to have the same solution set, they must have the same direction (both opening upwards or both downwards), and the same set of roots (or both have no real roots). Also, if they have real roots, the intervals where they are positive must coincide.Therefore, the coefficients must satisfy certain conditions. Let's suppose that the two quadratics are scalar multiples of each other. If Q is true, meaning a1/a2 = b1/b2 = c1/c2 = k for some constant k, then the second quadratic is just k times the first. But for the inequalities to have the same solution set, the scalar k must be positive. Because if k is negative, multiplying by k would reverse the inequality. So, if the coefficients are proportional with a positive constant, then the inequalities would be equivalent. So in that case, Q (with the added condition that the scalar is positive) would imply P. But the problem states Q as the ratios being equal, without specifying the sign. Wait, but if a1/a2 = b1/b2 = c1/c2, then the scalar k could be positive or negative? Wait, if the ratios are all equal, then k can be either positive or negative. However, if k is negative, then the quadratic a2x² + b2x + c2 would be k times a1x² + b1x + c1, which would flip the direction of the inequality. So, for the inequalities to have the same solution set, the coefficients must be proportional with a positive constant. So, if Q is just the ratios being equal, regardless of the sign, then Q is not sufficient, because if the scalar is negative, the inequalities would have opposite solution sets. Therefore, maybe Q as stated (just the equality of ratios, not considering the sign) is not sufficient. Wait, but in Q, the problem says "a1/a2 = b1/b2 = c1/c2". If the ratios are equal, but if k is negative, then a1 and a2 would have opposite signs. Then, the two quadratics would open in opposite directions. Therefore, their solution sets would not be the same. Therefore, in order for the inequalities to have the same solution set, the coefficients must be proportional with a positive constant. Therefore, Q is equivalent to the quadratics being scalar multiples, but if the scalar is positive, then P holds. But if the scalar is negative, then P does not hold. Therefore, if Q is given without any restriction on the sign of the scalar, then Q would include cases where the scalar is negative, which would make the inequalities have opposite solution sets. Therefore, Q is not sufficient for P unless we also know that the scalar is positive. So, perhaps Q is a necessary condition but not sufficient? Wait, let's check.Suppose that P is true, meaning the two inequalities have the same solution set. Then, does that imply Q? If the solution sets are the same, then the quadratics must be proportional with a positive scalar. Because otherwise, they could have different coefficients but still the same roots and direction. Wait, but is that possible?Wait, suppose we have two different quadratics that open in the same direction, have the same roots, but are not scalar multiples of each other. For example, take (x - 1)(x - 2) = x² - 3x + 2 and 2x² - 6x + 4. The second one is just twice the first one. They would have the same solution set for the inequality > 0. But if we take a different quadratic, say x² - 3x + 2 and x² - 3x + 3. These have different roots. Wait, but suppose we take two quadratics with the same roots and same leading coefficient. Then they are identical. So, maybe two quadratics with the same solution set for the inequality must be positive scalar multiples of each other. Wait, that seems plausible.Wait, let's think. Suppose two quadratics f(x) and g(x) satisfy f(x) > 0 and g(x) > 0 having the same solution set. Then, f(x) and g(x) must have the same roots (if any) and the same leading coefficient sign. But even more, they must be proportional. Let me see.Suppose f(x) = a1x² + b1x + c1 and g(x) = a2x² + b2x + c2. If they have the same solution set for > 0, then they must open in the same direction (so a1 and a2 have the same sign). Also, they must have the same discriminant, but even more, the same roots. Wait, if they have the same roots, then they are proportional. Because if they have the same roots r1 and r2, then they can be written as a1(x - r1)(x - r2) and a2(x - r1)(x - r2). Therefore, f(x) = a1/a2 * g(x). So, in this case, the coefficients are proportional. So, if they have the same roots and same direction, then they are proportional. Similarly, if there are no real roots, then for the inequality to hold for all x (if a1 > 0) or for no x (if a1 < 0). So, in that case, if both quadratics have no real roots and open in the same direction, then they are both always positive or always negative. However, in that case, their coefficients do not necessarily have to be proportional. Wait, but no. Wait, if a quadratic has no real roots, then it's always positive if a > 0, always negative if a < 0. So, for two quadratics with no real roots to have the same solution set, they just need to have the same sign on the leading coefficient. They don't necessarily have to be proportional. For example, take f(x) = x² + 1 and g(x) = 2x² + 3. Both are always positive, so their solution sets are all real numbers. But their coefficients are not proportional. So here, P is true (they have the same solution set) but Q is false because 1/2 ≠ 0/0 (wait, b1 and b2 in these examples are zero. Wait, in my example, f(x) = x² + 1 and g(x) = 2x² + 3. Then a1/a2 = 1/2, but b1/b2 is 0/0, which is undefined. Hmm. Wait, but if both b1 and b2 are zero, then technically the ratio is undefined, but maybe in that case, we can still say the ratios are equal if all the coefficients are proportional. But in this case, the coefficients are not proportional. a1/a2 = 1/2, c1/c2 = 1/3. So the ratios are not equal. Therefore, in this case, P holds (same solution set) but Q does not. Therefore, Q is not a necessary condition for P. Wait, that contradicts my earlier thought.Wait, so if two quadratics with no real roots have the same leading coefficient sign, then their inequalities have the same solution set (all real numbers or none). However, their coefficients don't need to be proportional. For example, x² + 1 > 0 and 2x² + 3 > 0 both have all real numbers as solutions. But the ratios of coefficients are 1/2, undefined (since b1 = b2 = 0), and 1/3. So, Q is not satisfied here, even though P is. Therefore, Q is not a necessary condition for P.Therefore, that suggests that Q is not necessary. But earlier, when quadratics have real roots, if they have the same solution set, they must be proportional. So, in that case, Q is necessary. But if there are no roots, Q is not necessary. Therefore, overall, Q is not necessary for P.Alternatively, maybe if we require that the quadratics have the same solution set, regardless of whether they have roots or not, then Q is only necessary in cases where they have roots, but not when they don't. Therefore, Q is not a necessary condition for P. Because there exist cases where P holds without Q.On the other hand, if Q holds, does that make P hold? Suppose Q holds, so a1/a2 = b1/b2 = c1/c2 = k. Then, the second quadratic is a2x² + b2x + c2 = (a1/k)x² + (b1/k)x + (c1/k). Therefore, the second inequality is (a1/k)x² + (b1/k)x + (c1/k) > 0. If k is positive, then multiplying both sides by k (which doesn't change the inequality direction) gives a1x² + b1x + c1 > 0. So, in that case, the inequalities are equivalent, so their solution sets are the same. But if k is negative, multiplying by k would reverse the inequality, so the second inequality would be equivalent to a1x² + b1x + c1 < 0. Therefore, if k is negative, then the solution sets would be opposites. Therefore, Q with a positive k implies P, but Q with a negative k does not. However, in the problem statement, Q is just the equality of the ratios, not specifying the sign. Therefore, if the ratios are equal but negative, Q is still technically true? For example, a1/a2 = b1/b2 = c1/c2 = -2. Then, the coefficients are proportional with a negative constant. So, in this case, Q is satisfied, but the inequalities would have opposite solution sets, so P is not satisfied. Therefore, Q is not sufficient for P, because if the proportionality constant is negative, then P fails. Therefore, Q is only sufficient if the proportionality constant is positive. But since Q allows for negative constants, it's not sufficient.However, in the problem statement, Q is defined as the ratios being equal. But if the ratios are equal to a negative number, then technically, Q is still satisfied. For example, a1 = -2, a2 = 1, so a1/a2 = -2. Similarly, b1 = -4, b2 = 2, so b1/b2 = -2, and c1 = -6, c2 = 3, so c1/c2 = -2. Therefore, Q is satisfied, but the second quadratic is -2 times the first quadratic. Then, the first inequality is -2x² -4x -6 > 0, which simplifies to -2(x² + 2x + 3) > 0, or x² + 2x + 3 < 0. Since x² + 2x + 3 has discriminant 4 - 12 = -8, it's always positive. Therefore, the inequality x² + 2x + 3 < 0 has no solution. The second inequality is 1x² + 2x + 3 > 0, which is always true. So, in this case, Q is satisfied, but the solution sets are different (one is empty, the other is all real numbers). Therefore, Q is not sufficient for P.Therefore, Q is not sufficient. What about necessity? As earlier, when quadratics have real roots and same solution set, they must be proportional. But when they don't have real roots, they can have the same solution set without being proportional. Therefore, Q is not necessary for P, because there are cases where P holds but Q doesn't. For example, x² + 1 > 0 and 2x² + 3 > 0 both have all real numbers as solutions, but their coefficients are not proportional. So, P holds, Q doesn't. Therefore, Q is not necessary.But wait, in the case where quadratics have real roots, if they have the same solution set, does Q necessarily hold? Let me check. Suppose two quadratics have the same roots and same leading coefficient sign. Then, they must be proportional. For example, take f(x) = (x - 1)(x - 2) = x² - 3x + 2 and g(x) = 2x² - 6x + 4. These are proportional (g(x) = 2f(x)), so their coefficients ratios are equal. So, Q holds. If I have another quadratic h(x) = 3x² - 9x + 6, same thing. But if I have a quadratic that is not a scalar multiple, but still has the same roots and same leading coefficient sign, is that possible? For example, suppose f(x) = x² - 3x + 2 and h(x) = x² - 3x + 3. Wait, h(x) has discriminant 9 - 12 = -3, so no real roots. So, not the same as f(x). If I try to create two quadratics with the same roots but different coefficients, it seems they have to be proportional. Because if they have the same roots r1 and r2, then they can be written as a1(x - r1)(x - r2) and a2(x - r1)(x - r2). Therefore, the coefficients must be proportional. So, if two quadratics have the same roots and same leading coefficient sign, they are proportional. Hence, Q holds. So, in that case, if P holds (same solution set) and the quadratics have real roots, then Q must hold. But if the quadratics don't have real roots, then P can hold without Q. Therefore, Q is necessary only when quadratics have real roots, but not otherwise. Since there exist cases where P holds without Q (the no real roots case), Q is not a necessary condition for P in general. Therefore, Q is neither necessary nor sufficient. Wait, but earlier when I thought about quadratics with real roots, Q is necessary, but in cases without real roots, Q is not necessary. Since the problem statement doesn't specify anything about the quadratics, we have to consider all possibilities.Therefore, putting it all together: If Q holds (with positive proportionality constant), then P holds. If Q holds with negative constant, P doesn't hold. So Q is not sufficient. If P holds, Q may or may not hold. For example, in the case where quadratics have no real roots, P can hold without Q. In cases with real roots, P implies Q. But overall, since there are cases where P holds without Q, Q is not necessary. Therefore, Q is neither necessary nor sufficient. So the answer should be D.But wait, let me verify again. If two quadratics have the same solution set for their inequalities, does that require their coefficients to be proportional in the case where they have no real roots? No, because as the example x² +1 and 2x² +3, they both are always positive but not proportional. Therefore, in such cases, P is true but Q is false. Therefore, Q is not necessary. Conversely, if Q is true with positive proportionality constant, then P is true. But if Q is true with negative constant, P is false. Therefore, Q is not sufficient. Therefore, Q is neither necessary nor sufficient. Therefore, the answer is D.But wait, let me check another angle. Suppose Q is satisfied with a positive constant. Then P is true. So Q with an additional condition (positive constant) is sufficient. But as the problem states Q without any condition on the sign, Q can be positive or negative. Therefore, Q is not sufficient. Similarly, as there are cases where P is true without Q, Q is not necessary. So yes, D is the correct answer.

✅:To determine the relationship between propositions ( Q ) and ( P ):1. Sufficiency of ( Q ): If ( Q ) holds, the quadratics are scalar multiples. However, if the proportionality constant ( k ) is negative, the inequalities ( a_1x^2 + b_1x + c_1 > 0 ) and ( a_2x^2 + b_2x + c_2 > 0 ) would have opposite solution sets (e.g., one positive definite, the other negative definite). Thus, ( Q ) alone (without ( k > 0 )) is not sufficient for ( P ).2. Necessity of ( Q ): If ( P ) holds, the quadratics must open in the same direction and share roots (or both lack real roots). For quadratics with real roots, proportionality (( Q )) is required. However, for quadratics without real roots (e.g., ( x^2 + 1 > 0 ) and ( 2x^2 + 3 > 0 )), ( P ) holds even if the coefficients are not proportional. Thus, ( Q ) is not necessary for ( P ).Conclusion: ( Q ) is neither necessary nor sufficient for ( P ). Answer: boxed{D}

🔑:Let's analyze the given problem and solution in detail.# Problem:We are given two quadratic inequalities:[ a_{1} x^{2} + b_{1} x + c_{1} > 0 ]and[ a_{2} x^{2} + b_{2} x + c_{2} > 0, ]and we need to determine the truth of statement (Q):[ frac{a_{1}}{a_{2}} = frac{b_{1}}{b_{2}} = frac{c_{1}}{c_{2}}, ]in relation to the solution sets of the given inequalities being the same.# Steps:# Step 1: Checking Sufficiency and Necessity1. Sufficiency Check: - Consider the case where (a_1 = a_2), (b_1 = b_2), and (c_1 = c_2). In this scenario, the two inequalities: [ a_{1} x^{2} + b_{1} x + c_{1} > 0 quad text{and} quad a_{2} x^{2} + b_{2} x + c_{2} > 0 ] are identical. Hence, their solution sets would be exactly the same. - Consider the general case where: [ frac{a_{1}}{a_{2}} = frac{b_{1}}{b_{2}} = frac{c_{1}}{c_{2}} = k, quad k neq 0 ] then we can express the second inequality as: [ a_{2}x^2 + b_{2}x + c_{2} > 0 quad text{which becomes} quad frac{a_{1}}{k}x^2 + frac{b_{1}}{k}x + frac{c_{1}}{k} > 0. ] Multiplying through by (k) (since (k neq 0)), we get: [ a_{1}x^2 + b_{1}x + c_{1} > 0. ] Hence, both inequalities are reduced to the same quadratic form, and thus their solution sets are indeed the same.2. Necessary Condition Check: - However, for necessity, consider other inequalities: [ x^2 - 3x + 2 > 0 quad text{and} quad -x^2 + 3x - 2 > 0. ] Solving these: [ x^2 - 3x + 2 > 0 ] Factoring: [ (x - 1)(x - 2) > 0, ] the solution is: [ x < 1 quad text{or} quad x > 2. ] [ -x^2 + 3x - 2 > 0 ] Negating and factoring: [ -(x^2 - 3x + 2) > 0 implies (x - 1)(x - 2) < 0, ] the solution is: [ 1 < x < 2. ] Hence, the solution sets are different despite (a_{1} / a_{2}), (b_{1} / b_{2}), and (c_{1} / c_{2}) not being equal which disproves necessity.# Step 2: Counterexamples1. Another Example to Prove Non-Necessity: Consider: [ x^2 + x + 1 > 0 quad text{and} quad x^2 + x + 3 > 0. ] For the first inequality: [ x^2 + x + 1 = 0 quad text{yields} quad Delta = 1^2 - 4 cdot 1 cdot 1 = -3 < 0, ] hence, always true since no real roots exist. For the second inequality: [ x^2 + x + 3 = 0 quad text{yields} quad Delta = 1^2 - 4 cdot 1 cdot 3 = -11 < 0, ] hence, always true since no real roots exist. The solution sets for both inequalities are all real numbers ( mathbb{R} ).Since both examples show both sufficiency without necessity and equivalence even without exact ratios, neither claim A, B, nor C is strictly true. Conclusion:The statement ( Q ) is neither a sufficient nor a necessary condition for the solution sets of the inequalities to be the same.[ boxed{D} ]