🧠:[<think>Okay, so I need to find all positive integers m between 1000 and 2021 inclusive such that there's a positive integer n where mn is a perfect square and m - n is a prime number. Hmm, let's break this down step by step.First, the key conditions here are:1. mn is a perfect square.2. m - n is prime.3. m is between 1000 and 2021.Starting with the first condition: mn is a perfect square. If the product of two numbers is a perfect square, then each prime factor in their combined prime factorization must have an even exponent. That means that the primes in m and n must complement each other to make all exponents even. In other words, n must be equal to m multiplied by some square number, but adjusted so that their combined prime factors are even. Wait, maybe I should think in terms of square factors.Another approach is that if mn is a perfect square, then m and n must share the same square-free part. Remember, every integer can be expressed as k² * s, where s is square-free (the square-free part). So, if m is k_m² * s and n is k_n² * s, then their product would be k_m² * s * k_n² * s = (k_m * k_n * s)², which is a perfect square. So, m and n must have the same square-free part s. That means n must be a multiple of s, just like m is.Therefore, n can be written as (m / s) * t², but actually, more accurately, since m = s * a² for some integer a, then n must be s * b² for some integer b. Wait, maybe let's formalize this.Let’s express m as s * a² where s is square-free. Then for mn to be a perfect square, n must be s * b², where b is an integer. Because then mn = s * a² * s * b² = s² * a² * b² = (sab)², which is a perfect square. So, n = s * b². Got it.So, m = s * a², n = s * b². Then, the difference m - n = s(a² - b²) = s(a - b)(a + b). But the problem states that m - n must be prime. Since primes have only two positive divisors, 1 and themselves, then s(a - b)(a + b) must be a prime. Therefore, this product can only be written as 1 * prime. Therefore, all factors except one must be 1. Let's think about this.Since s is square-free, and (a - b) and (a + b) are positive integers (since m > n, as m - n is prime and primes are positive, so m > n). So, s, (a - b), and (a + b) are all positive integers, and their product is prime. So, for their product to be prime, one of them must be the prime itself, and the others must be 1. Because primes have exactly two divisors, 1 and themselves. So, two of the factors s, (a - b), (a + b) must be 1, and the remaining one must be the prime.So, there are three possibilities:1. s = 1, (a - b) = 1, then (a + b) = prime.2. s = prime, (a - b) = 1, (a + b) = 1. But (a + b) cannot be 1 if a and b are positive integers since a > b (because a - b =1), so a and b are at least 1 and 0, but b must be positive. So, impossible.3. s = 1, (a + b) = prime, (a - b) =1. Wait, same as case 1.Wait, maybe I need to check all three cases where each factor is the prime:Case 1: s = 1, then (a - b)(a + b) = prime. Then (a - b) and (a + b) are two integers whose product is prime. Since primes are either 1 and itself, but (a - b) < (a + b), so (a - b) must be 1 and (a + b) must be prime. Then, this is possible.Case 2: (a - b) =1, so then s*(a + b) must be prime. Here, s is square-free and a + b is an integer. So, s*(a + b) is prime. Since s is square-free, and prime is only divisible by 1 and itself. So, s must be 1 or the prime itself. But if s is the prime, then (a + b) would have to be 1, but since a > b (as a - b =1), a + b must be at least (b +1) + b = 2b +1 ≥ 3 (since b ≥1). So (a + b) can't be 1. Therefore, s must be 1. Then this reduces to case 1.Case 3: (a + b) = prime, then s*(a - b) must be 1. Since s is a positive integer and (a - b) is a positive integer (since a > b), their product is 1. So, s =1 and (a - b)=1. Again, same as case 1.Therefore, the only possible case is when s =1, (a - b)=1, and (a + b)=prime. Thus, combining these, we have a - b =1 and a + b = p (prime). Then, solving these equations:Adding both equations: 2a = p +1 => a = (p +1)/2Subtracting: 2b = p -1 => b = (p -1)/2Since a and b must be integers, p must be an odd prime (since p +1 and p -1 must be even). Therefore, p is an odd prime.Therefore, m = s * a² =1 * [(p +1)/2]^2, and n = s * b² =1 * [(p -1)/2]^2. Therefore, m = [(p +1)/2]^2 and n = [(p -1)/2]^2. Then, m - n = [(p +1)/2]^2 - [(p -1)/2]^2. Let me compute that:[(p +1)/2 + (p -1)/2][(p +1)/2 - (p -1)/2] = [ (2p)/2 ][ 2/2 ] = [p][1] = p. So, indeed, m -n = p, which is prime. So this works.Therefore, all such m must be of the form [(p +1)/2]^2 where p is an odd prime. Therefore, m must be a square of the form [(prime +1)/2]^2.So, m = [(prime +1)/2]^2. Therefore, to find all m between 1000 and 2021, we need to find primes p such that [(p +1)/2]^2 is in that range.So, first, let's solve for p:1000 ≤ [(p +1)/2]^2 ≤ 2021Take square roots:sqrt(1000) ≤ (p +1)/2 ≤ sqrt(2021)Compute sqrt(1000): approx 31.622, so (p +1)/2 ≥ 31.622 => p +1 ≥ 63.245 => p ≥ 62.245. Since p is prime, p ≥ 67 (next prime after 63.245). Wait, wait, wait. Wait, let's compute exact values.Wait, m = [(p +1)/2]^2. So, [(p +1)/2]^2 ≥ 1000 => (p +1)/2 ≥ sqrt(1000) ≈ 31.622 => p +1 ≥ 63.245 => p ≥ 62.245. Since p is prime, the smallest prime greater than 62.245 is 67.Similarly, upper bound: [(p +1)/2]^2 ≤ 2021 => (p +1)/2 ≤ sqrt(2021) ≈ 44.94 => p +1 ≤ 89.88 => p ≤ 88.88. So primes p ≤ 88. Since primes less than or equal to 88.88 are up to 83, 89. Wait, 89 is prime but 89.88 is the upper limit. Wait, 89.88, so p ≤ 89.88 -1 = 88.88. So p ≤ 88. But 89 is prime and 89 is greater than 88.88, so p can be up to 83, which is the prime just below 88. Wait, let's check.Wait, let's solve for p exactly.From m = [(p +1)/2]^2, so (p +1)/2 must be an integer? Because m must be an integer. Wait, no. Wait, m is [(p +1)/2]^2. So, (p +1)/2 must be an integer because m is an integer. Therefore, p +1 must be even, so p must be odd. Which primes are odd? All primes except 2. But p has to be odd, so p is an odd prime. Therefore, p +1 is even, so (p +1)/2 is integer. So, m is a perfect square. Therefore, m is a square of the integer (p +1)/2.Therefore, m must be a perfect square in [1000, 2021], where m = k², and k = (p +1)/2, with p prime. So, k must be an integer such that k² is in [1000, 2021], and 2k -1 is prime.Therefore, first, find all integers k where k² is between 1000 and 2021. Then check if 2k -1 is prime.Compute the range for k:sqrt(1000) ≈ 31.622, so k starts at 32.sqrt(2025) = 45, so 45² = 2025, which is above 2021, so the maximum k is 44, since 44² = 1936.Wait, 44 squared is 1936, and 45 squared is 2025. So k can be from 32 to 44 inclusive. Let me confirm:32² = 102433² = 108934² = 115635² = 122536² = 129637² = 136938² = 144439² = 152140² = 160041² = 168142² = 176443² = 184944² = 193645² = 2025, which is over 2021. So k ranges from 32 to 44 inclusive. Therefore, m can be 32² to 44². But then, for each of these k, check if 2k -1 is prime. If so, then m = k² is a solution.Therefore, let's list all k from 32 to 44 and compute 2k -1:k: 32, 2k -1 = 63k: 33, 2k -1 = 65k:34, 2k -1=67k:35, 2k -1=69k:36, 2k -1=71k:37, 2k -1=73k:38, 2k -1=75k:39, 2k -1=77k:40, 2k -1=79k:41, 2k -1=81k:42, 2k -1=83k:43, 2k -1=85k:44, 2k -1=87Now, check which of these 2k -1 numbers are prime:63: 7*9, not prime.65: 5*13, not prime.67: prime.69: 3*23, not prime.71: prime.73: prime.75: 5*15, not prime.77: 7*11, not prime.79: prime.81: 9*9, not prime.83: prime.85: 5*17, not prime.87: 3*29, not prime.So primes are at k = 34 (67), 36 (71), 37 (73), 40 (79), 42 (83). Therefore, the corresponding m values are:k=34: 34²=1156k=36:36²=1296k=37:37²=1369k=40:40²=1600k=42:42²=1764So these are the m values. Therefore, the possible m between 1000 and 2021 are 1156, 1296, 1369, 1600, 1764.Wait, let's verify each of these:For m=1156, which is 34². Then p=2*34 -1=67, which is prime. Then n= [(67 -1)/2]^2= (66/2)^2=33²=1089. Then mn=1156*1089. Let's check if this is a perfect square. 34²*33²=(34*33)^2=1122². So yes, it is a perfect square. Then m-n=1156-1089=67, which is prime. So that's correct.Similarly, m=1296=36². p=2*36 -1=71, prime. n=(71-1)/2=35, so 35²=1225. mn=36²*35²=1260², perfect square. m-n=1296-1225=71, prime. Good.m=1369=37². p=73. n=(73-1)/2=36²=1296. mn=37²*36²=(37*36)^2=1332². m-n=1369-1296=73, prime. Correct.m=1600=40². p=79. n=(79-1)/2=39²=1521. mn=40²*39²=(40*39)^2=1560². m-n=1600-1521=79, prime. Correct.m=1764=42². p=83. n=(83-1)/2=41²=1681. Wait, 83-1=82, 82/2=41. So n=41²=1681. mn=42²*41²=(42*41)^2=1722². m-n=1764-1681=83, prime. Correct.So all these m satisfy the conditions. Now, are there any other m in [1000, 2021] outside of these squares? Wait, because in the initial analysis, we concluded that m must be a square because m = [(p +1)/2]^2, but is that the only possibility?Wait, going back to the original problem. mn is a perfect square, m -n is prime. The problem doesn't state that m itself must be a square. However, through the reasoning, we found that m must be a square. Wait, why?Because when we considered mn as a perfect square, we factored m as s*a² and n as s*b² where s is the square-free part. Then m -n = s(a² - b²). To have m -n be prime, we needed s*(a -b)(a +b) = prime. Then, the conclusion was that s must be 1, and a -b =1, a +b = prime. Hence, m = ( (prime +1)/2 )², which is a square. Therefore, in this case, m must be a square. So, in this setup, m must be a square. So, the answer should only include square numbers in [1000,2021] where (2√m -1) is prime. So, this seems to capture all possibilities.But let me just check if there's another case where s is not 1, but maybe s is prime. Wait, but earlier reasoning suggested that s must be 1 because otherwise, the product s*(a -b)*(a +b) would have more factors. Wait, suppose s is a prime. Then, s*(a -b)*(a +b) is a prime. Then, since s is prime, either (a -b) or (a +b) must be 1, and the other factor multiplied by s must be 1. But (a -b) and (a +b) are both positive integers, and a > b, so (a -b) ≥1, (a +b) ≥ (a -b) + 2b ≥1 + 2*1 =3. So if s is prime, then s*(a -b)*(a +b)=prime. Then, either:- s=prime, and (a -b)*(a +b)=1. But (a -b)*(a +b)=1 is only possible if a -b=1 and a +b=1, which is impossible because a +b > a -b ≥1. So this is invalid.- Or, one of the terms (a -b) or (a +b) is 1. But (a -b) can't be 1 unless s*(a +b)=prime. If (a -b)=1, then s*(a +b)=prime. Then s must be 1 or prime. If s is prime, then (a +b)=1, which is impossible. If s=1, then (a +b)=prime. Then we're back to the previous case where s=1, and (a -b)=1, (a +b)=prime, which gives m as a square. Alternatively, if (a +b)=1, which is impossible.Therefore, the only possible case is s=1. Therefore, m must be a square. Hence, our initial conclusion is correct. Therefore, the only possible m are the squares where 2k -1 is prime, with k from 32 to 44, leading to m=1156,1296,1369,1600,1764.But let me just check m=1089, which is 33 squared. Wait, m=1089 is below 1000? Wait, 33 squared is 1089, which is 1089, which is above 1000. Wait, but in our previous k list, k starts at 32, which is 32 squared=1024. Wait, but for k=32, 2k -1=63, which is not prime, so m=1024 is excluded. Similarly, k=33, 2k -1=65, not prime. So m=1089 is excluded. Similarly, k=34 is 1156, which is included. So the answer is the five m's listed above.Therefore, the answer should be those five numbers: 1156, 1296, 1369, 1600, 1764. Let me check each of these is between 1000 and 2021.1156: yes, between 1000 and 2021.1296: yes.1369: yes.1600: yes.1764: yes.All of these are within the range. 1764 is 42 squared, and 44 squared is 1936, which would be next, but 2*44 -1=87, which is not prime. So 1936 is not included.Wait, but 44 squared is 1936, and 2*44 -1=87, which is composite, so 1936 is excluded. Then the next is 43 squared=1849. For k=43, 2k-1=85, which is composite. So 1849 is excluded. Similarly, 42 squared=1764, which is included.Therefore, the list is correct.Therefore, the final answer is the numbers 1156, 1296, 1369, 1600, 1764. Let me check once more if there could be another m not a square in this range. For example, suppose m is not a square, but when multiplied by some n, becomes a square. But according to our earlier reasoning, since m-n is prime, s must be 1, so m is a square. So I don't think there are non-square m that satisfy the conditions. Therefore, these five numbers are the only solutions.Final AnswerThe values of ( m ) are (boxed{1156}), (boxed{1296}), (boxed{1369}), (boxed{1600}), and (boxed{1764}).</think>Given a positive integer ( m ) such that there exists a positive integer ( n ) where ( mn ) is a perfect square and ( m - n ) is a prime number, we need to find all ( m ) in the range ( 1000 leq m leq 2021 ).1. Key Conditions: - ( mn ) is a perfect square. - ( m - n ) is a prime number. - ( 1000 leq m leq 2021 ).2. Analysis: - If ( mn ) is a perfect square, then ( m ) and ( n ) must have the same square-free part. This leads to expressing ( m ) and ( n ) as ( m = s cdot a^2 ) and ( n = s cdot b^2 ) where ( s ) is square-free. - The difference ( m - n = s(a^2 - b^2) = s(a - b)(a + b) ) must be a prime number. For this product to be prime, ( s ), ( (a - b) ), and ( (a + b) ) must be such that their product is prime. This implies ( s = 1 ), ( a - b = 1 ), and ( a + b = p ) (a prime).3. Solving for ( m ): - From ( a - b = 1 ) and ( a + b = p ), we get ( a = frac{p + 1}{2} ) and ( b = frac{p - 1}{2} ). - Therefore, ( m = left( frac{p + 1}{2} right)^2 ).4. Range for ( p ): - ( 1000 leq left( frac{p + 1}{2} right)^2 leq 2021 ) - Solving for ( p ), we find ( 63.245 leq p + 1 leq 89.88 ), leading to primes ( p ) between 67 and 83.5. Checking Primes: - The primes ( p ) in this range are 67, 71, 73, 79, and 83. - Corresponding ( m ) values are ( left( frac{67 + 1}{2} right)^2 = 34^2 = 1156 ), ( left( frac{71 + 1}{2} right)^2 = 36^2 = 1296 ), ( left( frac{73 + 1}{2} right)^2 = 37^2 = 1369 ), ( left( frac{79 + 1}{2} right)^2 = 40^2 = 1600 ), and ( left( frac{83 + 1}{2} right)^2 = 42^2 = 1764 ).6. Verification: - Each ( m ) is verified to be within the range ( 1000 leq m leq 2021 ) and satisfies the conditions ( mn ) being a perfect square and ( m - n ) being prime. Final AnswerThe values of ( m ) are (boxed{1156}), (boxed{1296}), (boxed{1369}), (boxed{1600}), and (boxed{1764}).,<think>Okay, so I need to find all positive integers m between 1000 and 2021 inclusive such that there exists a positive integer n where mn is a perfect square and m - n is a prime number. Hmm, let's break this down step by step.First, mn being a perfect square means that the product of m and n has all even exponents in its prime factorization. Since m and n could share some prime factors, the exponents of primes in m and n combined must be even. So, n must supply the missing exponents to make each prime's exponent even. Therefore, n must be equal to m multiplied by some square number, but adjusted so that they together form a square. Wait, maybe n is m divided by a square factor? Or is it the other way around?Let me recall that if mn is a perfect square, then m and n must have the same square-free part. The square-free part of a number is the product of primes that appear an odd number of times in its prime factorization. So, for example, if m's square-free part is k, then n's square-free part must also be k. Therefore, n must be equal to k times a perfect square, and m must also be equal to k times another perfect square. So, m = k * a² and n = k * b² for some integers a and b. Then, mn = k² * a² * b² = (kab)², which is a perfect square. That makes sense.So, m = k * a² and n = k * b². Then, the condition m - n is prime. So, m - n = k(a² - b²) = k(a - b)(a + b). This must be a prime number. Now, primes are only divisible by 1 and themselves, so the product k(a - b)(a + b) must be equal to a prime. That implies that most of these terms must be 1 except one prime factor.Let's think. Since k, (a - b), and (a + b) are all positive integers (since m and n are positive integers and m > n because m - n is prime, which is at least 2), their product is prime. So, the only way this product is prime is if two of them are 1 and the third is the prime itself. Because a prime can't be written as a product of three numbers greater than 1. So, one of k, (a - b), or (a + b) must be the prime, and the other two must be 1.So, there are three cases to consider:1. k = prime, (a - b) = 1, (a + b) = 1. But (a + b) = 1 and (a - b) = 1 would imply that a = 1 and b = 0, but b has to be positive. So this case is impossible.2. (a - b) = prime, k = 1, (a + b) = 1. Again, (a + b) = 1 implies a and b are both less than 1, which is impossible since they are positive integers. So this case is also impossible.3. (a + b) = prime, k = 1, (a - b) = 1. Then, (a - b) = 1 and (a + b) is prime. Let's check this. If (a - b) = 1, then a = b + 1. Then, (a + b) = (b + 1 + b) = 2b + 1. So, 2b + 1 must be prime, and k = 1. Therefore, m = k * a² = a², and n = k * b² = b². Then, m - n = a² - b² = (a - b)(a + b) = 1 * (2b + 1) = 2b + 1, which is prime. So, in this case, m must be a perfect square, and n is the square just below it, such that their difference is a prime number.Alternatively, maybe there's another case where k is the prime and the other terms are 1? Wait, case 1: k = prime, (a - b) = 1, (a + b) = 1. But as we saw, (a + b) can't be 1 if a and b are positive integers. Similarly, case 2: (a - b) = prime, but then k and (a + b) would have to be 1. Since (a + b) can't be 1, the only possible case is case 3 where k = 1, (a - b) = 1, and (a + b) is prime.Therefore, m must be a square of an integer a, n must be the square of an integer b = a - 1, such that m - n = 2a - 1 is prime. Wait, hold on. Wait, m = a², n = (a - 1)². Then, m - n = a² - (a - 1)² = [a - (a - 1)][a + (a - 1)] = 1*(2a - 1). So, m - n = 2a - 1, which must be prime. Therefore, 2a - 1 is prime, so m = a² must satisfy that 2a - 1 is prime, and n = (a - 1)².But wait, in the previous analysis, we considered m = k * a² and n = k * b², with k = 1. But is that the only possibility?Wait, perhaps I made a mistake here. Let's re-examine.We had m = k * a² and n = k * b². Then, m - n = k(a² - b²) = k(a - b)(a + b). For this to be prime, since primes are positive integers greater than 1, the product k(a - b)(a + b) must be prime. So, the product can only have one prime factor and 1s. Hence, two of the terms k, (a - b), (a + b) must be 1, and the remaining one must be the prime. Let's check possible cases:Case 1: k = prime, (a - b) = 1, (a + b) = 1. As before, (a + b) = 1 is impossible because a and b are positive integers, so this case is invalid.Case 2: (a - b) = prime, k = 1, (a + b) = 1. Again, (a + b) = 1 is impossible.Case 3: (a + b) = prime, k = 1, (a - b) = 1. This is possible, as discussed.Case 4: k = prime, (a - b) = 1, (a + b) = prime. Wait, no, if k is prime and (a - b) = 1, then m - n = k * 1 * (a + b). For this to be prime, either k = 1 and (a + b) is prime (which is covered in case 3), or k is prime and (a + b) = 1, which is impossible, or (a - b) = 1 and (a + b) = prime multiplied by 1, but k is prime. Wait, this is getting confusing.Wait, perhaps the only possible case is when k=1, and (a - b)=1, and (a + b) is prime. Then, m - n = 1*(1)*(prime) = prime. So that works. Alternatively, if k is a prime and both (a - b) and (a + b) equal 1, but that's impossible. Alternatively, if (a - b) is prime and k*(a + b)=1, which can't happen since k and a + b are positive integers. So the only possible case is k=1, (a - b)=1, and (a + b)=prime. Therefore, m = a² and n = (a -1)^2, with m - n = 2a -1 is prime.Therefore, all m that are perfect squares such that 2a -1 is prime, where m = a². So m must be a square, and the difference between m and the previous square is a prime number. So, 2a -1 is prime. So, a must be such that 2a -1 is prime. So, a = (p +1)/2, where p is a prime number. Therefore, a must be an integer, so p must be an odd prime, so p is odd, so (p +1)/2 is an integer. Therefore, for each odd prime p, m = [(p +1)/2]^2. Then, m must be in [1000, 2021].Therefore, first, we can find all primes p such that [(p +1)/2]^2 is between 1000 and 2021. Let's write that as:1000 ≤ [(p +1)/2]^2 ≤ 2021Taking square roots:sqrt(1000) ≤ (p +1)/2 ≤ sqrt(2021)sqrt(1000) is approx 31.622, so (p +1)/2 ≥ 31.622 => p +1 ≥ 63.245 => p ≥ 62.245, so p ≥ 63, but p must be prime. The next prime after 63 is 67.sqrt(2021) is approx 44.955, so (p +1)/2 ≤ 44.955 => p +1 ≤ 89.91 => p ≤ 88.91, so p ≤ 89. So p must be between 67 and 89 inclusive, primes in that range.Primes between 67 and 89 are: 67, 71, 73, 79, 83, 89.So p = 67,71,73,79,83,89. Then, a = (p +1)/2:For p=67: a=(67+1)/2=34, so m=34²=1156p=71: a=(71+1)/2=36, m=36²=1296p=73: a=37, m=1369p=79: a=(79+1)/2=40, m=1600p=83: a=(83+1)/2=42, m=1764p=89: a=(89+1)/2=45, m=2025But wait, m needs to be ≤ 2021. 2025 is greater than 2021, so we exclude p=89. So possible m's are 1156,1296,1369,1600,1764. Now, check if these are within 1000-2021. All of them except 2025 are. So these are possible candidates.But wait, is this all? Because this approach assumes that k=1, but maybe there's another case where k is a prime and other factors. Wait, earlier analysis suggested that the only possible case is when k=1, but perhaps there are other cases where k is a prime and (a - b) or (a + b) is 1?Wait, let's re-examine. Suppose k is a prime, and (a - b)=1. Then m - n = k*(1)*(a + b) = k*(a + b). For this to be prime, since k is prime, then (a + b) must be 1. But (a + b)=1 is impossible because a and b are positive integers. Therefore, that case is invalid.Alternatively, if (a - b)=prime, and k*(a + b)=1. But k and (a + b) are positive integers, so their product can't be 1 unless k=1 and (a + b)=1, which is impossible. Similarly, if (a + b)=prime, and k*(a - b)=1. Then, k=1 and (a - b)=1. Wait, this is the same as case 3. So, the only possible case is k=1, (a - b)=1, (a + b)=prime.Therefore, indeed, the only possible m's are perfect squares where 2a -1 is prime. So the list above is the correct list. Therefore, the m's are 1156,1296,1369,1600,1764. Wait, but let's check each of these:Take m=1156, which is 34². Then n=(34 -1)²=33²=1089. m - n=1156-1089=67, which is prime. Correct.m=1296=36². n=35²=1225. 1296 - 1225=71, prime. Correct.m=1369=37². n=36²=1296. 1369 - 1296=73, prime. Correct.m=1600=40². n=39²=1521. 1600 - 1521=79, prime. Correct.m=1764=42². n=41²=1681. 1764 -1681=83, prime. Correct.m=2025=45². n=44²=1936. 2025 -1936=89, prime, but 2025 is above 2021, so excluded.So those are the candidates. But wait, are there other m's where k is not 1? Let me think again.Suppose m is not a perfect square, but when multiplied by some n, which is not necessarily a square, gives a perfect square. Then, m and n must have the same square-free part. So m = k * a², n = k * b², where k is square-free. Then, m - n = k(a² - b²) = k(a - b)(a + b). For m - n to be prime, since k is square-free, and (a - b)(a + b) is an integer. So, the product k(a - b)(a + b) is prime.Since k is square-free, it could be 1 or a prime. If k is a prime, then (a - b)(a + b) must be 1, but (a - b)(a + b)=1 implies a - b=1 and a + b=1, which is impossible. Therefore, k cannot be a prime unless (a - b)(a + b)=1, which is impossible. Therefore, k must be 1, so m must be a square, as before. Therefore, indeed, the only possible m's are perfect squares where 2a -1 is prime. Therefore, the answer should be those m's I listed: 1156,1296,1369,1600,1764.But wait, let me verify with another approach. Suppose m is not a square, but mn is a square, so m must be a square times a square-free number. Let m = s² * k where k is square-free. Then, n must be k * t² for some integer t. Then, m - n = s²k - kt² = k(s² - t²) = k(s - t)(s + t). For this to be prime, k(s - t)(s + t) must be prime. Since k is square-free, and (s - t)(s + t) is an integer, then similar to before, either k is prime and (s - t)(s + t)=1, which is impossible, or (s - t) or (s + t) is prime and the rest are 1.But since k is square-free, if k=1, then m is a square, which we already considered. If k is prime, then s² - t² must be 1. But s² - t² =1 implies (s - t)(s + t)=1. The solutions to that are s - t=1 and s + t=1, which gives s=1, t=0, but t must be positive. So no solution here.Alternatively, if k is composite square-free, say k = pq where p and q are primes, then m - n = pq(s - t)(s + t). For this to be prime, pq must divide the prime, which is impossible because primes can't be factored into multiple primes. Therefore, k must be 1, leading back to m being a square. Hence, only the case where k=1 gives valid solutions.Therefore, the only possible m's are perfect squares where the difference with the previous square is prime. So the answer is those m's: 1156,1296,1369,1600,1764.But wait, let me check m=1600. 1600-1521=79, which is prime. Similarly, 1764-1681=83, prime. So those are correct. Now, let's check if there are any other numbers in the range 1000-2021 that are not perfect squares but could satisfy the condition.Suppose someone picks m=1001, which is not a square. Then, mn must be a square. So, m=1001. The square-free part of 1001 is 1001 itself (since 1001=7×11×13, all primes, so exponents are 1). Therefore, n must be 1001 multiplied by a square. So n=1001×k². Then, m - n =1001 -1001×k²=1001(1 -k²). For this to be positive, 1 -k² must be positive, so k=0, but n must be positive. Therefore, impossible. So m=1001 cannot work.Similarly, take m=1002. Square-free part is 1002=2×3×167. Then n must be 1002×k². m -n=1002 -1002k²=1002(1 -k²). Again, same problem. Negative unless k=0, which is invalid. So, in general, if m is not a square, then n must be m multiplied by some square, leading to m -n negative unless the square is 0, which isn't allowed. Therefore, only when m is a square can n be smaller than m and positive. Therefore, only perfect squares m can satisfy the condition, and their corresponding n=(sqrt(m)-1)^2, leading to m -n=2sqrt(m)-1, which needs to be prime.Therefore, all m in the range [1000,2021] that are perfect squares, and 2sqrt(m)-1 is prime. So the steps are:1. Find all perfect squares m between 1000 and 2021.2. For each such m, compute a=sqrt(m). Check if 2a -1 is prime.3. Collect all m where 2a -1 is prime.So let's list all perfect squares between 1000 and 2021.32²=102433²=108934²=115635²=122536²=129637²=136938²=144439²=152140²=160041²=168142²=176443²=184944²=193645²=2025But 2025 is beyond 2021, so the list stops at 44²=1936.Now, for each of these m's, compute a=sqrt(m) and check if 2a -1 is prime.32²=1024. a=32. 2*32 -1=63, which is 7×9, composite. So discard.33²=1089. a=33. 2*33 -1=65=5×13. Composite. Discard.34²=1156. a=34. 2*34 -1=67, prime. Keep.35²=1225. a=35. 2*35 -1=69=3×23. Composite.36²=1296. a=36. 2*36 -1=71, prime. Keep.37²=1369. a=37. 2*37 -1=73, prime. Keep.38²=1444. a=38. 2*38 -1=75=3×5². Composite.39²=1521. a=39. 2*39 -1=77=7×11. Composite.40²=1600. a=40. 2*40 -1=79, prime. Keep.41²=1681. a=41. 2*41 -1=81=9×9. Composite.42²=1764. a=42. 2*42 -1=83, prime. Keep.43²=1849. a=43. 2*43 -1=85=5×17. Composite.44²=1936. a=44. 2*44 -1=87=3×29. Composite.So, the valid m's are 1156,1296,1369,1600,1764. These correspond to a=34,36,37,40,42, which give primes 67,71,73,79,83 respectively.Therefore, the answer should be these five numbers: 1156, 1296, 1369, 1600, 1764.Just to confirm, let's check if there could be other cases where k is not 1 but m -n is prime. For example, suppose m = 2 * a², n=2 * b². Then, m -n = 2(a² - b²)=2(a -b)(a + b). For this to be prime, 2(a -b)(a + b) must be prime. The only way is if one of the terms is 1 and the others are adjusted. For example, 2*(1)*(something) = prime. So, (a -b)=1, and 2*(1)*(a + b)=prime. Therefore, a + b must be prime/2. But since a and b are integers, a + b must be an integer, so prime must be even. The only even prime is 2. So 2*(a + b)=2 => a + b=1, which is impossible. Hence, no solutions here.Similarly, if k=2, m=2a², n=2b², m -n=2(a² - b²). For this to be prime, 2(a -b)(a + b)=prime. So prime must be 2, but 2(a -b)(a + b)=2 => (a -b)(a + b)=1. Which implies a -b=1 and a + b=1, impossible. So no solutions here.Similarly, if k is another prime, say 3. Then m=3a², n=3b². m -n=3(a² - b²)=3(a -b)(a + b). For this to be prime, 3*(a -b)*(a + b)=prime. So again, either 3=prime and (a -b)(a + b)=1, which is impossible, or (a -b)=1 and 3*(a + b)=prime. Then, (a + b)=prime/3. So prime must be divisible by 3, so prime=3. Then, a + b=1. Again impossible. Therefore, no solutions here.Therefore, no solutions exist when k is a prime other than 1. Similarly, for composite k. Hence, our initial conclusion that only perfect squares m with 2sqrt(m)-1 prime are the solutions is correct.Therefore, the final answer is m = 1156, 1296, 1369, 1600, 1764.Final AnswerThe values of ( m ) are (boxed{1156}), (boxed{1296}), (boxed{1369}), (boxed{1600}), and (boxed{1764}).</think>To find all positive integers ( m ) between 1000 and 2021 such that there exists a positive integer ( n ) where ( mn ) is a perfect square and ( m - n ) is prime, we start by noting that ( mn ) being a perfect square implies ( m ) and ( n ) must have the same square-free part. This leads us to consider ( m = k cdot a^2 ) and ( n = k cdot b^2 ) for some integers ( k ), ( a ), and ( b ). Given ( m - n ) must be prime, we analyze the expression ( k(a^2 - b^2) = k(a - b)(a + b) ). For this product to be prime, two of the factors ( k ), ( (a - b) ), and ( (a + b) ) must be 1, with the remaining factor being the prime. The only feasible case is when ( k = 1 ), ( (a - b) = 1 ), and ( (a + b) ) is prime. This leads to ( m = a^2 ) and ( n = (a - 1)^2 ), with ( m - n = 2a - 1 ) being prime.We then find all perfect squares ( m ) between 1000 and 2021 and check if ( 2sqrt{m} - 1 ) is prime. The valid values of ( m ) are:- ( 34^2 = 1156 ) (since ( 2 cdot 34 - 1 = 67 ) is prime)- ( 36^2 = 1296 ) (since ( 2 cdot 36 - 1 = 71 ) is prime)- ( 37^2 = 1369 ) (since ( 2 cdot 37 - 1 = 73 ) is prime)- ( 40^2 = 1600 ) (since ( 2 cdot 40 - 1 = 79 ) is prime)- ( 42^2 = 1764 ) (since ( 2 cdot 42 - 1 = 83 ) is prime)Thus, the values of ( m ) are (boxed{1156}), (boxed{1296}), (boxed{1369}), (boxed{1600}), and (boxed{1764}).]

🔑:1. Let ( m ) be a positive integer such that there exists a positive integer ( n ) for which ( mn ) is a perfect square and ( m - n ) is prime. We need to find all ( m ) in the range ( 1000 leq m leq 2021 ).2. Since ( mn ) is a perfect square, we can write ( m ) and ( n ) in the form ( m = s q^2 ) and ( n = s r^2 ), where ( s ) is a common factor and ( q ) and ( r ) are integers. This ensures that ( mn = s^2 q^2 r^2 ) is a perfect square.3. Given ( m - n ) is prime, we have: [ p = m - n = s q^2 - s r^2 = s (q^2 - r^2) = s (q - r)(q + r) ] Since ( p ) is prime, the factors ( s ), ( q - r ), and ( q + r ) must be such that one of them is 1. The only feasible solution is ( s = 1 ), ( q - r = 1 ), and ( q + r = p ). This simplifies to: [ q = frac{p + 1}{2} quad text{and} quad r = frac{p - 1}{2} ]4. Therefore, we have: [ m = q^2 = left( frac{p + 1}{2} right)^2 quad text{and} quad n = r^2 = left( frac{p - 1}{2} right)^2 ]5. We need to find ( m ) such that ( 1000 leq m leq 2021 ). This translates to: [ 1000 leq left( frac{p + 1}{2} right)^2 leq 2021 ]6. Taking the square root of the inequality: [ sqrt{1000} leq frac{p + 1}{2} leq sqrt{2021} ] [ 31.62 leq frac{p + 1}{2} leq 44.94 ]7. Multiplying through by 2: [ 63.24 leq p + 1 leq 89.88 ]8. Subtracting 1 from all parts: [ 62.24 leq p leq 88.88 ]9. Since ( p ) must be a prime number, we consider the prime numbers in the range ( 63 leq p leq 87 ): [ p in {67, 71, 73, 79, 83} ]10. For each ( p ), we calculate ( m ): - For ( p = 67 ): [ m = left( frac{67 + 1}{2} right)^2 = 34^2 = 1156 ] - For ( p = 71 ): [ m = left( frac{71 + 1}{2} right)^2 = 36^2 = 1296 ] - For ( p = 73 ): [ m = left( frac{73 + 1}{2} right)^2 = 37^2 = 1369 ] - For ( p = 79 ): [ m = left( frac{79 + 1}{2} right)^2 = 40^2 = 1600 ] - For ( p = 83 ): [ m = left( frac{83 + 1}{2} right)^2 = 42^2 = 1764 ]11. All these values of ( m ) lie within the range ( 1000 leq m leq 2021 ).The final answer is ( boxed{{1156, 1296, 1369, 1600, 1764}} )

🧠:[<think>Okay, so I need to figure out the "good order number" of the reversed sequence given that the original sequence has a "good order number" of 3. Let me start by understanding the problem again.First, the "good order" is defined as a pair of elements in the sequence where the later element (i_p) is greater than the earlier one (i_q) when p > q. The "good order number" is just the count of such pairs. So, in other words, it's the number of inverses of the usual inversion count? Wait, no. Wait, hold on. In the example given, the sequence (1, 3, 4, 2) has "good orders" listed as "1,3", "1,4", "1,2", "3,4". Let me count the inversions here. Wait, the usual inversion count is the number of pairs where p < q but i_p > i_q. So in this example, the usual inversion count would be the pairs (3,2), (4,2). So inversion count 2, but the "good order number" here is 4. So actually, the "good order number" is the number of pairs that are not inversions. Because in a sorted sequence, all pairs would be in "good order", so the maximum number. So if the sequence is sorted in increasing order, the "good order number" would be n(n-1)/2. If it's sorted in decreasing order, the "good order number" would be 0. So "good order number" is actually the number of pairs that are in the correct (increasing) order. So it's the total number of possible pairs minus the number of inversions. Because in any permutation, each pair is either an inversion or a "good order".Therefore, if the original sequence has a "good order number" of 3, then the number of inversions in that sequence is equal to the total number of pairs minus 3. The total number of pairs in a sequence of length n is C(n, 2) = n(n-1)/2. Here, n=7, so total pairs are 21. Therefore, if the "good order number" is 3, the number of inversions is 21 - 3 = 18.Now, when we reverse the sequence, the number of inversions in the reversed sequence will be equal to the number of "good orders" in the original sequence, which is 3. Wait, is that correct? Let me think. Reversing a permutation swaps the roles of inversions and non-inversions. Because if you have a pair (i, j) in the original sequence where i < j and a_i > a_j, that's an inversion. In the reversed sequence, the positions of i and j are swapped. So if in the original sequence, a_i > a_j and i < j, then in the reversed sequence, the element a_i is now at position n - i + 1, and a_j is at position n - j + 1. Since i < j in the original, n - i + 1 > n - j + 1. So in the reversed sequence, a_i is at a higher position than a_j, but a_i > a_j, so that pair would still be an inversion? Wait, no. Wait, in the original sequence, a_i > a_j with i < j is an inversion. In the reversed sequence, the positions of a_i and a_j are reversed. So in the reversed sequence, the element a_i is at position n - i + 1, and a_j is at position n - j + 1. Since i < j, n - i + 1 > n - j + 1. Therefore, in the reversed sequence, the position of a_i is greater than the position of a_j. However, a_i > a_j, so in the reversed sequence, the pair (a_i, a_j) is such that the element at the higher position is greater than the one at the lower position. Wait, so that would actually be a "good order" pair, not an inversion. Wait, maybe inversions in the original sequence correspond to non-inversions in the reversed sequence and vice versa.Let me clarify. Let's take an example. Suppose the original sequence is (1, 3, 4, 2). The inversions here are (3,2) and (4,2). The "good order number" is 4, as given. If we reverse the sequence, we get (2, 4, 3, 1). Now, let's compute the inversions in the reversed sequence. The pairs in reversed sequence:2 and 4: 2 < 4, positions 1 and 2: correct order, no inversion.2 and 3: 2 < 3, positions 1 and 3: correct order, no inversion.2 and 1: 2 > 1, positions 1 and 4: inversion.4 and 3: 4 > 3, positions 2 and 3: inversion.4 and 1: 4 > 1, positions 2 and 4: inversion.3 and 1: 3 > 1, positions 3 and 4: inversion.So total inversions in reversed sequence: 1 (2,1) + 1 (4,3) + 1 (4,1) + 1 (3,1) = 4 inversions. The original sequence had 2 inversions and 4 "good orders". The reversed sequence has 4 inversions. Wait, but the "good order number" of the reversed sequence would be total pairs (6) minus inversions (4) = 2. But in the original problem, the "good order number" of the original was 4, and the reversed sequence's "good order number" is 2. So in this case, the "good order number" of the reversed sequence is equal to the "good order number" of the original sequence minus 2. But that doesn't form a general rule. Wait, but in this example, original "good order" is 4, reversed is 2. But how does that relate to the inversions?Wait, original sequence had 2 inversions, reversed sequence had 4 inversions. So inversions in reversed sequence is total pairs (6) minus original "good order number" (4) = 2? Wait, no. Wait, original "good order number" is 4, which is total pairs (6) minus inversions (2). So inversions in original is 2. Then in reversed sequence, inversions is 4. Which is total pairs (6) minus original "good order number" (4). So maybe in general, inversions in reversed sequence is total pairs minus original "good order number".But wait, let's check another example. Suppose original sequence is (1,2,3,4). Then "good order number" is 6 (all pairs), inversions is 0. Reversed sequence is (4,3,2,1). Then inversions here are all 6 pairs, so "good order number" is 0. Which is total pairs (6) minus original inversions (0) = 6. But in this case, original "good order number" was 6, reversed sequence's "good order number" is 0. Hmm. Wait, so perhaps the "good order number" of the reversed sequence is equal to the original inversions. Wait, in first example, original inversions were 2, reversed "good order number" is 2? Wait no, in first example, original inversions were 2, reversed "good order number" is 2? Wait, reversed "good order number" in first example is 2? Wait, reversed sequence is (2,4,3,1). The "good order number" should be the number of pairs where later element is greater than earlier. So in reversed sequence:(2,4): 2 < 4, good order.(2,3): 2 < 3, good order.(2,1): 2 > 1, not.(4,3): 4 > 3, not.(4,1): 4 > 1, not.(3,1): 3 > 1, not.So "good order number" is 2. Which matches the original inversions count (2). So in that case, reversed "good order number" is original inversions. Similarly, in the second example, original inversions is 0, reversed "good order number" is 0. Wait, but reversed sequence "good order number" in second example was 0, which would be original inversions (0). Wait, but original "good order number" was 6. So in the first case, original "good order number" is 4, original inversions is 2. Then reversed "good order number" is 2, which is original inversions. So seems like the "good order number" of reversed sequence is equal to the number of inversions in the original sequence.So in general, if original sequence has "good order number" G, then the inversions in original is T - G, where T is total number of pairs. Then, when reversed, the inversions in reversed sequence is equal to G. Therefore, the "good order number" of reversed sequence is T - (T - G) = G. Wait, no. Wait, let's think again.Wait, suppose original sequence has G "good order" pairs, then inversions in original is T - G. Then, when reversed, the number of inversions in reversed sequence is equal to G. Because reversing swaps the roles of inversions and "good orders". Therefore, inversions in reversed sequence is G, so "good order number" of reversed sequence is T - G. Wait, but in the first example, original "good order" was 4, inversions was 2. Then reversed sequence's inversions should be 4, so "good order number" is T - 4 = 6 - 4 = 2, which matches. Similarly, in the second example, original "good order" was 6, inversions was 0. Reversed sequence's inversions is 6, so "good order number" is 6 - 6 = 0, which is correct. Another example: if original sequence's "good order number" is 3, then inversions is T - 3 = 21 - 3 = 18. Then reversed sequence's inversions would be 3, so "good order number" of reversed sequence is 21 - 3 = 18. Wait, but that contradicts the first example. Wait, in the first example, T = 6. Original G = 4, inversions = 2. Then reversed sequence's inversions is G = 4? But in that case, reversed "good order number" would be 6 - 4 = 2. Wait, but in reality, reversed sequence's inversions were 4, and "good order number" was 2, which matches. So seems like the formula is:Original sequence: G = good order number, I = inversions = T - G.Reversed sequence: inversions = G, so good order number = T - G.Wait, but that would mean that reversed sequence's good order number is T - G. Wait, in the original example, G was 4, T was 6. Then reversed "good order number" would be 6 - 4 = 2, which matches.But in the problem given here, the original sequence has "good order number" 3. So G = 3. Then reversed sequence's "good order number" would be T - G = 21 - 3 = 18. But wait, T is the total number of pairs for n=7, which is 21. So 21 - 3 = 18. Therefore, the answer is 18. But wait, let me verify again.Wait, perhaps there's confusion in the reasoning. Let me rephrase.Let’s denote:For a permutation, the number of "good orders" (ascending pairs) is G.The number of inversions (descending pairs) is I.Since each pair is either ascending or descending, we have G + I = T, where T = C(n, 2).Now, when we reverse the permutation, each ascending pair in the original becomes a descending pair in the reversed permutation, and vice versa.Therefore, the number of ascending pairs in the reversed permutation (G') is equal to the number of descending pairs in the original permutation (I). Hence, G' = I.But since I = T - G, substituting, G' = T - G.Therefore, the "good order number" of the reversed sequence is T - G.Given that the original sequence has G = 3, then the reversed sequence has G' = 21 - 3 = 18.Therefore, the answer is 18.But let me check with the example provided.Original sequence: (1, 3, 4, 2). G = 4, T = 6. Then reversed sequence's G' = 6 - 4 = 2. Which matches the manual calculation. So the formula holds.Another check: original sequence (1,2,3,4). G = 6, T = 6. Then reversed G' = 6 - 6 = 0. Correct.Another check: original sequence (4,3,2,1). G = 0, reversed G' = 6 - 0 = 6. Correct.Therefore, in the problem, given original G = 3, reversed G' = 21 - 3 = 18. So the answer should be 18.But wait, the problem says "the sequence of distinct positive integers (a_1, a_2, a_3, a_4, a_5, a_6, a_7)" has "good order number" 3. Then the reversed sequence is (a_7, a_6, a_5, a_4, a_3, a_2, a_1). The "good order number" of the reversed sequence is 21 - 3 = 18.Yes, that seems to be the case. Therefore, the answer is 18, which is 21 - 3. Therefore, boxed answer is boxed{18}.Final Answerboxed{18}</think>Given a sequence of distinct positive integers ((a_1, a_2, a_3, a_4, a_5, a_6, a_7)) with a "good order number" of 3, we need to determine the "good order number" of the reversed sequence ((a_7, a_6, a_5, a_4, a_3, a_2, a_1)).1. Understanding Good Order Number: The "good order number" of a sequence is the number of pairs ((i_p, i_q)) such that (i_p > i_q) when (p > q). This is equivalent to the number of ascending pairs in the sequence.2. Total Number of Pairs: For a sequence of length (n), the total number of pairs is (binom{n}{2}). Here, (n = 7), so the total number of pairs is (binom{7}{2} = 21).3. Inversions and Good Order Relationship: The number of inversions (pairs where (i_p < i_q) when (p > q)) is the complement of the good order number. If the good order number is (G), the number of inversions (I) is (I = binom{n}{2} - G).4. Reversed Sequence: Reversing a sequence swaps the roles of ascending and descending pairs. The number of inversions in the reversed sequence is equal to the good order number of the original sequence. Therefore, the good order number of the reversed sequence is the total number of pairs minus the number of inversions in the original sequence.5. Calculations: - Original good order number (G = 3). - Number of inversions in the original sequence (I = 21 - 3 = 18). - The reversed sequence's number of inversions is equal to the original good order number (3). - The good order number of the reversed sequence is (21 - 3 = 18).Thus, the "good order number" of the reversed sequence ((a_7, a_6, a_5, a_4, a_3, a_2, a_1)) is (boxed{18}).,<think>Okay, so I need to figure out the good order number of the reversed sequence given that the original sequence has a good order number of 3. Let me start by recalling the definition of a "good order." It says that in a sequence of distinct positive integers, if there's a pair where the later element is larger than the earlier one (i_p > i_q when p > q), then that pair is considered a good order. The total number of such pairs is the good order number of the sequence. For example, in the sequence (1, 3, 4, 2), the good orders are (1,3), (1,4), (1,2), (3,4), so 4 in total.So, the problem here is that we have a sequence (a1, a2, a3, a4, a5, a6, a7) with a good order number of 3. Then we reverse the sequence to (a7, a6, a5, a4, a3, a2, a1) and need to find its good order number.First, let me think about what reversing the sequence does to the order of elements. If in the original sequence, a certain pair (a_p, a_q) with p < q is a good order (i.e., a_p < a_q), then in the reversed sequence, the positions of these elements would be (a_{8 - p}, a_{8 - q}) since reversing a 7-element sequence maps position 1 to 7, 2 to 6, etc. So, if originally a_p was at position p and a_q at position q, after reversing, a_p will be at position 8 - p and a_q at position 8 - q. Now, since p < q in the original sequence, 8 - p > 8 - q in the reversed sequence. So, in the reversed sequence, the element that was originally a_p is now at a higher position than the element originally a_q. But in the original sequence, since (a_p, a_q) is a good order, that means a_p < a_q. In the reversed sequence, the positions are reversed: the element a_q is now at position 8 - q, and a_p is at position 8 - p. But since 8 - p > 8 - q (because p < q), in the reversed sequence, the pair (a_q, a_p) would be considered for a good order. However, since a_p < a_q, this would mean that a_q > a_p, but in the reversed sequence, their positions are such that 8 - q < 8 - p. Wait, no, hold on: if p < q, then 8 - p > 8 - q. So in the reversed sequence, a_p is now at position 8 - p, and a_q is at position 8 - q. Since p < q, 8 - p > 8 - q, so in the reversed sequence, the position of a_p is after the position of a_q. Therefore, the pair (a_p, a_q) in the original sequence corresponds to (a_q, a_p) in terms of their elements in the reversed sequence, but their positions are reversed. So, in the reversed sequence, we have a_q at position 8 - q and a_p at position 8 - p. Since 8 - p > 8 - q, the pair in the reversed sequence would be (a_q, a_p) with the position of a_p being later. But since in the original sequence, a_p < a_q, then in the reversed sequence, a_p is at a higher position than a_q, so if a_p < a_q, then in the reversed sequence, since it's (a_q, a_p), which would mean that the later element (a_p) is smaller than the earlier element (a_q). Therefore, this pair is not a good order in the reversed sequence. Conversely, if in the original sequence, a pair is not a good order (i.e., a_p > a_q), then in the reversed sequence, the later element (a_p) is now at a higher position and a_q is at a lower position. So in the reversed sequence, the pair would be (a_q, a_p) with a_p being later. If a_p > a_q, then in the reversed sequence, since a_p is at a higher position than a_q, but a_p > a_q, then this would be a good order in the reversed sequence.Wait, this seems a bit confusing. Let me try to rephrase. Let me denote the original sequence as S = (s1, s2, s3, s4, s5, s6, s7), where s1 = a1, s2 = a2, ..., s7 = a7. Then the reversed sequence S' = (s7, s6, s5, s4, s3, s2, s1). In the original sequence S, a good order is a pair (s_i, s_j) where i < j and s_i < s_j. The total number of such pairs is 3. Now, in the reversed sequence S', each element s_k in S is now at position 8 - k. So, for the reversed sequence, the positions of s_i and s_j (where i < j in S) are now 8 - i and 8 - j in S', respectively. But since i < j, 8 - i > 8 - j, so in S', s_i comes after s_j. Therefore, in S', the pair (s_j, s_i) is considered. Whether this is a good order in S' depends on whether s_j < s_i. But in the original sequence S, since i < j, if (s_i, s_j) was a good order (i.e., s_i < s_j), then in S' the pair (s_j, s_i) would have s_j at position 8 - j and s_i at position 8 - i (with 8 - i > 8 - j). So, in S', the pair is (s_j, s_i) with s_j at a lower position than s_i, but s_j > s_i (since s_i < s_j). Therefore, this is not a good order in S'. Conversely, if in the original sequence S, (s_i, s_j) is not a good order (i.e., s_i > s_j), then in S', the pair (s_j, s_i) would be s_j < s_i, which would be a good order in S' because s_j is at a lower position than s_i (since 8 - j > 8 - i). Wait, no: if i < j in S, then in S', s_i is at position 8 - i, and s_j is at position 8 - j. So since i < j, 8 - i > 8 - j, so s_i is after s_j in S'. Therefore, the pair (s_j, s_i) in S' is considered for a good order. If s_i < s_j in S (which would be a good order in S), then in S' we have s_j at position 8 - j, s_i at position 8 - i. Since 8 - i > 8 - j, the positions of s_i is after s_j. Therefore, in S', the pair (s_j, s_i) is considered where s_j is at a lower position than s_i. Since s_j > s_i (because s_i < s_j in S), this would mean in S' the pair (s_j, s_i) is a decreasing pair, hence not a good order. On the other hand, if in S, s_i > s_j (not a good order), then in S', the pair (s_j, s_i) would have s_j < s_i (since s_i > s_j), but their positions in S' are s_j at 8 - j and s_i at 8 - i. Wait, no: s_i in S is at position i, s_j at position j. In S', s_i is now at position 8 - i, s_j at 8 - j. If i < j, then 8 - i > 8 - j, so in S', s_i comes after s_j. So in S', the pair (s_j, s_i) is considered. If in S, s_i > s_j, then in S', s_j (which is at position 8 - j) and s_i (position 8 - i). So in S', s_j is at a lower position than s_i, and s_j < s_i (since in S, s_i > s_j). Therefore, this would be a good order in S'.Therefore, the number of good orders in the reversed sequence S' is equal to the number of pairs that were not good orders in the original sequence S. Because each pair (i, j) in S where i < j and s_i > s_j (which is a bad order in S) corresponds to a pair (j, i) in S' which is a good order because in S', j is now at a lower position than i, and s_j < s_i. Conversely, pairs that were good orders in S (i < j and s_i < s_j) become bad orders in S'.So, the total number of possible pairs in a sequence of length n is C(n, 2) = n(n - 1)/2. For n = 7, this is 21. The original sequence has 3 good orders. Therefore, the number of bad orders (inversions) in the original sequence is 21 - 3 = 18. Therefore, the reversed sequence S' will have 18 good orders. But wait, the answer is 18? Let me check this logic again.Total number of pairs: 7*6/2 = 21. Original sequence has 3 good orders. Therefore, the number of inversions (bad orders) is 21 - 3 = 18. When reversed, each inversion becomes a good order, and each original good order becomes an inversion. Therefore, the reversed sequence should have 18 good orders. Therefore, the answer is 18. Is this correct?Wait, but let me verify with the example given. The example sequence (1, 3, 4, 2) has 4 good orders. Let's compute the number of inversions: total pairs is 6 (since 4 elements). 6 - 4 = 2 inversions. If we reverse the sequence, we get (2, 4, 3, 1). Now, let's count the good orders here. The pairs are:(2,4): good(2,3): good(2,1): bad(4,3): bad(4,1): bad(3,1): badSo in reversed sequence, there are 2 good orders. Which is equal to the number of inversions in the original sequence. So yes, in this case, the reversed sequence's good order number is equal to the original sequence's number of inversions. Similarly, the original sequence's good order number is 4, inversions 2, reversed sequence's good order number is 2. So the formula holds here.Therefore, in general, reversing a sequence swaps the number of good orders with the number of inversions. Since the total number of pairs is C(n,2), then if original sequence has G good orders, the reversed sequence has C(n,2) - G good orders.Therefore, in our problem, the original sequence has 3 good orders. The total number of pairs is 21. Therefore, the reversed sequence should have 21 - 3 = 18 good orders. Therefore, the answer is 18.But wait, let me check with another example to be sure. Take n=2. Suppose the original sequence is (1,2). Good orders: 1. Reversed sequence is (2,1). Good orders: 0. Total pairs: 1. 1 - 1 = 0. Correct.Another example: n=3. Suppose original sequence is (1, 3, 2). Good orders: (1,3), (1,2), (3,2) is bad. So 2 good orders. Inversions: 1 (3,2). Reversed sequence is (2, 3, 1). Good orders: (2,3), (2,1), (3,1). So (2,3) and (3,1) are good? Wait, (2,3) is good, (2,1) is bad (since 2 > 1), and (3,1) is bad (3 > 1). Wait, no: in reversed sequence (2, 3, 1). The pairs are (2,3): 2 < 3, good; (2,1): 2 > 1, bad; (3,1): 3 > 1, bad. So only 1 good order. But original inversions were 1. Hmm, 1 inversion in original corresponds to 1 good order in reversed. But the total number of pairs is 3. Original had 2 good orders, reversed has 1. 3 - 2 = 1. Correct. So in this case, reversed sequence's good order number is 1, which is 3 - 2 = 1. So the formula holds.Wait, but in the original problem, the example given was (1,3,4,2) with good orders 4. Then reversed sequence (2,4,3,1) has good orders 2. Total pairs is 6. 6 - 4 = 2. So formula holds. So it seems the formula is that reversed good orders = total pairs - original good orders.Therefore, in the problem at hand, since original sequence has good order number 3, total pairs is 21. Then reversed sequence's good order number is 21 - 3 = 18. Therefore, the answer should be 18.But just to be thorough, let's think if there's any possibility that reversing doesn't exactly invert the count. But in all examples I checked, it's true. Because every pair that was a good order in the original becomes a bad order in the reversed, and vice versa. Since the total number of pairs is fixed, the reversed sequence's good order number is total pairs minus original's good order number.Therefore, unless there's some edge case I'm missing, the answer should be 18. Therefore, the reversed sequence's good order number is 18.Final Answerboxed{18}</think>Given a sequence of distinct positive integers ((a_1, a_2, a_3, a_4, a_5, a_6, a_7)) with a "good order number" of 3, we need to determine the "good order number" of the reversed sequence ((a_7, a_6, a_5, a_4, a_3, a_2, a_1)).1. Understanding Good Orders: A "good order" in a sequence is a pair ((i_p, i_q)) where (p > q) and (i_p > i_q). The "good order number" is the count of such pairs.2. Total Pairs Calculation: For a sequence of length (n), the total number of pairs is (binom{n}{2}). For (n = 7), this is (binom{7}{2} = 21).3. Inversions and Good Orders: The "good order number" of the original sequence is 3. The number of inversions (pairs that are not good orders) in the original sequence is (21 - 3 = 18).4. Reversed Sequence: Reversing the sequence converts each inversion in the original sequence into a good order in the reversed sequence. Therefore, the number of good orders in the reversed sequence is equal to the number of inversions in the original sequence.Thus, the "good order number" of the reversed sequence ((a_7, a_6, a_5, a_4, a_3, a_2, a_1)) is (boxed{18}).]

🔑:Since the "good order number" of the sequence of distinct positive integers (a_1, a_2, a_3, a_4, a_5, a_6, a_7) is 3, we consider the sequence (a_7, a_6, a_5, a_4, a_3, a_2, a_1). We can choose any 3 elements from this sequence, and there are C_6^3 = 20 such combinations.Let's analyze the "good order number" of (a_7, a_6, a_5, a_4, a_3, a_2, a_1):1. First, we find all possible pairs of numbers in the sequence. Since there are 7 numbers, we have C_7^2 = 21 pairs.2. Next, we observe that the sequence (a_7, a_6, a_5, a_4, a_3, a_2, a_1) is a reverse ordering of (a_1, a_2, a_3, a_4, a_5, a_6, a_7). Therefore, every "good order" in the original sequence becomes a "bad order" (reverse order) in the reversed sequence, and vice versa.3. Since there are 3 "good orders" in the original sequence, there will be 3 "bad orders" in the reversed sequence. This means that there are 21 - 3 = 18 "good orders" in the reversed sequence.Thus, the "good order number" of (a_7, a_6, a_5, a_4, a_3, a_2, a_1) is boxed{18}.

🧠:[<think>Okay, so I need to show that for all positive integers n, the value of n^(1/n) is between 1 and 2, and then find the smallest k that satisfies this inequality. Hmm, let's start with the first part: proving 1 ≤ n^(1/n) ≤ 2 for all positive integers n. First off, let's consider what n^(1/n) actually represents. For any positive integer n, taking the nth root of n. When n is 1, 1^(1/1) is 1. For n=2, it's 2^(1/2) which is sqrt(2) ≈ 1.414. For n=3, it's 3^(1/3) ≈ 1.442, and so on. Wait, but when n increases, does this value increase or decrease? Let me check a few more values. For n=4, 4^(1/4) is sqrt(2) ≈ 1.414 again. For n=5, 5^(1/5) ≈ 1.379. Hmm, so it seems like after a certain point, the value starts decreasing. Maybe there's a maximum somewhere?So, if I can find the maximum value of the function f(n) = n^(1/n) for positive integers n, then that maximum would be the smallest k such that 1 ≤ n^(1/n) ≤ k for all n. The problem also asks to first show that k=2 is an upper bound, and then find the smallest such k. Let me tackle these one at a time.Starting with the first part: proving 1 ≤ n^(1/n) ≤ 2. For all positive integers n. Let's check the lower bound first. Since n is a positive integer, n ≥ 1. Taking the nth root of n, when n=1, 1^(1/1)=1. For n > 1, n ≥ 2, so n^(1/n) must be greater than 1. Because any number greater than 1 raised to a positive power is greater than 1, but here it's the nth root. Wait, actually, if n > 1, then n^(1/n) is the number that when multiplied by itself n times gives n. Since n > 1, but how does that relate to the root? For example, 2^(1/2) is sqrt(2) ≈1.414, which is greater than 1. Similarly, 3^(1/3) ≈1.442, also greater than 1. So yes, for all n ≥1, n^(1/n) ≥1. The minimum value is achieved at n=1.Now, the upper bound. We need to show that n^(1/n) ≤2 for all positive integers n. Let's test for n=1: 1^(1/1)=1 ≤2, which holds. For n=2: sqrt(2)≈1.414 ≤2. For n=3: cube root of 3≈1.442 ≤2. Continuing, n=4: 4^(1/4)=sqrt(2)≈1.414, still ≤2. As n increases, what's the behavior of n^(1/n)? Let's see for larger n. Let's take n=10: 10^(1/10)≈1.258. n=100: 100^(1/100)≈1.047. n approaching infinity, the limit of n^(1/n) as n→∞ is 1, since log(n)/n approaches 0. So, the function tends to 1 as n becomes large. Therefore, the maximum value of n^(1/n) must be somewhere for smaller n. Wait, but from the earlier examples, when n=3, it was around 1.442, which is higher than n=2 and n=4. So maybe n=3 is the maximum? Let's check n=5: 5^(1/5)≈1.379, which is less than n=3. Hmm. Let's compute a few more terms to see where the maximum occurs.n=1: 1n=2: ~1.414n=3: ~1.442n=4: ~1.414n=5: ~1.379n=6: 6^(1/6). Let's compute that. 6^(1/6) is the same as (6^(1/2))^(1/3) which is sqrt(6)≈2.449, then take the cube root: approx 1.348.n=7: 7^(1/7). Hmm, 7^(1/7). Let's compute log(7)/7. ln(7) ≈1.9459, divided by 7≈0.277. So exponentiating e^0.277≈1.320. So 7^(1/7)≈1.320.Wait, so n=3 gives the highest so far. Let's check n=3 and maybe n=4. Wait, n=3 is higher than n=2 and n=4. Let's check n=3 in more detail. Is 3^(1/3) the maximum? Let's check n=3 and n=4.Wait, 3^(1/3)=e^(ln3/3)=e^(1.0986/3)=e^0.366≈1.442. For n=4: 4^(1/4)=e^(ln4/4)=e^(1.386/4)=e^0.346≈1.414. So yes, n=3 is higher. What about n=5? As above, 1.379. So it's decreasing after n=3. Let's check n=1,2,3,4,5, etc. So seems like the maximum occurs at n=3. Wait, but the problem statement says to show that n^(1/n) is ≤2, which is obviously true, but the actual maximum is around 1.442. Therefore, the smallest k would be approximately 1.442, but we need to find the exact value. Wait, but 3^(1/3) is the maximum. Wait, maybe the maximum is e^(1/e)? Wait, let's recall calculus. If we consider the function f(x)=x^(1/x) for real numbers x>0, the maximum occurs at x=e, approximately 2.718. Therefore, the maximum value is e^(1/e)≈1.444. Which is close to 3^(1/3)≈1.442. Since n must be an integer, the maximum of n^(1/n) occurs at n=3. Wait, but when x=e≈2.718, the function f(x)=x^(1/x) reaches its maximum. So since e is approximately 2.718, between 2 and 3. Therefore, we check the integer values around e, which are n=2 and n=3. So, since f(2)=sqrt(2)≈1.414, f(3)=3^(1/3)≈1.442, f(4)=4^(1/4)=sqrt(2)≈1.414. So the maximum is at n=3. Therefore, the smallest k is 3^(1/3). But is that correct? Wait, but 3^(1/3) is approximately 1.442, but is there a higher value at some integer n? Wait, let's check n=1:1, n=2:1.414, n=3:1.442, n=4:1.414, n=5:1.379, etc. So yes, n=3 is the maximum. Therefore, the smallest k such that 1 ≤n^(1/n) ≤k for all positive integers n is 3^(1/3). So the answer would be k=3^(1/3), which is the cube root of 3. However, the problem first asks to show that 1 ≤n^(1/n) ≤2, which is obviously true since all the values we computed are less than 2, but the tighter upper bound is cube root of 3.Wait, but the problem is structured in two parts: first to show that 1 ≤n^(1/n) ≤2 for all positive integers n, and then to find the smallest such k. So first part is straightforward. For the first part, since n is a positive integer, n ≥1. Then n^(1/n) is 1 when n=1. For n ≥2, n^(1/n) ≥1, since n ≥1 and the nth root of a number ≥1 is ≥1. For the upper bound, 2. Let's see. For n=1, 1 ≤2. For n=2, sqrt(2)≈1.414 ≤2. For n=3, cube root of 3≈1.442 ≤2. For n=4, sqrt(2)≈1.414 ≤2. For n=5, fifth root of 5≈1.379 ≤2. As n increases, n^(1/n) decreases towards 1. Therefore, 2 is an upper bound. So that's the first part done.Now, the second part: find the smallest k such that 1 ≤n^(1/n) ≤k for all positive integers n. As we saw, the maximum of n^(1/n) occurs at n=3, so k is 3^(1/3). Therefore, the smallest k is the cube root of 3. But let's verify this. Suppose k=3^(1/3). Then for all positive integers n, n^(1/n) ≤3^(1/3). Let's check for n=1:1 ≤3^(1/3)≈1.442. Yes. For n=2: sqrt(2)≈1.414 ≤1.442. Yes. For n=3: exactly 3^(1/3). For n=4: sqrt(2)≈1.414. Still less. And for n≥5, it's even smaller. So yes, 3^(1/3) is the maximum value of the function over positive integers, so it's the minimal k. Therefore, the answer is k=3^(1/3). To write it in boxed notation, that's boxed{sqrt[3]{3}}.But just to be thorough, let's make sure there isn't a higher value at some larger n. For example, could there be another n where n^(1/n) is larger than 3^(1/3)? Let's check n=4, which is 4^(1/4)=sqrt(2)≈1.414 <1.442. n=5:≈1.379. n=6:≈1.348. n=7:≈1.320, n=8: 8^(1/8)=2^(3/8)=approx 1.296, n=9:9^(1/9)=3^(2/9)≈1.291, n=10:10^(1/10)≈1.258. So no, after n=3, it's decreasing. Therefore, the maximum occurs at n=3. Therefore, the minimal k is 3^(1/3). Another way to confirm is by calculus. If we treat the function f(x)=x^(1/x) for real x>0, take the natural logarithm: ln(f(x))= (ln x)/x. Take derivative: d/dx (ln x /x)= (1/x *1/x) - (ln x)/x^2 = (1 - ln x)/x^2. Set derivative to zero: 1 - ln x=0 → ln x=1 →x=e≈2.718. So the maximum occurs at x=e. Since e is between 2 and 3, we check the integer points around e, which are n=2 and n=3. Since f(2)=sqrt(2)≈1.414 and f(3)=3^(1/3)≈1.442, so n=3 gives a higher value. Therefore, the maximum over integers is at n=3. Hence, the minimal k is 3^(1/3).Therefore, the conclusion is that the smallest k is the cube root of 3.Final AnswerThe smallest ( k ) is boxed{sqrt[3]{3}}.</think>To show that (1 le n^{1/n} le 2) for all positive integers (n):1. Lower Bound: For any positive integer (n), (n ge 1). Taking the (n)-th root of (n), we get (n^{1/n} ge 1). This is because any number greater than or equal to 1 raised to any positive power is still greater than or equal to 1.2. Upper Bound: For (n = 1), (1^{1/1} = 1 le 2). For (n ge 2), we observe that (n^{1/n}) decreases as (n) increases. The maximum value of (n^{1/n}) for small (n) occurs at (n = 3), which is approximately (1.442). However, all values of (n^{1/n}) for positive integers (n) are less than or equal to 2. For example: - (2^{1/2} = sqrt{2} approx 1.414) - (3^{1/3} approx 1.442) - (4^{1/4} = sqrt{2} approx 1.414) - As (n) increases, (n^{1/n}) approaches 1, which is always less than 2.To find the smallest (k) such that (1 le n^{1/n} le k) for all positive integers (n):1. We consider the function (f(n) = n^{1/n}) and analyze its behavior. The maximum value of (n^{1/n}) occurs at (n = 3) for positive integers, as shown by: - (3^{1/3} approx 1.442) - For (n = 2), (2^{1/2} approx 1.414) - For (n ge 4), the values decrease and approach 1 as (n) increases.2. Using calculus for the continuous function (f(x) = x^{1/x}), we find the maximum occurs at (x = e). However, since (e) is approximately 2.718, the closest integers are 2 and 3. Evaluating these, we find (3^{1/3}) is larger than (2^{1/2}).Thus, the smallest (k) such that (1 le n^{1/n} le k) for all positive integers (n) is the cube root of 3.[boxed{sqrt[3]{3}}],<think>Okay, so I need to show that for all positive integers n, the expression n^(1/n) is between 1 and 2, inclusive. Then, I have to find the smallest k such that this inequality holds. Hmm, let's start by understanding what n^(1/n) actually means. First, n is a positive integer, so n can be 1, 2, 3, and so on. The expression n^(1/n) is the nth root of n. For example, when n=1, it's 1^(1/1) = 1. When n=2, it's 2^(1/2) which is the square root of 2, approximately 1.414. For n=3, it's 3^(1/3) which is the cube root of 3, approximately 1.442. Wait, so as n increases, does this value increase or decrease? Let me check a few more values. n=4: 4^(1/4) is the fourth root of 4, which is sqrt(2) ≈ 1.414. Then n=5: 5^(1/5) is approximately 1.379. Hmm, so it seems like after n=3, the value starts decreasing. Let me confirm that. For n=10: 10^(1/10) is approximately 1.258. For n=100: 100^(1/100) is around 1.047. And as n approaches infinity, the limit of n^(1/n) is known to approach 1. So, it seems like the function n^(1/n) increases up to some point and then decreases towards 1. Therefore, the maximum value of n^(1/n) must be somewhere when n is small. Wait, so when I calculated n=3, it's about 1.442, n=4 is about 1.414, so maybe the maximum is at n=3? But let's check n=2: 1.414, n=3:1.442, so yes, n=3 gives a higher value. Then, starting from n=3 onwards, the value decreases. Therefore, the maximum value of n^(1/n) is at n=3, which is approximately 1.442. But the problem statement says to show that 1 ≤ n^(1/n) ≤ 2 for all positive integers n. So, since 1.442 is less than 2, that upper bound is safe, but it's not tight. Then, the second part of the question is to find the smallest k such that 1 ≤ n^(1/n) ≤ k for all positive integers n. That would be the maximum value of n^(1/n), which seems to be at n=3. So, the answer would be k= cube root of 3, which is approximately 1.442. But let's verify this more carefully. Let's start with n=1: 1^(1/1)=1. For n=2: sqrt(2) ≈1.414. For n=3: cube root of 3 ≈1.442. For n=4: 4^(1/4)=sqrt(2)≈1.414. Then n=5: 5^(1/5)≈1.379, n=6:6^(1/6)≈1.348, and so on. So, the maximum is indeed at n=3. Therefore, the smallest k is cube root of 3. But the problem wants me to first show that 1 ≤ n^(1/n) ≤2 for all positive integers n. Let me think about how to prove that. First, the lower bound. For any positive integer n, n ≥1. Therefore, n^(1/n) ≥1^(1/n)=1. Since 1 raised to any power is 1, and n ≥1, so n^(1/n) ≥1. That's straightforward. Now, the upper bound. We need to show that n^(1/n) ≤2 for all positive integers n. Let's check for n=1: 1^1=1 ≤2. For n=2: sqrt(2)≈1.414 ≤2. For n=3: cube root of 3≈1.442 ≤2. For n=4: sqrt(2)≈1.414 ≤2, and so on. But how do we prove this in general? Let's consider that for n≥1, n ≤2^n. If that's true, then taking both sides to the power of 1/n, we get n^(1/n) ≤2. Is n ≤2^n for all positive integers n? Let's check for n=1:1 ≤2^1=2. True. For n=2:2 ≤4. True. For n=3:3 ≤8. True. For n=4:4 ≤16. True. And for n≥1, 2^n grows exponentially, which is much faster than linear growth of n. Therefore, n ≤2^n is always true. Therefore, n^(1/n) ≤2. Hence, 1 ≤n^(1/n) ≤2. But the problem also asks to find the smallest k such that 1 ≤n^(1/n) ≤k for all positive integers n. As we observed earlier, the maximum of n^(1/n) occurs at n=3. Let's verify this more rigorously. To find the maximum of the function f(n)=n^(1/n). Since n is a positive integer, we can consider n as a real variable and find its maximum. Let’s take natural logarithm to make differentiation easier. Let’s set f(n)=n^(1/n). Then ln(f(n))= (1/n) ln n. Let’s define g(n)= (ln n)/n. To find the maximum of g(n), take derivative with respect to n: g’(n)= (1/n *1/n) - (ln n)/n^2 = (1 - ln n)/n^2. Setting derivative equal to zero: (1 - ln n)/n^2=0 =>1 - ln n=0 => ln n=1 =>n=e≈2.718. So, the maximum occurs at n=e. Since n must be a positive integer, check n=2 and n=3. Compute g(2)=ln2 /2≈0.3466, g(3)=ln3 /3≈0.3662. So, g(3) >g(2). Therefore, the maximum of g(n) occurs at n=3, so the maximum of f(n)=n^(1/n) is at n=3, which is e^(1/e)≈1.444, but since n must be integer, the actual maximum is at n=3, which is 3^(1/3)≈1.442. Hence, the smallest k is 3^(1/3). Therefore, the answer is k=3^(1/3), which is the cube root of 3. But let me check for n=3, is there any integer where n^(1/n) exceeds cube root of 3? For example, if we take n=4, it's 4^(1/4)=sqrt(2)≈1.414 <1.442. For n=5, it's even smaller, so no. For n=1, it's 1. For n=2, sqrt(2)≈1.414. So, n=3 gives the maximum. Therefore, the minimal k is cube root of 3. So, the steps are: 1. Show that n^(1/n) ≥1 because n≥1 and 1^(1/n)=1. 2. Show that n^(1/n) ≤2 by proving n ≤2^n for all positive integers n, which can be done via induction or recognizing exponential growth. 3. To find the smallest k, find the maximum of n^(1/n) which occurs at n=3, so k=3^(1/3). Therefore, the problem is solved.Final AnswerThe smallest such ( k ) is (boxed{sqrt[3]{3}}).</think>To show that (1 le n^{1/n} le 2) for all positive integers (n):1. Lower Bound: For any positive integer (n), since (n ge 1), we have (n^{1/n} ge 1^{1/n} = 1). Thus, (1 le n^{1/n}).2. Upper Bound: We need to show (n^{1/n} le 2). For all positive integers (n), (n le 2^n). Taking both sides to the power of (1/n), we get (n^{1/n} le 2^{n/n} = 2). This holds because (2^n) grows exponentially, which is faster than the linear growth of (n).To find the smallest (k) such that (1 le n^{1/n} le k) for all positive integers (n):1. Consider the function (f(n) = n^{1/n}). Taking the natural logarithm, we get (ln(f(n)) = frac{ln n}{n}). Define (g(n) = frac{ln n}{n}).2. To find the maximum of (g(n)), take the derivative with respect to (n): (g'(n) = frac{1 - ln n}{n^2}). Setting the derivative to zero gives (1 - ln n = 0 implies ln n = 1 implies n = e approx 2.718). Since (n) must be an integer, check (n = 2) and (n = 3).3. Calculate (g(2) = frac{ln 2}{2} approx 0.3466) and (g(3) = frac{ln 3}{3} approx 0.3662). Since (g(3) > g(2)), the maximum occurs at (n = 3). Thus, the maximum value of (n^{1/n}) is (3^{1/3}).Therefore, the smallest (k) is (boxed{sqrt[3]{3}}).]

🔑:1. Establishing the bounds for ( n^{1/n} ): - We need to show that ( 1 leq n^{1/n} leq 2 ) for all positive integers ( n ). - For ( n = 1 ), ( n^{1/n} = 1^{1/1} = 1 ), which satisfies the inequality. - For ( n geq 2 ), we need to show that ( n^{1/n} leq 2 ).2. Finding the smallest ( k ) such that ( 1 leq n^{1/n} leq k ) for all positive integers ( n ): - We observe that ( n^{1/n} ) decreases as ( n ) increases. - We need to find the smallest ( k ) such that ( n^{1/n} leq k ) for all ( n ).3. Evaluating ( n^{1/n} ) for small values of ( n ): - For ( n = 1 ), ( 1^{1/1} = 1 ). - For ( n = 2 ), ( 2^{1/2} = sqrt{2} approx 1.414 ). - For ( n = 3 ), ( 3^{1/3} approx 1.442 ).4. Claiming ( k = sqrt[3]{3} ): - We claim that ( k = sqrt[3]{3} approx 1.442 ) is the smallest value such that ( 1 leq n^{1/n} leq k ) for all positive integers ( n ).5. Proving the claim by induction: - Base case: For ( n = 1, 2, 3 ), we have already verified that ( n^{1/n} leq sqrt[3]{3} ). - Inductive step: Assume ( n^{1/n} leq sqrt[3]{3} ) for some ( n = k geq 3 ). We need to show that ( (k+1)^{1/(k+1)} leq sqrt[3]{3} ).6. Inductive step calculation: - From the inductive hypothesis, ( k leq (sqrt[3]{3})^k ). - We need to show ( k+1 leq (sqrt[3]{3})^{k+1} ). - Consider the inequality ( (sqrt[3]{3})^k cdot sqrt[3]{3} = (sqrt[3]{3})^{k+1} ). - Since ( sqrt[3]{3} approx 1.442 ), we have ( k+1 < sqrt[3]{3} cdot k ) for ( k geq 3 ).7. Verification: - For ( k = 3 ), ( 3 + 1 = 4 ) and ( sqrt[3]{3} cdot 3 approx 1.442 cdot 3 approx 4.326 ), which is greater than 4. - Hence, ( k+1 < sqrt[3]{3} cdot k ) holds for ( k geq 3 ).8. Conclusion: - By induction, we have shown that ( 1 leq n^{1/n} leq sqrt[3]{3} ) for all positive integers ( n ).The final answer is ( boxed{ sqrt[3]{3} } ).

🧠:[<think>Okay, let's try to figure out this problem. So, we have the set S = {1, 2, ..., 100}, and we need to find the maximum size of a subset X where no two distinct elements in X multiply to give another element in X. Hmm, that sounds a bit tricky. Let me think through it step by step.First, let me restate the condition to make sure I understand it. If I pick any two different numbers from X, say a and b, then their product a*b should not be in X. So, for example, if X contains 2, 3, and 6, that's bad because 2*3=6, which is also in X. So we have to avoid such triples. The goal is to make X as large as possible without any such triples.I need to find the largest possible subset X of S where this condition holds. Let me consider how to approach this. Maybe start by thinking about the structure of numbers in S and how their products relate to each other.One idea is to split the numbers into different categories or groups where the product of two numbers in a group might land in the same group. For example, if we have small numbers, their products can still be small, but larger numbers might multiply to something outside of S (which is up to 100). Wait, actually, the product of two numbers in S can be up to 100*100=10,000, which is way larger than 100. So only products that are less than or equal to 100 could be in S, and hence in X. So perhaps larger numbers, when multiplied, will result in numbers outside of S, so they don't pose a problem? Let me check.For numbers greater than 10, their product will exceed 100. For example, 11*10=110, which is already over 100. Wait, 10*10=100, which is in S. So actually, numbers from 1 to 10 can multiply together to get up to 100, but numbers above 10, when multiplied by any number greater than or equal to 10, will give over 100. But numbers above 10 multiplied by numbers less than 10 might still land in S. For example, 11*9=99, which is in S. So even numbers above 10 can still multiply with smaller numbers to give elements in S. So maybe we need to be careful with numbers both above and below 10.Wait, maybe it's better to think in terms of prime numbers, but I'm not sure. Let me try a different approach. Let's consider the numbers in S and see which ones can be included in X without violating the condition.If I include 1 in X, then any other number in X multiplied by 1 will give that number, which is in X. But wait, the problem states "no two distinct elements in X multiply to an element in X". So, if I have 1 and a in X, then 1*a = a, which is in X. But since a is not distinct from itself? Wait, the problem says "two distinct elements". So, if I have two distinct elements, their product should not be in X. So 1 and a are distinct, their product is a, which is in X. So that's not allowed. Therefore, if X contains 1, then X cannot contain any other elements, because 1 multiplied by any other element is that element, which would be in X. Therefore, including 1 in X would limit X to just {1}. Since we want the maximum size, we probably shouldn't include 1. So maybe 1 is excluded from X. That's a crucial point.So, X cannot contain 1 and any other number. Therefore, the maximum subset X either includes 1 alone or excludes 1 and includes as many numbers as possible. Since 1 alone is only 1 element, and excluding 1 might allow us to include many more, it's better to exclude 1.Therefore, we can assume 1 is not in X. Now, we need to construct X such that for any two distinct elements a and b in X, a*b is not in X.Next, let's consider prime numbers. If we include a prime number p in X, then we need to ensure that p isn't multiplied by any other number in X to get another element in X. For example, if we include 2 in X, we need to make sure that there's no other number q in X such that 2*q is also in X. So if we have 2 and 3 in X, then 6 should not be in X. Similarly, if we have 2 and 5, then 10 should not be in X, and so on.Alternatively, if we include composite numbers, we have to check if they can be formed by multiplying two other numbers in X. For instance, if we include 6 in X, we need to make sure that 2 and 3 are not both in X. Because 2*3=6. Similarly, if we include 12, we need to avoid having pairs like 3 and 4, or 2 and 6, etc., in X.So, perhaps the problem is similar to constructing a set where no element is the product of two others. Wait, but the condition is slightly different: it's not that no element is the product of two others, but that the product of any two distinct elements is not in the set. So even if two elements multiply to something not in the set, that's okay. But if their product is in the set, that's not allowed. So the difference is that in the standard "product-free" set, you usually have that no two elements multiply to give another element in the set. Wait, actually, that's exactly the condition here. Wait, the problem states: "no two distinct elements in X multiply to an element in X." So that's the definition of a product-free set. So X is a product-free subset of S. Therefore, the maximum size of a product-free subset of {1,2,...,100}.Therefore, maybe I can recall that in the set {1,2,...,n}, the maximum product-free subset has size approximately n/2, but maybe there's a specific construction for it. Wait, but with n=100, perhaps the maximum product-free subset is the set of numbers from 51 to 100. Let's check.If we take numbers from 51 to 100, then any two numbers in that range, when multiplied, will give at least 51*51=2601, which is way larger than 100, so their product is not in S. Therefore, such a set is product-free. The size of this set is 50. But is this the maximum possible?Alternatively, perhaps we can take a larger set. For example, numbers from 26 to 100. But 26*2=52, which is in 26 to 100. But wait, if we include 26 and 2, then their product is 52, which is in the set. But if we exclude the smaller numbers, then maybe it's okay. Wait, but if we take numbers from 26 to 100, but if 26 is in X, then 26*2=52, so if both 26 and 52 are in X, and 2 is not in X, then is that okay? Wait, 2 is not in X, so even if 26 and 52 are in X, since 2 is not in X, there's no two elements in X whose product is 52. Wait, but 26 and 2 would need to be in X for their product to be 52, but since 2 is not in X, 26 and 52 can coexist. So maybe numbers from 26 to 100 can still form a product-free set? Wait, let's check with some examples.Take 50 and 2. If both are in X, then 50*2=100. If 100 is in X, then that's a problem. But if we include 50 and 100 in X, but exclude 2, then 50 and 100 can be in X because their product is 5000, which is not in X. Wait, but 50 and 2: if 2 is not in X, then even if 50 is in X, 50*2=100. If 100 is in X, then the problem is if both 50 and 2 are in X, but if 2 is excluded, then 50 and 100 can coexist because there's no element in X (other than 50 and 100) such that their product is 100. Wait, but 10*10=100. If 10 is in X, then 10*10=100 would be a problem. But 10 is not in X if we take numbers from 26 to 100. Wait, 10 is in 1 to 25. So if we take X as numbers from 26 to 100, then 10 is not in X. So, in that case, 100 can be in X, because the only pairs that multiply to 100 would be (2,50), (4,25), (5,20), (10,10). But all of these pairs involve numbers not in X (since 2,4,5,10,20,25 are all ≤25). Therefore, 100 is safe in X.Similarly, let's check 50. The factors of 50 in S are (1,50), (2,25), (5,10). Again, 1,2,5,10,25 are all ≤25, so not in X. Therefore, 50 is okay. So maybe numbers from 26 to 100 form a product-free set. The size of that set is 75 elements. Wait, 100-26+1=75. But wait, earlier when I thought of 51-100, that's 50 elements, but 26-100 is 75 elements. But is 26-100 actually product-free?Wait, let's test with numbers in 26-100. Take 26 and 2. 2 is not in X, so 26*2=52. 52 is in X. But since 2 is not in X, there is no pair in X that multiplies to 52. Because 26 is in X, but 2 is not. 52 is in X, but if there's another number in X such that some pair multiplies to 52. The factors of 52 are (1,52), (2,26), (4,13). All of these factors except 52 are not in X (since 1,2,4,13 are ≤25). Therefore, 52 is okay in X. Similarly, 26*3=78. 3 is not in X, so 78 can be in X. The factors of 78 are (1,78), (2,39), (3,26), (6,13). Again, all factors except 78 are not in X. So 78 is okay. So actually, even though numbers in 26-100 can be products of smaller numbers, since the smaller numbers are not in X, their product is allowed to be in X. Because the condition is only that two elements in X cannot multiply to give another element in X. So even if an element in X can be expressed as a product of two numbers not in X, that's fine. Therefore, the set from 26 to 100 is indeed product-free, and its size is 75.But wait, is that true? Let's check another example. Take 27 and 3. 3 is not in X, so 27*3=81. 81 is in X. But since 3 is not in X, 27 and 81 can coexist because there is no element in X that when multiplied by 27 gives another element in X. The only possible element would be 3, which isn't in X. Therefore, 27 and 81 can both be in X. Similarly, 30 and 10. 10 is not in X, so 30*10=300, which is not in X. Wait, 30 is in X, but 10 is not. So 30 can be in X. If 30 and 5 were both in X, then 30*5=150, which is not in X. But 5 is not in X. So maybe numbers from 26 to 100 are okay.But wait, wait. Let's check if there are any two numbers in X (26-100) whose product is also in X. For example, take 50 and 2. 2 is not in X, so 50*2=100. 100 is in X. But since 2 is not in X, this is allowed. Now, take two numbers in X: 50 and 50. 50*50=2500, not in X. But they are the same element. The problem says "two distinct elements", so duplicates aren't considered. So even if we have two copies, but in the set S, all elements are unique, so in X, all elements are unique. Therefore, multiplying two same elements is not considered here because they are not distinct. So, 50*50 is 2500, not in X, so that's okay.Another example: 25 is not in X (since X is 26-100), so 25 is excluded. What about 26 and 4? 4 is not in X, so 26*4=104, which is not in X. So no problem. Hmm, seems like all products of two elements in X (26-100) are either not in X or require a factor not in X. Therefore, the set from 26 to 100 is product-free and has 75 elements. But wait, can we do even better?Wait, 26-100 is 75 elements, but perhaps we can include some smaller numbers as well, as long as they don't create a product within X. For example, suppose we include 25. But 25 is in 1-25, so if we include 25, we have to check if there are any elements a and b in X such that a*b=25. Since 25 is prime squared (5*5), but 5 is not in X (since X is 26-100). So 25 can be included? Wait, if we include 25 in X, which is originally not in X (as we had X from 26-100), then adding 25 would require that there are no two elements in X whose product is 25. Since the only way to get 25 is 1*25, 5*5. But 1 and 5 are not in X, so 25 can be safely added. Wait, but if we add 25 to X, then we need to ensure that for any a in X, a and 25 don't multiply to another element in X. So 25*a should not be in X. But 25*a, where a is in 26-100, would be 25*26=650, which is way above 100, so not in X. Therefore, adding 25 to X is safe. So, perhaps we can include 25 as well, increasing the size to 76.Similarly, maybe 24. If we include 24 in X, we need to check if there are any elements a and b in X such that a*b=24. The factors are 1*24, 2*12, 3*8, 4*6. Since 1,2,3,4,6,8,12 are not in X (as X is 26-100 plus 25). So including 24 is safe. Also, multiplying 24 by any element in X (26-100) would give 24*26=624, which is not in X. Therefore, 24 can be added. Similarly, 23: factors are 1*23. Since 1 is not in X, 23 can be added. 22: factors 1*22, 2*11. 1,2,11 not in X. So 22 can be added. Continuing this way, perhaps we can add all numbers from 25 down to, say, 2? Wait, but adding 2 would cause problems. Wait, let's think.Suppose we try to add 25, 24, 23, ..., 2 into X. Let's check if adding these smaller numbers would introduce any pair whose product is in X. For example, adding 2 to X. Then, if 2 is in X, we need to ensure that 2 does not multiply with any other element in X to give another element in X. For instance, 2*26=52. If 52 is in X, then 2 and 26 cannot both be in X. But if we are adding 2 to X, and 26 is already in X, then 2*26=52. If 52 is in X, then we have a problem. But 52 is in X (since X is 26-100 plus added smaller numbers). Therefore, if we add 2, we have to remove 52 from X. But that complicates things. Similarly, adding 3 would require checking if 3*26=78 is in X, which it is, so we would have to remove 78. So if we add smaller numbers, we have to remove some elements from X to maintain the product-free condition.Therefore, the strategy of adding smaller numbers might not be straightforward because each smaller number added could conflict with multiple existing elements in X. For example, adding 2 would conflict with 26 (2*26=52), 27 (2*27=54), ..., up to 50 (2*50=100). So adding 2 would require removing 52, 54, ..., 100, which is a lot. Therefore, the trade-off is probably not worth it. It might be better to keep the upper half (26-100) and not include the smaller numbers.But let's see. Suppose we try to include some smaller numbers without conflicting. For example, if we add 25 to X, we don't need to remove anything because 25*26=650, which is not in X. Similarly, adding 24: 24*26=624, not in X. So adding 24,25 is safe. How about 23? 23*26=598, not in X. Safe. Similarly, 22: 22*26=572, safe. Continue down to, say, 51. Wait, 51 is already in X (if X is 26-100). Wait, X is 26-100. Wait, 51 is in 26-100. Hmm. Wait, maybe adding numbers from 25 down to 2 would be possible, but as mentioned, adding numbers like 2 would cause conflicts with existing numbers. So maybe if we exclude some numbers from 26-100 when adding smaller numbers, we might be able to get a larger set.But this seems complicated. Let's try to think of another approach. The standard result for product-free sets in the integers {1,2,...,n} is that the maximum size is achieved by taking the upper half of the numbers, which gives a product-free set of size ⌈n/2⌉. But wait, in our case, n=100, so the upper half would be 51-100, size 50. But earlier, we considered 26-100, which is 75 elements. So why is there a discrepancy?Wait, maybe the standard result is different. Let me recall. In a group, a product-free set is a set with no three elements a,b,c such that a*b=c. In the integers under multiplication, this is similar. The usual construction for the largest product-free set in {1,2,...,n} is indeed taking the upper half, as their products exceed n. For example, in "Product-Free Sets in Groups", the maximum size is often around n/2. However, in our case, the set from 26-100 (75 elements) seems to satisfy the product-free condition, but according to standard results, the maximum product-free set should be around 50 elements. So perhaps there's a mistake in my reasoning.Wait, let me check again. If we take numbers from 26 to 100, is there any pair of numbers in that range whose product is also in the range? For example, take 26 and 26. 26*26=676, which is not in S. 26*27=702, not in S. 50*2=100, but 2 is not in X. Wait, no. All products of two numbers in 26-100 will be at least 26*26=676, which is way above 100, so none of these products are in S. Therefore, the set 26-100 is indeed product-free, since the product of any two elements is not in S at all. Therefore, such a set is product-free. Therefore, the size is 75. But why does this contradict the standard result?Wait, perhaps the standard result is for the multiplicative semigroup, where the operation is multiplication modulo n or something else? Wait, no. Wait, actually, in the set {1,2,...,n}, a product-free set is a set where no two elements multiply to give a third. The upper half (n/2 to n) is product-free because their products exceed n. So for example, in {1,2,...,100}, the numbers from 51 to 100, any two numbers in that range multiply to more than 100, so their product is not in the set. Therefore, that's a product-free set of size 50. However, if we take a larger range, like 26 to 100, the products of two numbers in that range (26*26=676) are still outside of S, so even larger. Therefore, why is 26-100 also product-free? Because the products of any two elements are outside S. Therefore, actually, any subset of numbers where all elements are greater than sqrt(n) will be product-free, since a*b > sqrt(n)*sqrt(n) = n. Therefore, in our case, sqrt(100)=10, so numbers from 11 to 100. Wait, 11*11=121 > 100. Therefore, numbers from 11 to 100, size 90, would form a product-free set. But wait, 11*9=99, which is in S. But 9 is not in X (if X is 11-100). So 11 and 9 would not both be in X, so 99 can be in X. Wait, but 99 is in X (11-100), and 9 is not. So since 11 is in X and 9 is not, 11*9=99 is allowed, because only one of them is in X. Wait, the problem states that no two distinct elements in X multiply to an element in X. So if 11 and 9 are in X, then 11*9=99 would be a problem if 99 is in X. But if 9 is not in X, then even if 11 and 99 are in X, there's no issue because 11*9=99, but 9 isn't in X. Therefore, 11 and 99 can coexist in X.Wait, so actually, numbers from 11 to 100, size 90, can form a product-free set because any two numbers in that set multiply to a number greater than 100, except when one of the numbers is small enough such that their product is within 100. For example, 11*9=99, but 9 is not in X. 11*10=110, which is over 100. So if X is 11-100, then 11 is in X, but 10 is not, so 11*10=110 not in X. Similarly, 11*11=121 not in X. So perhaps numbers from 11 to 100 form a product-free set. Wait, but 12*8=96. If 8 is not in X, then 12 and 96 can be in X. Since 8 is not in X, there's no two elements in X that multiply to 96. Wait, 96 is in X. The factors of 96 within X would require two numbers in X such that their product is 96. For example, 12*8=96, but 8 is not in X. 16*6=96, 6 not in X. 24*4=96, 4 not in X. So even though 96 can be factored into smaller numbers, those smaller numbers are not in X, so 96 is safe. Therefore, numbers from 11 to 100 can be included, and their pairwise products either require a factor not in X or are outside of S. Therefore, the set 11-100 is product-free and has 90 elements.Wait, this seems even larger than 26-100. But then, why didn't I think of this earlier? Let me verify with an example. Take 12 and 8. 8 is not in X (11-100), so 12*8=96. 96 is in X, but since 8 is not in X, there is no pair in X multiplying to 96. Similarly, 15*7=105, which is over 100. 15 is in X, 7 is not. 15*10=150, over 100. So even numbers like 15, which can multiply with numbers not in X to reach elements in X, but since the other factor is not in X, it's okay. Therefore, numbers from 11-100, size 90, seem to satisfy the product-free condition.But wait, 10 is not in X. 10 is in 1-10. So numbers from 11-100 are 90 elements. But 10 is excluded. So, if I take X as 11-100, that's 90 elements. Then, any two elements in X multiply to a number either not in S or in S but not in X (since the other factor is less than 11). For example, 11*11=121, not in S. 11*12=132, not in S. 12*12=144, not in S. However, 11*9=99. 99 is in X, but 9 is not in X. So 11 and 99 can coexist because 9 is not in X. Similarly, 12*8=96. 96 is in X, but 8 is not in X. So 12 and 96 are okay. Therefore, X from 11-100 is product-free.Wait, but actually, this seems too good. The problem is from the CCA Math Bonanza, so maybe the answer is higher than 50. But according to this reasoning, 11-100 gives 90 elements, which is quite large. However, I need to verify if there are any pairs within X that multiply to give another element in X.Wait, take 25 and 4. 4 is not in X, so 25*4=100. 100 is in X, but 4 is not. Therefore, 25 and 100 can coexist. Similarly, 50 and 2. 2 is not in X, so 50 and 100 can coexist. So even if a number in X can be expressed as a product involving a number not in X, it's okay. The only restriction is that two numbers in X cannot multiply to a third number in X. So as long as for any a, b in X, a*b is not in X, regardless of whether a*b is in S or not. Wait, the problem says "no two distinct elements in X multiply to an element in X". Therefore, even if a*b is in S but not in X, that's fine. The only thing not allowed is a*b being in X. Therefore, if we take X as 11-100, then for any a, b in X, a*b is either not in S (if a*b > 100) or in S but not in X (if a*b is between 11 and 100). Wait, but a*b can be between 11 and 100? For example, 11*2=22, but 2 is not in X, so 11 is in X, but 2 is not. 11*3=33, 3 not in X. Wait, but if both factors are in X, like 11*11=121, which is not in X. 12*12=144, not in X. So actually, for any a, b in X (11-100), a*b is either greater than 100 or, if between 11-100, at least one of the factors is not in X. Wait, but if a and b are both in X (11-100), then a*b is at least 11*11=121, which is greater than 100. Therefore, all products of two elements in X are outside of S. Therefore, X=11-100 is product-free, and its size is 90 elements. Therefore, this must be the maximum possible.But wait, hold on. Let me confirm with specific numbers. Take 20 and 5. 5 is not in X, so 20*5=100. 100 is in X, but since 5 is not in X, 20 and 100 can coexist. 20 is in X, 5 is not. Therefore, 20*5=100 is allowed because 5 is not in X. Wait, but actually, the problem states that "no two distinct elements in X multiply to an element in X". So if 20 and 5 are both in X, then 20*5=100 would be a problem. But since 5 is not in X, even if 20 is in X, 20*5 is not considered because 5 is not in X. Therefore, 20 and 100 can be in X as long as 5 is not. Therefore, the set 11-100 is safe.Therefore, according to this reasoning, the maximum size of X is 90. However, this seems too straightforward, and given that the problem is from a competition, the answer might not be 90. Let me check if there's a mistake here.Wait, actually, if we take X as 11-100, then for any a, b in X, a*b is at least 11*11=121, which is greater than 100, so a*b is not in S. Therefore, the product is outside of S, hence not in X. Therefore, X=11-100 is indeed a product-free set. Therefore, the maximum size is 90. But why does the standard result say that the maximum product-free subset is around n/2? Wait, perhaps the standard result is for different structures, like in groups, where inverses exist, or for abelian groups. In the multiplicative semigroup of integers, perhaps the maximum product-free set can be larger.Wait, let me check some references. For the set {1,2,...,n}, the largest product-free subset is indeed achieved by taking the numbers greater than n/2, which has size ⌈n/2⌉. But in our case, n=100, so that would be 50 numbers (51-100). However, here, we have that numbers from 11-100 (90 numbers) are product-free because their pairwise products exceed 100. Wait, that seems contradictory.Wait, perhaps the discrepancy arises from the definition. In some definitions, a product-free set requires that no three elements a, b, c satisfy a*b=c, regardless of whether a, b, c are distinct or not. But in our problem, the condition is that no two distinct elements multiply to an element in X. So, the standard product-free set definition might include cases where a*b=c even if a=b, but in our problem, duplicates aren't allowed since the set has unique elements, and the multiplication is of two distinct elements.But in either case, in the standard literature, the largest product-free subsets of {1,2,...,n} are often considered to be the interval [⌊n/2⌋+1, n], which has size ⌈n/2⌉. For example, in the paper "Product-Free Sets in Groups" by Gowers, or other references. However, in our case, we are observing that a much larger interval [11,100] is product-free because the product of any two elements is outside of the set S. So why is that?Wait, perhaps the key is that in the standard problem, they consider the operation to be closed, i.e., within the set, but here, we are allowing multiplication to go outside of the set. Wait, no. The problem states that "no two distinct elements in X multiply to an element in X". So even if the product is in S but not in X, that's fine. Therefore, actually, the maximum product-free subset can be larger. For example, in our case, the numbers from 11 to 100, their pairwise products are either in S but not in X or outside S. However, in reality, for numbers from 11 to 100, any product of two numbers in that range is at least 11*11=121, which is outside of S (since S only goes up to 100). Therefore, actually, all products of two elements in X are outside of S. Therefore, X=11-100 is product-free and has size 90.Therefore, perhaps the answer is 90. But I need to confirm this.Wait, let me take a smaller example to test. Suppose S={1,2,3,4,5}. Then, according to the same logic, the product-free subset would be {3,4,5}, size 3, which is the upper half (⌈5/2⌉=3). However, if we take numbers greater than sqrt(5)≈2.236, so numbers from 3-5, which is the same as upper half. So in this case, both methods give the same result.But if S={1,2,...,10}, then sqrt(10)≈3.16. So numbers from 4-10, which is 7 elements. However, the upper half would be 6-10, size 5. Which is smaller. But according to our previous logic, numbers from 4-10 would be product-free. Let's check. For example, 4*4=16>10, so okay. 4*5=20>10, okay. All products of two numbers in 4-10 are greater than 10, so not in S. Therefore, X=4-10 is product-free, size 7, which is larger than the upper half (5 elements). Therefore, in this case, the maximum product-free subset is larger than the upper half.Therefore, it seems that taking all numbers greater than sqrt(n) gives a larger product-free set. For n=10, sqrt(n)=3.16, so numbers from 4-10, size 7. For n=100, sqrt(n)=10, so numbers from 11-100, size 90.This seems to be a better approach. Therefore, the maximum size is n - floor(sqrt(n)). For n=100, that would be 100 - 10 = 90. Therefore, the answer should be 90.But let me check for another n. For n=16, sqrt(n)=4. So numbers from 5-16, size 12. The upper half is 9-16, size 8. The products of numbers in 5-16: 5*5=25>16, so all products are outside. Therefore, X=5-16 is product-free, size 12. However, the upper half is smaller. Therefore, the general formula would be floor(n) - floor(sqrt(n)).Therefore, applying this to n=100, the maximum product-free subset has size 100 - 10 = 90. Therefore, the answer should be 90.But wait, let me check if there is a way to have a larger set. Suppose we include some numbers below 11. For example, let's include 10. If we include 10 in X, we need to ensure that 10*a is not in X for any a in X. Since X is 11-100 including 10. So 10*11=110, which is not in X. 10*12=120, not in X. All products of 10 with numbers in X are over 100, so not in X. Therefore, adding 10 is safe. Similarly, adding 9: 9*11=99. 99 is in X. But 9 is now in X, so 9*11=99, which is in X. Therefore, if we add 9 to X, then we have 9 and 11 in X, and their product 99 is also in X. This violates the condition. Therefore, we cannot add 9.Similarly, adding 10 is okay. Let's check:If X = {10} ∪ {11,...,100}, size 91. Now, check if any two distinct elements multiply to another element in X. 10*11=110, not in X. 10*12=120, not in X. ..., 10*100=1000, not in X. 11*11=121, not in X. All products are outside of X. Therefore, X = {10,11,...,100} is product-free, size 91.Similarly, can we add 9? If we add 9, then 9*11=99. 99 is in X, so 9 and 11 cannot coexist. Therefore, adding 9 would require removing 99 from X. So net gain zero. Similarly, adding 8: 8*12=96. If 96 is in X, we have a conflict. So we'd have to remove 96. So adding 8 and removing 96: size remains 91. But then, maybe add multiple numbers and remove multiple numbers. For example, add 10,9,8,..., but for each added number, we have to remove some numbers from X.But perhaps it's not worth it. For instance, if we add 10, we don't need to remove anything. If we add 9, we have to remove 99. If we add 8, remove 96. Add 7, remove 77 (7*11=77). Wait, 77 is in X. So adding 7 would require removing 77. Similarly, add 6, remove 66 (6*11=66). Add 5, remove 55 (5*11=55). Add 4, remove 44. Add 3, remove 33. Add 2, remove 22. Add 1, can't do anything. So if we add numbers from 2 to 10, each addition requires removing one number from X. So adding 10 gives +1 (no removal). Adding 9 gives +1 but -1 (remove 99). So net 0. Similarly, adding 8: +1 -1. Etc. So total, adding 10 gives +1, and adding 2-9 gives 8 additions and 8 removals, net +0. So total size would be 91 (original 90 +10) +8 -8=91.But is this correct? Let's see. Original X is 11-100 (90 elements). Add 10: now 91 elements. Now, to add 9, we have to remove 99. So X becomes {10,9,11,...,98,100}, size 91. Then add 8: remove 96. X becomes {10,9,8,11,...,95,97,98,100}, size 91. Continuing this way, we can add 2-10 but remove one element each time. So adding 9 elements (2-10) and removing 9 elements (22,33,...,99), so net 0. But adding 10 didn't require removal. So overall, adding 10 and then adding 2-9 with removals would result in 91 elements. But is this allowed?Wait, let's check with specific numbers. If we add 10,9,8,...,2 and remove 99,96, etc. Let's say we have X = {2,3,4,5,6,7,8,9,10, 11-100 except 22,33,44,55,66,77,88,99}. So size is 90 (original) +9 (numbers 2-10) -8 (removed numbers 22,33,...,99). Wait, 22 is removed because 2*11=22. 33 is 3*11=33, etc. So for each added number k (2-10), we remove k*11 from X. Therefore, total added elements: 9 (2-10). Total removed elements: 8 (k*11 for k=2-10, but 10*11=110 is not in X). Wait, 10*11=110 is not in X, so when adding 10, we don't need to remove anything. So when adding 2-10:- Add 2: remove 22- Add 3: remove 33- ...- Add 9: remove 99- Add 10: no removalSo total added elements: 9 (2-10)Total removed elements:8 (22,33,...,99)Thus, new size: 90 +9 -8=91Therefore, the new X would have 91 elements. But we need to verify that this set is indeed product-free.Let's check if there are any two distinct elements in X whose product is also in X. Consider two cases:1. Both elements are in the original X (11-100 except 22,33,...,99). Their product is either over 100 (safe) or in 11-100. If it's in 11-100, it must be formed by multiplying two numbers in 11-100. However, the product of two numbers ≥11 is at least 121, which is over 100. So their products are safe.2. One element is in the added set {2-10} and the other is in the original modified X.For example, take 2 and 11. 2*11=22. But 22 was removed from X, so 22 is not in X. Similarly, 3 and 11=33, which was removed. Continue this pattern: for any k in {2-10} and m in X (modified), k*m is either:- If m=11: k*11 was removed from X, so k*11 not in X.- If m>11: then k*m ≥2*12=24. But since m is in the modified X (11-100 except 22,33,...,99), m is not equal to any k*11. So k*m could be in X only if k*m is in 11-100 and not removed. But since k is 2-10 and m is ≥12 (since 11 is in X but 2*11=22 is removed, etc.), k*m ≥2*12=24. But for k=2 and m=12: 2*12=24. If 24 is in X, then we need to check if 2 and 12 are both in X. 2 is added, 12 is in modified X (since 12≠22,33,...,99). But 2*12=24. Is 24 in X? Yes, because 24 wasn't removed. Wait, but 24=2*12. If both 2 and 12 are in X, then their product is 24, which is in X. This violates the condition. Therefore, this approach has a flaw.Oh no, this is a problem. If we add 2 to X and keep 12 in X, then 2*12=24. Since 24 is in X (assuming we didn't remove it), then this violates the product-free condition. Therefore, our previous reasoning was incorrect. Therefore, simply removing k*11 for k=2-10 is not sufficient. Because when we add smaller numbers like 2, they can multiply with numbers in X (modified) to give products still in X. For example, 2*12=24, which is in X if we didn't remove it. But we didn't remove 24. We only removed 22,33,...,99. So 24 remains in X. Therefore, 2 and 12 are in X, and their product 24 is also in X, which is not allowed. Therefore, this construction doesn't work.Therefore, adding smaller numbers requires more careful removal of elements. For example, adding 2 requires not only removing 22 but also any element in X that can be formed by multiplying 2 with another element in X. But since X contains numbers from 11-100 (excluding some), multiplying 2 by those numbers can result in many elements in X. For example, 2*12=24, 2*13=26, ..., 2*50=100. All of these products are in X (24,26,...,100). Therefore, adding 2 would require removing all even numbers from 24 to 100. Which is a lot (about 39 numbers). So, the trade-off of adding 2 and removing 39 elements is not beneficial. Similarly, adding 3 would require removing 33,36,...,99. Which is multiple elements. Therefore, the strategy of adding smaller numbers and removing some elements does not lead to a larger set.Therefore, the initial approach of taking X=11-100 (90 elements) is product-free, and adding any smaller number would require removing multiple elements from X, resulting in a net loss or minimal gain. Therefore, the maximum size is likely 90.But wait, let's check the problem statement again. It says "no two distinct elements in X multiply to an element in X". So if we have X=11-100, then any two distinct elements in X, their product is not in X because it's either over 100 or in 11-100. Wait, if two numbers in X multiply to a number in X, but we just established that for X=11-100, the product of any two elements is at least 11*11=121>100, which is outside of S. Therefore, their product is not in X. Therefore, X=11-100 is indeed product-free. Therefore, the answer should be 90.But I need to verify this with an example. Let's take two numbers in X=11-100, say 20 and 30. 20*30=600, which is not in X. 15 and 20: 15*20=300, not in X. 11 and 11: 121, not in X. 50 and 2: 100, but 2 is not in X. Wait, 50 is in X, 2 is not. 50*2=100, which is in X. But since 2 is not in X, it's allowed. The condition is that two distinct elements in X should not multiply to an element in X. Since 2 is not in X, 50 and 2 are not both in X. Therefore, 50 and 100 can coexist because there's no element in X that can be paired with 50 to get 100. 100 is in X, but the only possible pairs are (2,50), (4,25), (5,20), (10,10). None of 2,4,5,10,20,25 are in X, so 100 is safe.Similarly, 25 is not in X (since X is 11-100). Wait, 25 is in X=11-100. Wait, 25 is between 11 and 100. So 25 is in X. Then, 4 and 25: 4 is not in X, 25 is in X. 4*25=100. 100 is in X. But since 4 is not in X, it's okay. 25 and 100 can coexist.Another example: 20 and 5. 5 is not in X, 20 is in X. 20*5=100, which is in X. But since 5 is not in X, this is allowed. So 20 and 100 can be in X.Therefore, all potential products that land in X involve at least one factor not in X, so they don't violate the condition. Therefore, X=11-100 is indeed product-free with 90 elements. Therefore, this must be the maximum possible.But to be thorough, let's check if we can add any more elements to X=11-100. The only numbers not in X are 1-10. Adding 1 would prevent adding any other numbers, as discussed earlier. Adding 2-10 would each require removing multiple elements from X, as their products with numbers in X would land in X. For example, adding 2 would require removing all even numbers from 24 to 100 (since 2*12=24, 2*13=26, ..., 2*50=100). That's 50-12+1=39 numbers. So adding 2 and removing 39 numbers would result in a net loss of 38. Similarly, adding 3 would require removing 3*11=33, 3*12=36, ..., up to 3*33=99. That's (33-11)/3 +1=8 numbers. So adding 3 and removing 8 numbers gives a net loss of 7. Therefore, it's not beneficial.Therefore, the maximum size of X is indeed 90.However, according to the problem's source (2022 CCA Math Bonanza Individual Round #3), the answer might not be 90. Maybe I made a mistake. Let me check another angle.Wait, the problem says "no two distinct elements in X multiply to an element in X". So, if X contains a number which is a square, say 25, then 25*25=625, which is not in X, so that's okay. But if X contains a number like 16, and 4, then 4*4=16 would be a problem. But 4 is not in X (since X is 11-100). So 16 can be in X even if 4 is not. Therefore, squares are okay as long as their square roots are not in X.But wait, in X=11-100, the square roots of numbers in X would be from sqrt(11)≈3.316 to 10. So the square roots are not in X (since X starts at 11). Therefore, numbers like 16,25, etc., can be in X without issues.Another case: prime numbers. For example, 11 is prime. If 11 is in X, and any multiple like 22,33,..., but 11*k for k≥2 would be 22,33,..., which are in X only if k is in X. But k=2,3,... are not in X. Therefore, 11 is safe. Similarly, 13 is safe, etc.Therefore, I think my reasoning is correct. The maximum size of X is 90. But given that this is a competition problem, and the answer is expected to be an integer in a box, I need to confirm if 90 is indeed the correct answer.Wait, let me check with n=10. If we take X=4-10 (7 elements), is this product-free? For example, 4*5=20 not in X. 5*5=25 not in X. All products are outside X. So yes, it's product-free. And the size is 7, which is larger than the upper half (5-10, size 6). Therefore, in that case, the maximum product-free set is larger. So the general formula is numbers greater than sqrt(n). For n=100, sqrt(n)=10, so numbers 11-100. Therefore, the answer is 90.Therefore, I believe the maximum number of elements is 90.Final Answerboxed{90}</think>Given the set ( S = {1, 2, ldots, 100} ), we need to find the maximum size of a subset ( X ) such that no two distinct elements in ( X ) multiply to give another element in ( X ).1. Understanding the Problem: - The subset ( X ) must be product-free, meaning no two distinct elements in ( X ) can multiply to another element in ( X ). - The product of two numbers in ( X ) should not be in ( X ).2. Key Observations: - Including 1 in ( X ) would restrict ( X ) to only {1} because 1 multiplied by any other element is that element itself. - Numbers greater than ( sqrt{100} = 10 ) when multiplied by any number in ( X ) will result in a product greater than 100, which is outside ( S ).3. Constructing the Subset: - Consider the subset ( X ) containing all numbers from 11 to 100. This range is chosen because the product of any two numbers in this range will be at least ( 11 times 11 = 121 ), which is outside ( S ). - Numbers from 11 to 100 are 90 elements in total.4. Verification: - For any two distinct elements ( a ) and ( b ) in ( X ), the product ( a times b ) is either greater than 100 (and thus not in ( S )) or involves a factor not in ( X ) (since factors less than or equal to 10 are excluded). - All products of two elements in ( X ) are outside the set ( S ), ensuring the product-free condition.5. Conclusion: - The subset ( X ) containing numbers from 11 to 100 satisfies the product-free condition and has the maximum size of 90 elements.Thus, the maximum number of elements in ( X ) is (boxed{90}).,<think>Alright, so the problem is about finding the maximum size of a subset X of the set S = {1, 2, ..., 100} such that no two distinct elements in X multiply to give another element in X. Hmm, okay, interesting. Let me try to break this down.First, let me restate the problem in my own words to make sure I understand. We have the numbers from 1 to 100, and we need to pick as large a subset X as possible. The catch is that if I pick two different numbers from X, their product shouldn't be in X. So, for example, if I include 2 and 3 in X, then 6 shouldn't be in X. Similarly, if I include 4 and 5, then 20 can't be in X. Got it.The goal is to maximize the number of elements in X while avoiding such triples where two elements multiply to a third. This seems similar to some kind of graph problem where edges represent forbidden triples, but maybe I can approach it more combinatorially.Let me think about possible strategies. One approach might be to categorize numbers in S based on their size, maybe considering large numbers versus small numbers. Since multiplying two larger numbers would result in a product that's way beyond 100, right? For example, 50*50 is 2500, which is way over 100. So maybe if we take numbers above a certain threshold, their products won't be in S, so they can coexist without violating the condition.Wait, that's a good point. Let's consider numbers greater than 50. Any two numbers greater than 50, say 51 and 52, their product is 51*52 = 2652, which is way larger than 100. So if we take all numbers from 51 to 100, we can include them all in X because their products are outside of S, so there's no conflict. That gives us 50 numbers right there.But then we can maybe add some smaller numbers to X as well, provided that they don't multiply with each other or with the larger numbers to give another element in X. Hmm, so we need to check if adding a smaller number would create a product that's already in X.Wait, but X currently includes numbers from 51 to 100. If I want to add a smaller number, say k, to X, I need to ensure that for any number m in X (which could be from 51 to 100), k*m is not in X. Since X includes numbers up to 100, k*m must be greater than 100 or not in X. But if k is, say, 2, then 2*51 = 102, which is outside S, so that's okay. Similarly, 2*50 would be 100, but 50 isn't in X (since X starts at 51). Wait, but 50 isn't in X, so 2*50=100 isn't in X either. Wait, 100 is in X. Wait, if we include 100 in X, then 2*50 = 100. But 50 isn't in X, so as long as neither 2 nor 50 is in X, then 100 can be in X. But in our current setup, X includes 51 to 100, so 100 is already in X. So if we add 2 to X, we need to check if 2* any element in X (51-100) gives something in X. But 2*51=102, which is outside. Similarly, 2*52=104, etc. So all those products are outside S, so that's safe. However, we also have to check if adding 2 would create a situation where two elements in X multiply to get another element in X. Since X already has 51-100, and we are adding 2, we need to check if 2 multiplied by any other element in X (other than itself, since the problem says "two distinct elements") would land in X. Since 2*51=102, which is outside X, that's okay. Also, 2*2=4, but since we're only adding 2, we need to check if 2 is multiplied by any other element. But since 2 is the only new element being added, and all existing elements are 51-100, their products with 2 are all over 100. So adding 2 seems safe.Wait, but if we add 2, then 2 itself is in X. So if we also add another number like 3, we have to check if 2*3=6 is in X. If 6 is not in X, then it's okay. But since X currently is 51-100 plus 2 and 3, then 6 isn't in X, so that's okay. Wait, but 6 is not in X, so 2 and 3 can coexist. But then, if we add 2, 3, 4, etc., we have to check that none of their pairwise products are in X. For example, 2 and 4: 2*4=8. If 8 is in X, that's a problem, but 8 is not in X (since X is 51-100 plus some small numbers). So as long as we don't add numbers whose pairwise products are in X.Wait, but X is 51-100 plus some small numbers. So if we add small numbers, we need to make sure that no two of them multiply to give a number in X. But X includes numbers up to 100. So, for example, if we add 10 and 11, their product is 110, which is outside S, so that's okay. But if we add 10 and 10, but the problem says "two distinct elements", so duplicates aren't considered. Wait, but the set X has distinct elements, so we can't have two 10s. So as long as any two distinct elements in X, when multiplied, don't give another element in X. So if we add small numbers like 2, 3, 4, etc., we need to ensure that none of their products are in X. Since X includes 51-100, the products of small numbers need to not be in 51-100. For example, 2*25=50. But 50 isn't in X (since X is 51-100), so that's okay. Wait, but if we add 2 and 25 to X, then 2*25=50. But 50 isn't in X, so that's okay. Wait, but if 50 were in X, then having 2 and 25 in X would be a problem. But since 50 is not in X (because X starts at 51), that's okay. So actually, maybe we can add some small numbers as long as their products are either less than 51 or greater than 100. Since X only includes numbers from 51 to 100 and some small numbers. So if two small numbers multiply to something less than 51, that's okay because those products wouldn't be in X (since X's small numbers are only those we added). Wait, but if we add 5 and 10, their product is 50, which isn't in X, so that's okay. But if we add 7 and 8, their product is 56. If 56 is in X (which it is, since X is 51-100), then that's a problem. Wait, hold on, if we add both 7 and 8 to X, and 56 is already in X (as part of 51-100), then 7*8=56 is in X, which violates the condition. Therefore, adding both 7 and 8 would be bad because their product is in X.Therefore, the problem is that if we add two small numbers whose product is in X (i.e., between 51 and 100), then we can't have both in X. So if we want to add small numbers, we need to ensure that their pairwise products are either less than 51 or greater than 100. But since the products of small numbers can't exceed 100 unless the numbers are large enough. Wait, for example, 10*10=100, which is in X. So even adding 10 and 10 is a problem, but since they are the same element, the problem specifies "two distinct elements", so maybe 10*10 isn't considered. Wait, the problem says "no two distinct elements in X multiply to an element in X". So if X contains two 10s (but since it's a set, it can't), so actually, sets don't have duplicates. So each element is unique. So multiplying two distinct elements. So 10 can only be multiplied by other numbers. So if we have 10 in X, and another number k in X, then 10*k must not be in X. So if we add 10 to X, we need to ensure that 10*k is not in X for any other k in X. But X includes 51-100. So 10*51=510, which is outside S, so that's fine. However, 10*10=100. But since 10 is in X and 100 is in X, but 10*10=100, but they are not distinct. Wait, the problem says "two distinct elements", so 10 and 10 are not distinct, so that's okay. So 10 can coexist with 100. Wait, but 10*10=100, but since they are the same element, it's allowed. Therefore, if we add 10 to X, the only potential issue is 10 multiplied by other elements in X. Since 10*51=510, etc., which are outside, so that's okay. So maybe 10 can be added safely.Wait, but if we add 10, then 10 and any other number in X (51-100) multiply to numbers outside S, so that's okay. But if we add multiple small numbers, we need to check if their products are in X. For example, if we add 2 and 3, then 2*3=6, which is not in X (since X is 51-100 plus 2 and 3). So that's okay. But if we add 2 and 25, their product is 50, which is not in X, so that's okay. But if we add 5 and 10, their product is 50, which is not in X. Wait, but if we add 7 and 8, their product is 56, which is in X. So that's a problem. So 7 and 8 can't both be in X if 56 is in X. Since 56 is in X (as part of 51-100), we can't have both 7 and 8 in X. Therefore, if we want to add small numbers, we have to avoid adding pairs whose product is in X (51-100).Therefore, the problem reduces to: starting with X as 51-100 (50 elements), how many small numbers (1-50) can we add to X such that no two of them multiply to a number in 51-100. Additionally, each small number individually, when multiplied by any number in 51-100, gives a product outside of S (which it does, since 51*2=102 > 100). So the only constraint is that among the small numbers we add, no two of them multiply to give a number in 51-100.Therefore, the problem now is to find the maximum subset Y of {1, 2, ..., 50} such that no two distinct elements in Y multiply to a number in {51, 52, ..., 100}. Then, the total size of X would be 50 (from 51-100) plus the size of Y.So, the task is to find the maximum size of Y ⊂ {1, 2, ..., 50} where for any a, b ∈ Y with a ≠ b, a*b ∉ {51, 52, ..., 100}.Therefore, our problem is now to maximize |Y| where Y is a subset of {1, ..., 50} with the mentioned condition.This seems like a classic problem in combinatorics, perhaps similar to a graph where each number is a node, and edges connect numbers whose product is in 51-100. Then Y must be an independent set in this graph. So the problem reduces to finding the maximum independent set in this graph. But maximum independent set is NP-hard in general, but maybe for this specific graph, we can find a pattern or structure.Alternatively, maybe we can partition the numbers 1-50 in such a way that we can include as many as possible without having two numbers whose product is in 51-100.Let me think. For two numbers a and b in 1-50, their product is between 1*2=2 and 50*49=2450. But we are concerned with products in 51-100. So we need to avoid pairs (a, b) where 51 ≤ a*b ≤ 100.Therefore, for a given a in 1-50, which b's would cause a*b to be in 51-100? For each a, the forbidden b's would be those in [ceil(51/a), floor(100/a)]. So for each a, the range of b's that are forbidden is between ceil(51/a) and floor(100/a), inclusive. Therefore, to construct Y, we need to choose numbers such that for any a in Y, none of the numbers in [ceil(51/a), floor(100/a)] are in Y.This seems a bit complex, but maybe we can approach it by considering the ranges for each a. Let me see:For a = 1: Then b must be between 51 and 100, but b is in 1-50, so no such b. So 1 can be included without conflict.For a = 2: b must be between ceil(51/2) = 26 and floor(100/2) = 50. So if we include 2 in Y, we cannot include any numbers from 26 to 50.For a = 3: ceil(51/3) = 17, floor(100/3) ≈ 33. So b in 17-33. So including 3 forbids 17-33.For a = 4: ceil(51/4) = 13, floor(100/4) = 25. So b in 13-25.a = 5: ceil(51/5) = 11, floor(100/5) = 20. So b in 11-20.a = 6: ceil(51/6) = 9, floor(100/6) ≈ 16. So b in 9-16.a = 7: ceil(51/7) ≈ 8, floor(100/7) ≈14. So b in 8-14.a = 8: ceil(51/8) ≈7, floor(100/8)=12.5 →12. So b in 7-12.a = 9: ceil(51/9) ≈6, floor(100/9)≈11. So b in 6-11.a = 10: ceil(51/10) =6, floor(100/10)=10. So b in 6-10.a = 11: ceil(51/11)≈5, floor(100/11)≈9. So b in 5-9.a = 12: ceil(51/12)=5, floor(100/12)≈8. So b in 5-8.a = 13: ceil(51/13)≈4, floor(100/13)≈7. So b in 4-7.a = 14: ceil(51/14)≈4, floor(100/14)≈7. So b in 4-7.a = 15: ceil(51/15)=4, floor(100/15)≈6. So b in 4-6.a = 16: ceil(51/16)=4, floor(100/16)=6.25→6. So b in 4-6.a = 17: ceil(51/17)=3, floor(100/17)≈5. So b in 3-5.a = 18: ceil(51/18)=3, floor(100/18)≈5. So b in 3-5....Continuing up to a=50. But this seems tedious. Maybe there's a pattern here.Alternatively, note that for a ≥ 17, ceil(51/a) is 3 or lower. For example, a=17: 51/17=3. So ceil(51/17)=3. For a=25, ceil(51/25)=3, floor(100/25)=4. So a=25 would have b in 3-4.Wait, but as a increases, the lower bound of b decreases, and the upper bound also decreases. So for larger a, the forbidden b's are smaller numbers.Therefore, the conflict regions for a and b are as follows:- For small a (e.g., a=2 to around a=16), the forbidden b's are in the middle to upper range (e.g., 26-50 for a=2, 17-33 for a=3, etc.)- For larger a (e.g., a=17 to 50), the forbidden b's are smaller numbers (e.g., 3-5 for a=17, 3-4 for a=25, etc.)Therefore, perhaps if we partition the set {1, ..., 50} into two parts: the lower half and the upper half. Let's say we take all numbers from 1 to k and avoid numbers from k+1 to 50. Alternatively, maybe the maximum Y can be constructed by taking either the lower numbers or the upper numbers but not both.But the problem is that even lower numbers can have conflicts among themselves. For example, 2 and 26-50 are conflicting for a=2. If we exclude 26-50, then 2 can be included. But 3 conflicts with 17-33. So if we include 3, we have to exclude 17-33, but if we've already excluded 26-50, then 17-33 is partially overlapping. This seems complicated.Alternatively, since the forbidden ranges for different a's overlap, perhaps we need to model this as a graph and find the maximum independent set.But given that this is a math competition problem, there's likely a clever construction or an optimal partition.Wait, another idea: if we take all primes greater than 50/2 = 25. Wait, primes greater than 25 in 1-50 are primes from 26 to 50. But primes in 26-50 are 29, 31, 37, 41, 43, 47. But these primes, when multiplied by any number greater than 1, would be at least 29*2=58, which is within 51-100. So if we include these primes in Y, then multiplying them by any number ≥2 in Y would result in a product in 51-100. Therefore, we can't have two such primes in Y. Wait, but if we include a prime like 29, we can't include any number b such that 29*b is in 51-100. So b must be ≤ floor(100/29)=3.44, so b ≤3. So if we include 29 in Y, we can only include numbers ≤3 in Y. But if we have multiple primes like 29, 31, etc., each would restrict Y to numbers ≤ floor(100/prime). But since these primes are all >25, floor(100/prime) is ≤3 (for prime=29, floor(100/29)=3; for prime=31, floor(100/31)=3, etc.) So if we include any of these primes, we can only include numbers from 1-3 in Y. But if we don't include these primes, maybe we can include more numbers.Alternatively, perhaps focusing on numbers that are too big to have products in 51-100 when multiplied by others. Wait, but numbers in 1-50 can multiply with others to reach 51-100. For instance, 10*5=50, which is not in 51-100, but 10*6=60, which is in 51-100. So if we include 10 and 6, their product is 60, which is in X (since X includes 51-100). Therefore, we can't include both 10 and 6 in Y.This seems quite involved. Maybe another approach is to consider that if we take all numbers from 1 to 50, and remove numbers that are factors of numbers in 51-100. Wait, but numbers in 51-100 can have factors in 1-50. For example, 51=3*17, so 3 and 17 are factors. If we include both 3 and 17 in Y, their product is 51, which is in X. So we can't have both 3 and 17 in Y.Therefore, for each number in X (which includes 51-100), its factors in 1-50 cannot both be in Y. So if we have a number like 51 in X, then 3 and 17 cannot both be in Y. Similarly, 52=4*13, so 4 and 13 can't both be in Y. And so on.Therefore, the problem reduces to selecting a subset Y of {1, ...,50} such that for every number n in 51-100, Y does not contain both factors of n (other than n itself, which is not in Y). This seems similar to a conflict graph where each edge represents a pair of numbers that are factors of some n in 51-100. Then Y must be an independent set in this graph.But constructing such a graph and finding the maximum independent set is non-trivial. However, maybe we can use the concept of matchings here. Each number in 51-100 corresponds to a pair of factors in 1-50, so we can model this as a graph where each edge connects two numbers whose product is in 51-100. Then, the problem is to find the maximum independent set in this graph.Alternatively, maybe we can model this as a bipartite graph. Let me think. If we partition the numbers 1-50 into two sets, say small and large, then perhaps edges go from small to large. But I'm not sure.Alternatively, perhaps we can use the concept that each number in 51-100 is a product of two numbers in 1-50, so each such number corresponds to an edge between its factors. Therefore, the graph is the union of all such edges, and Y must be an independent set. So the maximum size of Y is equal to the maximum independent set in this graph.However, maximum independent set is difficult in general, but perhaps this graph has a particular structure we can exploit.Alternatively, note that numbers in 1-50 can be divided into those that are <= sqrt(100) and those that are > sqrt(100). Since sqrt(100)=10, so numbers 1-10 and 11-50.Wait, if we take numbers greater than 10, say 11-50, then multiplying two numbers in 11-50 would give at least 11*11=121, which is greater than 100. Therefore, any two numbers in 11-50 can be included in Y because their product is outside 51-100. Therefore, if we take all numbers from 11-50, that's 40 numbers, and none of them multiply to give a product in 51-100. Because 11*11=121 >100. So that's safe.Additionally, we can consider adding numbers from 1-10, but we have to check that adding them doesn't create a conflict with the existing numbers in Y (11-50) or with other numbers in 1-10.First, numbers from 11-50 are safe among themselves. Now, adding a number k from 1-10 to Y. For k in 1-10, we need to check that k multiplied by any number in Y (11-50) is not in X (51-100). But 11*k is 11-110. However, since k is at most 10, 11*10=110, which is greater than 100. So 11*k is either in 11-110. But 51-100 is within that range. For example, 11*5=55, which is in X. Therefore, if we include 5 in Y, and 11 is also in Y, their product is 55, which is in X. Therefore, that's a conflict. Wait, but 11 is in Y (from 11-50), so if we add 5 to Y, we have a problem because 5*11=55 is in X. Therefore, we can't add 5 to Y if 11 is in Y. Therefore, the numbers from 1-10 can't be added freely.Therefore, to add a number k from 1-10 to Y, we must ensure that k multiplied by any number in 11-50 is not in X (51-100). However, k*11 could be in 51-100. For example, k=5: 5*11=55, which is in X. Therefore, 5 can't be added to Y if 11 is in Y. Similarly, 6*11=66, which is in X, so 6 can't be added if 11 is in Y. Wait, but 11 is in Y (since Y is 11-50). Therefore, any k from 1-10 where k*11 ≥51 would be problematic. Let's compute for each k in 1-10, the minimum m in 11-50 such that k*m ≥51.For k=1: 1*11=11 <51. So 1*11=11, which is in Y, but 11 is already in Y. Wait, but 1*11=11. The product is 11, which is in Y. Wait, the problem states that no two distinct elements in X multiply to an element in X. So if 1 and 11 are in Y, then 1*11=11 is in X. But 11 is in Y, which is part of X. Wait, but the problem says "no two distinct elements in X multiply to an element in X". So 1 and 11 are distinct elements in X, their product is 11, which is in X. Therefore, this is not allowed. Therefore, even 1 cannot be added to Y if 11 is in Y, because 1*11=11 is in X. Similarly, 2*11=22, which is in Y (since Y is 11-50). So 2*11=22 is in X, so 2 and 11 cannot both be in Y. Similarly, 3*11=33 is in X, so 3 and 11 can't both be in Y. Wait, but Y is 11-50. So adding any k from 1-10 would result in k*11 being in X (since 11 is in Y, and k*11 is between 11 and 110, but 11*1=11, which is in Y). So actually, adding any number from 1-10 would conflict with 11 in Y. For example, 1*11=11, which is in Y, so 1 and 11 can't both be in Y. Similarly, 2*11=22, which is in Y, so 2 and 11 can't both be in Y. Wait, but 11 is in Y, so any k in 1-10 multiplied by 11 would be in Y (if k*11 is in 11-50). For k=1, 1*11=11. For k=2, 2*11=22. Up to k=4, 4*11=44. For k=5, 5*11=55, which is in X (the original X, which includes 51-100). So for k=5, 5*11=55, which is in X. Therefore, even if we don't consider the product being in Y, but in X (which includes 51-100). So actually, the problem is that if we include k in Y and m in Y (where m is from 11-50), then k*m must not be in X. But X includes 51-100. So k*m is in X (i.e., 51-100) only if k*m ≥51 and ≤100.Therefore, for each k in 1-10, and m in 11-50, we need to check if k*m is in 51-100. If yes, then k and m cannot both be in Y. Since m is in Y (11-50), then k cannot be in Y if k*m is in X (51-100).Therefore, for k in 1-10:- For k=1: 1*m = m. Since m is in 11-50, 1*m is in 11-50, which is in Y. But the problem states that two distinct elements in X multiply to an element in X. So 1 and m are in X, their product is m, which is in X. Therefore, 1 and m cannot both be in X. Therefore, if we include 1 in Y, we have to remove all m in Y where 1*m is in X. But 1*m = m, which is in X. Therefore, adding 1 to Y would require removing all elements from Y. But Y is 11-50, so we can't have 1 and any other element. Therefore, 1 cannot be added.Similarly, for k=2: 2*m. For m in 11-50, 2*m ranges from 22 to 100. So when is 2*m in 51-100? When m ≥26 (since 2*26=52). Therefore, if we include 2 in Y, we cannot include any m from 26 to 50 in Y. But Y currently is 11-50. So adding 2 would require removing 26-50 from Y. Then Y would be 11-25 plus 2. The size would be 15 (11-25) +1=16, which is much smaller than the original 40. So not beneficial.For k=3: 3*m in 51-100 when m ≥17 (3*17=51). So adding 3 would require removing 17-50 from Y. Y would be 11-16 plus 3. Size=6+1=7. Not beneficial.For k=4: 4*m in 51-100 when m ≥13 (4*13=52). So adding 4 requires removing 13-50. Y becomes 11-12 plus 4. Size=2+1=3.k=5: 5*m in 51-100 when m ≥11 (5*11=55). So adding 5 requires removing 11-20 (since 5*20=100). Wait, 5*m ≥51 ⇒ m ≥51/5=10.2 ⇒ m≥11. And 5*m ≤100 ⇒ m ≤20. So 11 ≤m ≤20. Therefore, adding 5 would require removing 11-20 from Y. Y would be 21-50 plus 5. Original size 30 (21-50) +1=31. Original Y was 40, so 31 is less. Not beneficial.k=6: 6*m in 51-100. 51/6≈8.5 ⇒ m≥9. But m is in 11-50. 6*9=54, but m starts at 11. So 6*11=66. So m=11-16 (6*16=96). Therefore, adding 6 requires removing 11-16 from Y. Y becomes 17-50 plus 6. Size=34 +1=35. Less than 40.k=7: 7*m in 51-100. m ≥8 (7*8=56). But m starts at 11. 7*14=98. So m=8-14. But m ≥11. So m=11-14. Adding 7 requires removing 11-14. Y becomes 15-50 plus 7. Size=36 +1=37.k=8: 8*m in 51-100. m≥7 (8*7=56). m starts at 11. 8*12=96. So m=7-12. But m ≥11. So m=11-12. Adding 8 requires removing 11-12. Y becomes 13-50 plus 8. Size=38 +1=39.k=9: 9*m in 51-100. m≥6 (9*6=54). m starts at 11. 9*11=99. So m=6-11. But m ≥11. So m=11. Adding 9 requires removing 11. Y becomes 12-50 plus 9. Size=39 +1=40. Wait, original Y was 40 (11-50). If we remove 11 and add 9, the size remains 40. But does this help? Wait, because if we do this, we can have 9 in Y instead of 11. But can we do this for multiple k's? For example, replacing multiple numbers in 11-50 with numbers from 1-10. But each replacement requires removing certain numbers. However, replacing 11 with 9 might allow us to add 9, but we have to check if 9 conflicts with other numbers in Y. For example, 9 and 12: 9*12=108, which is outside, so okay. But 9*13=117, outside. So 9 can coexist with 12-50. However, when we remove 11 and add 9, we have to ensure that 9 doesn't conflict with other elements in Y. Since 9* anything in 12-50 is 108 or more, which is outside, so it's safe. Therefore, swapping 11 for 9 keeps the size the same but allows us to include 9. However, this doesn't increase the size. Similarly, we might be able to swap multiple numbers. For example, swap 11 for 9, 12 for 8, etc. Let's see:If we remove 11 and add 9: Y becomes {9, 12-50} (size 40).Remove 12 and add 8: Y becomes {8,9,13-50} (size 40).Remove 13 and add 7: Y becomes {7,8,9,14-50} (size 40).Continue this way: remove 14 and add 6: {6,7,8,9,15-50} (size 40).Remove 15 and add 5: but 5*15=75 which is in X. But we removed 15, so adding 5 is okay as long as 5 doesn't conflict with others. However, 5*21=105, which is outside. So 5 can coexist with 21-50. But 5*11=55, but 11 is already removed. So adding 5 is okay. So Y becomes {5,6,7,8,9,16-50} (size 40).Similarly, remove 16 and add 4: Y becomes {4,5,6,7,8,9,17-50} (size 40).Remove 17 and add 3: Y becomes {3,4,5,6,7,8,9,18-50} (size 40). But 3*18=54, which is in X. So if 3 and 18 are both in Y, their product is 54, which is in X. Therefore, we can't have both 3 and 18 in Y. But we added 3 and kept 18-50. Therefore, this is invalid. So this approach doesn't work because adding 3 creates a conflict with 18-50. Specifically, 3*18=54 in X, so 3 and 18 can't both be in Y. Therefore, replacing 17 with 3 is problematic because 3 and 18 are now both in Y, leading to a conflict. Therefore, this replacement is invalid.Therefore, swapping higher numbers with lower ones may introduce new conflicts. Therefore, this approach isn't straightforward.Alternatively, instead of trying to swap, maybe focus on numbers in 1-10 that can be added without conflicting with 11-50.From earlier analysis, to add a number k in 1-10, we must ensure that k*11 is not in 51-100. Wait, but k*11 is in 51-100 for k=5 (55), 6 (66), etc. Wait, for k=1: 1*11=11 (in Y, so conflict). For k=2: 2*11=22 (in Y, conflict). Similarly up to k=9: 9*11=99 (in X, conflict). Only k=10: 10*11=110 (outside, no conflict). Wait, 10*11=110, which is outside S. Therefore, 10 can be added without conflicting with 11. Wait, let's verify:If we add 10 to Y, then 10*11=110, which is outside X (since X is 51-100 plus Y). Wait, X is the union of Y (subset of 1-50) and 51-100. So 110 is not in X. Therefore, adding 10 to Y is safe in terms of multiplying with 11. But we also need to check if 10 conflicts with other elements in Y. For example, 10 and 12: 10*12=120, which is outside. So no conflict. Similarly, 10*10=100, which is in X. But since we're adding 10 as a single element, the problem only considers distinct elements. So 10*10=100 is allowed because it's the same element. Therefore, 10 can be safely added to Y.Similarly, check other k's:For k=10: 10*11=110 (safe), 10*12=120 (safe), etc. So 10 can be added without conflict. Therefore, adding 10 to Y increases the size by 1. So Y would be 11-50 plus 10, total 41.Wait, but wait. If we add 10 to Y, we need to check if 10 multiplied by any other element in Y (11-50) is in X. For example, 10*6=60, but 6 isn't in Y (since Y is 11-50 plus 10). So 10 is in Y, and 6 is not in Y, so 10*6=60 is in X (since 60 is in 51-100). But 60 is in X, and 10 and 6 are in Y. Wait, but 6 isn't in Y. Therefore, only 10 is in Y, and 6 is not. Therefore, 10*6=60 is in X, but since only 10 is in Y, there's no conflict. The problem states that no two distinct elements in X multiply to an element in X. So if 10 and 6 are both in X, then 10*6=60 (in X) would be a conflict. But if only 10 is in X and 6 is not, then there's no conflict. Therefore, adding 10 is safe.Therefore, adding 10 is safe. So Y could be 11-50 plus 10, making 41 elements. But wait, 10 is in 1-10, and 11-50. Let's confirm:Original Y is 11-50 (40 elements). Adding 10 gives 41 elements. Then X would be 51-100 (50 elements) plus Y (41 elements), total 91. But wait, no. Wait, X is a subset of S =1-100. X is the union of Y (subset of 1-50) and 51-100. Wait, no: the problem says "X is a subset of S such that no two distinct elements in X multiply to an element in X." So initially, we thought to take Y as 51-100 and some small numbers. But the above approach is taking Y as 10 and 11-50, which would make X = {10, 11-100}. But 10 is in X, and 11-100 are in X. Then, for example, 10 and 11 are in X, and their product is 110, which is not in X, so that's okay. 10 and 50: 10*50=500, not in X. 10*10=100, which is in X, but since they are the same element, it's allowed. So X would be {10} ∪ {11-100}, which is 91 elements. Wait, but 11-100 is 90 elements, plus 10 is 91. But is this valid?Wait, let's check if there are any two distinct elements in X whose product is in X. Take two elements from X: they can be both from 11-100, or one from 11-100 and 10.For two elements from 11-100: their product is at least 11*11=121, which is outside X (since X is up to 100). So no conflict there.For one element 10 and another element m from 11-100: their product is 10*m, which is at least 10*11=110, which is outside X. Therefore, no conflict.The only possible conflict is within the added element 10 and itself, but since duplicates aren't allowed and the problem specifies "two distinct elements", 10*10=100 is allowed because they are not distinct. Therefore, X = {10} ∪ {11-100} is a valid set with 91 elements.Wait, but this seems too good. The answer might be 91. But that seems very high. Let me check this carefully.If X = {10, 11, 12, ..., 100}, then:- Any two numbers in 11-100 multiply to at least 11*11=121 >100, so their products are not in X.- Any number in 11-100 multiplied by 10 is at least 10*11=110 >100, so their products are not in X.- The only product involving 10 is 10*10=100, which is in X, but since it's the same element, it's allowed.Therefore, this X satisfies the condition and has 91 elements. Is this possible? The problem asks for the maximum size. If this is valid, then 91 is the answer. But the problem is from a math competition, and usually, such problems have a more involved solution. Maybe I missed something.Wait, let's check if 10 is in X and 100 is in X. 10*10=100, but they are the same element. The problem states "no two distinct elements in X multiply to an element in X." So since 10 and 10 are not distinct, it's allowed. Therefore, X can include both 10 and 100. Therefore, the set {10, 11-100} is valid.But then why wasn't this considered earlier? Because initially, I thought of starting with 51-100, but actually, taking 11-100 plus 10 gives a larger set. Wait, 51-100 is 50 elements, but 11-100 is 90 elements. Adding 10 gives 91. But that seems correct.Wait, but S is {1, 2, ..., 100}. So 11-100 is 90 numbers, plus 10 is 91. But the problem says "X is a subset of S such that no two distinct elements in X multiply to an element in X." So if we take X = {10, 11-100}, then indeed, as checked, there are no two distinct elements whose product is in X. Therefore, this seems valid.But why did the initial analysis suggest starting with 51-100? Because perhaps I misread the problem. Wait, the problem says "X is a subset of S such that no two distinct elements in X multiply to an element in X." Therefore, if we take X as 10, 11-100, this satisfies the condition, as shown. Therefore, the maximum size would be 91.But this seems too straightforward. Maybe I'm missing a conflict.Wait, let's take two numbers in X: say 10 and 20. 10*20=200, which is not in X. 10 and 50: 500, not in X. 20 and 30: 600, not in X. Any two numbers in 11-100 multiply to over 100. 10 and any number in 11-100 multiply to over 100. The only product within X is 10*10=100, which is allowed because they are the same element. Therefore, X = {10, 11-100} is indeed valid and has 91 elements.But the problem is from a math competition, and if the answer is 91, that would be the case. However, perhaps there's a mistake here.Wait, let's check if 10 and 100 are both in X. 10*10=100, which is in X. But since they are the same element, it's allowed. Therefore, no issue. But if there were two distinct elements multiplying to 100, like 20 and 5, but 5 is not in X. So no problem.Therefore, it seems that X = {10, 11-100} works and has 91 elements. But wait, 10 is in X, and 100 is in X. 10*10=100 is allowed. Therefore, 91 is the maximum. But this seems too large compared to the initial thought of 50 elements.But perhaps the correct answer is indeed 91. Let me verify with smaller cases.Suppose S = {1,2,...,10}. Find the maximum X such that no two distinct elements multiply to an element in X.Following the same logic: take numbers from 5 to 10 (6 elements), and add 4. Then check:X = {4,5,6,7,8,9,10}. Check products:4*5=20 not in X.4*6=24 not in X....5*6=30 not in X.All products are outside. So size 7. If we try to add 3, 3*5=15 not in X, 3*4=12 not in X, etc. But 3*3=9 is in X, but same element. So X could be {3,4,5,6,7,8,9,10} size 8. But 3*5=15 not in X. Wait, but in this case, the maximum X is all numbers from 5-10 (6 elements) plus 4 and 3 and 2 and 1? Wait, no, adding 2 would conflict with 5 (2*5=10, which is in X). So can't add 2. Similarly, adding 1 would conflict with all elements (1*5=5 in X). So 1 can't be added. 3 can be added if no element multiplied by 3 is in X. 3*5=15 not in X, 3*4=12 not in X, etc. So X could be {3,4,5,6,7,8,9,10}, size 8. Whereas 5-10 plus 4 gives 7, which is smaller. So the maximum would be 8. However, according to our previous logic, we would take {sqrt(10)=3.16, so numbers above 3.16, i.e., 4-10} plus 3? Not sure. Anyway, the point is that in some cases, adding lower numbers can increase the total.But returning to the original problem, according to this analysis, X = {10, 11-100} is valid and has 91 elements. However, I need to make sure there isn't a hidden conflict I missed.Wait, another example: take 15 and 10 in X. 15*10=150, not in X. Okay. Take 20 and 5: but 5 isn't in X. 20 and 6: 6 isn't in X. So no conflicts.Wait, what about 25 and 4: 25 is in X, but 4 isn't. So no conflict.Therefore, all products of distinct elements in X are either:- Both in 11-100: product is ≥121, not in X.- One is 10 and the other is in 11-100: product is ≥110, not in X.- Both are 10: but they are the same element, allowed.Therefore, X = {10, 11-100} satisfies the condition and has 91 elements. Therefore, the maximum size is 91.But this seems surprisingly large. Let me check the problem statement again: "X is a subset of S such that no two distinct elements in X multiply to an element in X." Yes, that's exactly what we have. So unless there's a restriction I'm missing, this should be correct.But maybe I made a mistake in considering the elements from 11-100. For example, 50 is in X. 50*2=100, but 2 isn't in X. 25*4=100, but 4 isn't in X. So no conflict. Therefore, the only product within X is 10*10=100, which is allowed.Therefore, the answer should be 91. But why did the original thought process start with 51-100? Because of miscalculating the permissible elements. But upon realizing that numbers above 10 have products above 100 when multiplied together, and adding 10 doesn't conflict with them, we can achieve a much larger set.Wait, but numbers from 11-50: multiplying two numbers from 11-50 gives a product ≥121, which is outside S (1-100). Therefore, in X = {10, 11-100}, any two distinct elements from 11-100 multiply to ≥121, not in X. 10 multiplied by any element in 11-100 is ≥110, not in X. The only product in X is 10*10=100, which is allowed. Therefore, this set satisfies all conditions. Therefore, the maximum size is 91.But I'm still in doubt because this seems too straightforward for a competition problem. Let me check if there are any other constraints.Wait, for example, 50 is in X. 50*2=100, but 2 isn't in X. So no conflict. 25*4=100, but neither 25 nor 4 are in X (wait, 25 is in X (11-100), but 4 is not. 25*4=100, but since 4 isn't in X, there's no conflict. Similarly, 10*10=100 is allowed.Therefore, the set X = {10, 11-100} is valid with 91 elements. Therefore, the answer should be 91.But let's see if we can add more numbers from 1-9. For example, can we add 9? If we add 9 to X, then we have to check if 9* anything in X is in X. 9*10=90, which is in X. So 9 and 10 are distinct elements in X, and their product is 90, which is in X. Therefore, this violates the condition. Therefore, we cannot add 9.Similarly, adding 8: 8*10=80, which is in X. Conflict. Similarly, 7*10=70, conflict. 6*10=60, conflict. 5*10=50, which is in X. Conflict. 4*10=40, which is in X. Wait, 40 is in X (since X is 11-100 plus 10). Wait, no: X is {10, 11-100}. So 40 is in X. Therefore, 4*10=40, which is in X. Therefore, adding 4 would conflict with 10. Therefore, can't add 4.Similarly, adding 3: 3*10=30, which is in X. Conflict.Adding 2: 2*10=20, which is in X. Conflict.Adding 1: 1*10=10, which is in X. But 1 and 10 are distinct elements, so their product is 10, which is in X. Therefore, conflict.Therefore, the only number we can add from 1-10 is 10 itself. Adding any other number from 1-9 would create a conflict with 10 or with numbers in 11-50.Therefore, the maximum size is indeed 91.But I'm still surprised because this seems like a large jump from the initial thought of 50. Maybe the answer is 50, but according to this analysis, it's 91. Let me check with another approach.Suppose we partition the numbers into "large" and "small". Large numbers are those for which their product with any other number in the set is outside S. For numbers greater than sqrt(100)=10, their product with another number greater than 10 is greater than 100. Therefore, any number greater than 10 can be included in X without conflicting with others of the same ilk. Then, we can add 10 as well, since 10*11=110 >100. However, 10*10=100 is allowed. Therefore, X can be {10,11-100} with size 91.Yes, this seems to align. Therefore, the maximum size is 91.But let's verify with the problem's source. The problem is from the 2022 CCA Math Bonanza Individual Round #3. If I recall correctly, the answer might be 67, but I'm not sure. Wait, no, I shouldn't rely on memory. Let me think again.Another way: perhaps there is a conflict with numbers like 25 and 4. 25*4=100. But 25 is in X (11-100), and 4 is not in X. Therefore, no conflict. Similarly, 20*5=100, but 5 isn't in X. So no problem.Alternatively, numbers like 50 and 2. 50*2=100. 50 is in X, 2 is not. No conflict. Therefore, the set X = {10, 11-100} is safe.Therefore, I think the correct answer is 91. However, given that this is a competition problem, perhaps there's a catch I'm missing. Let me think again.Wait, the problem says "no two distinct elements in X multiply to an element in X". So if two elements multiply to something in X, they cannot both be in X. Therefore, the forbidden pairs are those (a,b) where a*b is in X. Therefore, in constructing X, we need to ensure that for any a, b in X, a*b ∉ X.In the set X = {10, 11-100}, let's check if there are any such a, b.Take a=10 and b=10: a*b=100 ∈ X, but they are the same element, allowed.Take a=10 and b=11: a*b=110 ∉ X.Take a=11 and b=11: a*b=121 ∉ X.Take a=11 and b=12: a*b=132 ∉ X.In general, any two distinct elements in X will multiply to a number not in X, since:- If both are ≥11, their product is ≥121 >100.- If one is 10 and the other is ≥11, their product is ≥110 >100.- The only case where the product is in X is when multiplying 10 by itself, which is allowed.Therefore, the set X = {10, 11-100} is valid and has 91 elements. Therefore, the answer should be boxed{91}.But I'm still hesitant because this seems too straightforward. Maybe the original problem had a different condition, but according to the problem statement provided, this seems correct.Alternatively, perhaps I misread the problem. Let me check again:"X is a subset of S such that no two distinct elements in X multiply to an element in X."Yes, so two distinct elements can't multiply to any element in X, including elements not in the original pair. So in our construction, there are no such pairs. Therefore, the answer should indeed be 91.However, given that this is a competition problem, it's possible that the answer is different. But based on the analysis, 91 seems correct. Maybe the error is in assuming that we can take all numbers from 11-100. Let me check if there are any pairs within 11-100 that multiply to another element in 11-100.For example, take 12 and 8: 12*8=96. But 8 is not in X (since X is 11-100 plus 10). Wait, in our construction, X is {10, 11-100}. So 12 and 8: 8 is not in X, so no conflict. 12 and 10: 12*10=120 ∉ X. Therefore, no problem.Wait, but if X includes 16 and 6, but 6 isn't in X. So no conflict. Therefore, the only products to consider are those within X. Since all products are ≥121 or involve 10 and ≥11, which are ≥110, there are no products within X.Therefore, the construction holds. Therefore, the maximum number of elements is 91.But I'm still unsure because this answer seems high. Let me check smaller cases:Suppose S = {1,2,3,4,5}. What's the maximum X?If we take X = {4,5}, size 2. Check products: 4*5=20 ∉ X. Valid.Alternatively, X = {3,4,5}. 3*4=12 ∉ X, 3*5=15 ∉ X, 4*5=20 ∉ X. Valid. Size 3.Alternatively, X = {2,4,5}: 2*4=8 ∉ X, 2*5=10 ∉ X, 4*5=20 ∉ X. Valid. Size 3.Alternatively, X = {1,4,5}: 1*4=4 ∈ X. Conflict. Can't do that.Maximum seems to be 3.Using the previous method: numbers greater than sqrt(5)≈2.236. So numbers 3,4,5. Size 3. Correct.Similarly, in S= {1,...,10}, numbers greater than sqrt(10)=3.16, so 4-10 plus 3. Which would be 8 elements. As we saw earlier, possible.Therefore, following this pattern, in S=1-100, take numbers greater than sqrt(100)=10, i.e., 11-100, plus 10. Total 91. This matches our previous result.Therefore, the answer should be 91. Therefore, the maximum number of elements is boxed{91}.Final Answerboxed{91}</think>To solve the problem of finding the maximum number of elements in a subset ( X ) of ( S = {1, 2, ldots, 100} ) such that no two distinct elements in ( X ) multiply to an element in ( X ), we can use the following reasoning:1. Identify Large Numbers: Numbers from 51 to 100 are considered. Any two distinct numbers in this range multiply to a product greater than 100, so they do not conflict with each other. This gives 50 elements.2. Consider Smaller Numbers: We need to check if smaller numbers (1 to 50) can be added to the subset without causing conflicts. A key observation is that numbers greater than 10 (i.e., 11 to 50) when multiplied by any number in 11 to 50 result in a product greater than 100. Therefore, these numbers can coexist with each other.3. Check Adding 10: The number 10 is special. When multiplied by any number from 11 to 50, the product is at least 110, which is outside the set ( S ). Additionally, 10 multiplied by itself (100) is allowed since it is not a product of two distinct elements. Thus, 10 can be safely added.4. Final Subset Construction: Combining the numbers from 11 to 100 (90 elements) with the number 10 gives a subset ( X ) of size 91. This subset satisfies the condition that no two distinct elements multiply to another element in ( X ).Thus, the maximum number of elements in ( X ) is (boxed{91}).]

🔑:To solve this problem, we need to find the maximum number of elements in a subset ( X ) of ( S = {1, 2, ldots, 100} ) such that no two distinct elements in ( X ) multiply to an element in ( X ).1. Understanding the Problem: - We need to ensure that for any ( a, b in X ) (where ( a neq b )), the product ( ab notin X ).2. Constructing a Candidate Set: - Consider the set ( X = {10, 11, 12, ldots, 100} ). This set contains 91 elements.3. Verifying the Condition: - We need to check if for any ( a, b in X ) (where ( a neq b )), the product ( ab notin X ). - For ( a, b in {10, 11, 12, ldots, 100} ), the smallest possible product is ( 10 times 11 = 110 ), which is greater than 100. - Therefore, no product of two distinct elements in ( X ) will be in ( X ).4. Optimality: - To determine if this is the maximum possible size, consider the following: - If we include any number less than 10 in ( X ), say ( k ), then ( k times 10 leq 100 ), which would force ( k times 10 ) to be in ( X ), violating the condition. - Thus, including any number less than 10 would reduce the size of ( X ) while maintaining the condition.5. Conclusion: - The set ( X = {10, 11, 12, ldots, 100} ) is optimal because it maximizes the number of elements while satisfying the given condition.The final answer is ( boxed{91} ).

🧠:[<think>Okay, so I need to find the range of the real number a such that the function f(x) = 4x³ - 3x has a maximum value on the interval (a, a+2). Hmm, let's start by recalling some calculus. To find maxima or minima of a function on an interval, we usually check critical points and endpoints. But the problem here is that the interval is open, (a, a+2), so endpoints aren't included. That means the maximum must occur at a critical point inside the interval. First, let's find the critical points of f(x). Critical points are where the derivative is zero or undefined. Since f(x) is a polynomial, its derivative will exist everywhere, so we just need to find where the derivative is zero.Calculating the derivative: f'(x) = d/dx [4x³ - 3x] = 12x² - 3. Setting this equal to zero to find critical points:12x² - 3 = 0 12x² = 3 x² = 3/12 x² = 1/4 x = ±1/2So the critical points are at x = 1/2 and x = -1/2. These are potential points where the function could have local maxima or minima. Since we're dealing with a maximum on the interval (a, a+2), the interval must contain at least one of these critical points where the function has a local maximum. Now, let's analyze the behavior around these critical points to determine which one is a maximum. Let's take the second derivative to test for concavity:f''(x) = d/dx [12x² - 3] = 24x.At x = 1/2: f''(1/2) = 24*(1/2) = 12 > 0, which means the function is concave up here, so x = 1/2 is a local minimum.At x = -1/2: f''(-1/2) = 24*(-1/2) = -12 < 0, so the function is concave down here, meaning x = -1/2 is a local maximum.Therefore, the function f(x) has a local maximum at x = -1/2. So, in order for the interval (a, a+2) to contain a maximum, it must include x = -1/2. However, since the interval is open, we need to ensure that -1/2 is within (a, a+2). So, the condition is that a < -1/2 < a + 2. Let's write that as inequalities:1. a < -1/22. -1/2 < a + 2Let's solve the second inequality:-1/2 < a + 2 Subtract 2 from both sides: -1/2 - 2 < a -5/2 < aSo combining both inequalities:-5/2 < a < -1/2Therefore, the range of a is (-5/2, -1/2). But wait, let me check again. The interval is (a, a+2), and we need the local maximum at -1/2 to lie within this interval. So indeed, a must be less than -1/2, and a + 2 must be greater than -1/2. But let me verify with an example. Suppose a is -2, which is within the range -5/2 (-2.5) and -1/2 (-0.5). Then the interval would be (-2, 0). The critical point at -1/2 is inside this interval, so the function should have a maximum there. If a is -3, which is less than -5/2, the interval would be (-3, -1). But -1/2 is not in this interval, so there would be no maximum there. Similarly, if a is -1, which is greater than -1/2, the interval would be (-1, 1). Here, -1/2 is in the interval, but wait, a is -1, which is greater than -5/2, so according to our range, that should be okay. Wait, no, if a is -1, which is within (-2.5, -0.5), no, wait: -1 is between -2.5 and -0.5? Wait, -2.5 is -5/2, so -1 is greater than -2.5 but less than -0.5? Wait, no. -1 is greater than -2.5 (since it's to the right on the number line) but less than -0.5. So yes, a = -1 is within the range (-5/2, -1/2). Then the interval is (-1, 1). The critical point at -1/2 is inside (-1, 1), so the maximum is there. However, if a is, say, -0.6, then the interval is (-0.6, 1.4). The point -1/2 is -0.5, which is greater than a = -0.6, so it is inside (-0.6, 1.4). Wait, but according to our earlier conclusion, a must be less than -1/2. Wait, that's a contradiction. Wait, if a is -0.6, which is greater than -1/2 (-0.5), then according to our range, a should be in (-2.5, -0.5). But -0.6 is less than -0.5, right? Wait, no: -0.6 is less than -0.5? No, -0.6 is more negative than -0.5, so -0.6 < -0.5. So, if a is -0.6, which is less than -0.5, then a is in (-2.5, -0.5). Wait, but -0.6 is between -2.5 and -0.5, so that's okay. Then the interval would be (-0.6, 1.4). The critical point at -1/2 is -0.5, which is inside (-0.6, 1.4). So yes, the maximum is there. Wait, but if a is, say, -0.4, which is greater than -0.5, then the interval is (-0.4, 1.6). The critical point at -0.5 is not inside this interval because -0.5 is less than a = -0.4. So in this case, the interval doesn't include the local maximum. Therefore, our original conclusion that a must be between -5/2 and -1/2 is correct. Wait, but hold on: if a is -2.5, then the interval is (-2.5, -0.5). The critical point at -0.5 is the upper endpoint, but the interval is open, so -0.5 is not included. Therefore, a cannot be -2.5, only greater than -2.5. Similarly, if a is exactly -0.5, the interval is (-0.5, 1.5), but -0.5 is not included. So the critical point at -0.5 is not in the interval. Therefore, a must be strictly greater than -5/2 and strictly less than -1/2. Another test case: a = -2. Then the interval is (-2, 0). The critical point at -1/2 is -0.5, which is inside (-2, 0). So the function would attain a maximum there. If a = -3, interval (-3, -1). -0.5 is not in that interval. So no maximum. Similarly, if a is -1, interval (-1, 1), which includes -0.5, so the maximum is there. Therefore, the range of a is all real numbers between -5/2 and -1/2, not including the endpoints. So in boxed form, that would be boxed{(-dfrac{5}{2}, -dfrac{1}{2})}.Wait, just to confirm once more, since the critical point is at -1/2, for this point to lie within (a, a+2), the following must hold:a < -1/2 < a + 2Solving for a:First inequality: a < -1/2Second inequality: -1/2 < a + 2 Subtract 2: -1/2 - 2 < a Which is -5/2 < aTherefore, combining both inequalities: -5/2 < a < -1/2Yes, that's correct. So the answer should be the open interval from -5/2 to -1/2.Final AnswerThe range of the real number ( a ) is boxed{left( -dfrac{5}{2}, -dfrac{1}{2} right)}.</think>To find the range of the real number ( a ) such that the function ( f(x) = 4x^3 - 3x ) has a maximum value on the interval ( (a, a+2) ), we start by identifying the critical points of the function. The critical points are found by setting the derivative ( f'(x) = 12x^2 - 3 ) to zero:[12x^2 - 3 = 0 implies x^2 = frac{1}{4} implies x = pm frac{1}{2}]Next, we determine which of these critical points is a maximum by using the second derivative test. The second derivative ( f''(x) = 24x ) gives:- ( f''left(frac{1}{2}right) = 12 > 0 ), indicating a local minimum at ( x = frac{1}{2} ).- ( f''left(-frac{1}{2}right) = -12 < 0 ), indicating a local maximum at ( x = -frac{1}{2} ).For the function to have a maximum on the open interval ( (a, a+2) ), the critical point ( x = -frac{1}{2} ) must lie within this interval. This requires:[a < -frac{1}{2} < a + 2]Solving these inequalities:1. ( a < -frac{1}{2} )2. ( -frac{1}{2} < a + 2 implies -2 - frac{1}{2} < a implies -frac{5}{2} < a )Combining these inequalities, we get:[-frac{5}{2} < a < -frac{1}{2}]Thus, the range of the real number ( a ) is (boxed{left( -dfrac{5}{2}, -dfrac{1}{2} right)}).,<think>Okay, let's see. The problem is about finding the range of the real number ( a ) such that the function ( f(x) = 4x^3 - 3x ) has a maximum value on the interval ( (a, a+2) ). Hmm, so I need to figure out for which values of ( a ) the function actually attains a maximum on that open interval. First, I remember that for continuous functions on a closed interval, the Extreme Value Theorem guarantees both a maximum and a minimum. However, here the interval is open, ( (a, a+2) ), so the function might not attain its maximum or minimum there. Therefore, the maximum must be attained somewhere inside the interval, not at the endpoints. But since it's an open interval, the endpoints aren't included, so even if the function had maxima there, they wouldn't count. Wait, but the problem is saying that there is a maximum on the interval. So maybe the function reaches a critical point inside the interval? Because critical points are where the function could have local maxima or minima.So, the function ( f(x) = 4x^3 - 3x ). Let's find its critical points first. To do that, take the derivative:( f'(x) = 12x^2 - 3 ).Set the derivative equal to zero to find critical points:( 12x^2 - 3 = 0 )Divide both sides by 3:( 4x^2 - 1 = 0 )So,( x^2 = 1/4 )Therefore, the critical points are at ( x = 1/2 ) and ( x = -1/2 ).Okay, so the function has critical points at ( x = frac{1}{2} ) and ( x = -frac{1}{2} ). These are the points where the function could have local maxima or minima.Now, since we're looking for the maximum value on the interval ( (a, a+2) ), the function must attain a maximum there. Since the interval is open, the maximum can't be at the endpoints. Therefore, the maximum must be at one of the critical points. So, for the function to have a maximum on ( (a, a+2) ), one of the critical points must lie within this interval. That is, either ( frac{1}{2} ) or ( -frac{1}{2} ) must be inside ( (a, a+2) ).But wait, maybe both critical points could be inside the interval? Let me check.So, we need that either ( a < frac{1}{2} < a + 2 ) or ( a < -frac{1}{2} < a + 2 ). So, the interval ( (a, a+2) ) must contain at least one critical point. If it contains both, that's okay, but as long as it contains at least one. Because if there's a critical point inside the interval, then depending on the behavior around that critical point, the function could attain a maximum there. Wait, but we need to make sure that the critical point is a maximum. So, maybe we need to check if the critical point within the interval is a local maximum.Wait, let's check the second derivative to see the concavity at those critical points. The second derivative is ( f''(x) = 24x ). At ( x = 1/2 ), ( f''(1/2) = 24*(1/2) = 12 > 0 ), so that's a local minimum. At ( x = -1/2 ), ( f''(-1/2) = 24*(-1/2) = -12 < 0 ), so that's a local maximum. Ah, so the function has a local maximum at ( x = -1/2 ) and a local minimum at ( x = 1/2 ).Therefore, the only critical point that is a local maximum is at ( x = -1/2 ). The other critical point is a local minimum. Therefore, in order for the function ( f(x) ) to have a maximum on the interval ( (a, a+2) ), the interval must contain the local maximum at ( x = -1/2 ). Because the local minimum at ( x = 1/2 ) would not contribute to a maximum value on the interval. Unless, perhaps, the interval doesn't include the local maximum, but the function still attains a maximum at another point. Wait, but if the interval doesn't include any critical points, then the function would be strictly increasing or decreasing on the interval. Let's think.Suppose the interval ( (a, a+2) ) does not contain either critical point. Then, since the function is a cubic, it's going to have end behaviors going to positive infinity as ( x to infty ) and negative infinity as ( x to -infty ). However, on a finite open interval without critical points, the function would be either increasing or decreasing throughout the interval. Therefore, it wouldn't attain a maximum or minimum on the open interval because it would approach the endpoints but not reach them. Hence, in such a case, the function would not have a maximum on ( (a, a+2) ).Therefore, the key is that the interval must contain the local maximum at ( x = -1/2 ). Because if the interval contains ( x = -1/2 ), then the function will have that local maximum within the interval. However, if the interval is such that the function is increasing throughout the interval, then the supremum would be approaching ( a+2 ), but since it's open, there's no maximum. Similarly, if the function is decreasing throughout the interval, the supremum would approach ( a ), again no maximum. So only when the interval includes the local maximum point ( x = -1/2 ), then the function actually attains a maximum there.Therefore, the condition is that ( -1/2 ) must lie within the interval ( (a, a+2) ). So, we need:( a < -1/2 < a + 2 )Which translates to:Left inequality: ( a < -1/2 )Right inequality: ( -1/2 < a + 2 )Subtract 2 from both sides:( -1/2 - 2 < a )( -5/2 < a )Therefore, combining the two inequalities:( -5/2 < a < -1/2 )So the range of ( a ) is ( (-frac{5}{2}, -frac{1}{2}) ).Wait, but let me check if this is correct. Let me verify with examples.Suppose ( a = -2 ). Then the interval is ( (-2, 0) ). Does this interval contain ( -1/2 )? Yes. So the maximum is at ( -1/2 ), and the function should have a maximum there. Now, if ( a = -3 ), the interval is ( (-3, -1) ). Then ( -1/2 ) is not in the interval. So the function on ( (-3, -1) ). Since ( f(x) = 4x^3 - 3x ), let's compute the derivative on that interval. The derivative is ( 12x^2 - 3 ). If ( x ) is in ( (-3, -1) ), then ( x^2 ) is between 1 and 9, so ( 12x^2 - 3 ) is between ( 12*1 -3 = 9 ) and ( 12*9 -3 = 105 ), which are positive. Therefore, the function is increasing on ( (-3, -1) ). Therefore, on the open interval ( (-3, -1) ), the function is increasing, so it doesn't attain a maximum; the supremum would be approaching ( -1 ), but since the interval is open, there's no maximum. Similarly, if ( a = 0 ), the interval is ( (0, 2) ). The derivative in this interval: ( x ) is positive, so ( 12x^2 - 3 ). At ( x = 0 ), derivative is -3, but for ( x > sqrt{1/4} = 1/2 ), derivative is positive. So in (0,2), the derivative starts negative at 0 and becomes positive after ( x = 1/2 ). Therefore, the function first decreases from 0 to 1/2 and then increases from 1/2 to 2. But the critical point at 1/2 is a local minimum. However, the interval here is (0, 2). So the function decreases from 0 to 1/2, reaching a minimum at 1/2, then increases towards 2. Therefore, on the open interval (0, 2), the maximum would be approached as ( x ) approaches 2 from the left, but since the interval is open, there's no maximum. Similarly, near 0, the function approaches ( f(0) = 0 - 0 = 0 ). Therefore, the function doesn't attain a maximum on (0, 2). Therefore, if the interval does not include -1/2, then the function doesn't have a maximum on the interval.Therefore, the conclusion is that ( a ) must satisfy ( -5/2 < a < -1/2 ). Let me check with another example. Let’s take ( a = -2.5 ). Then the interval is ( (-2.5, -0.5) ). Wait, ( a + 2 = -0.5 ). So the interval is ( (-2.5, -0.5) ). But ( -1/2 = -0.5 ), so the interval is up to but not including -0.5. Therefore, the critical point at -0.5 is not included in the interval. Hence, if ( a = -2.5 ), the interval is ( (-2.5, -0.5) ), which does not include the critical point at -0.5. So in that interval, the function's behavior: let's see. For ( x ) near -2.5, ( f(x) = 4x^3 - 3x ). Let's compute ( f(-2.5) ): ( 4*(-2.5)^3 - 3*(-2.5) = 4*(-15.625) + 7.5 = -62.5 + 7.5 = -55 ). As ( x ) approaches -0.5 from the left, ( x ) approaches -0.5. Let's compute the limit as ( x ) approaches -0.5 from the left: ( f(-0.5) = 4*(-0.5)^3 - 3*(-0.5) = 4*(-0.125) + 1.5 = -0.5 + 1.5 = 1.0 ). So on the interval ( (-2.5, -0.5) ), the function goes from around -55 approaching 1.0 as ( x ) approaches -0.5. The derivative in this interval: ( 12x^2 - 3 ). For ( x ) in ( (-2.5, -0.5) ), ( x^2 ) is between ( 0.25 ) and ( 6.25 ). Therefore, ( 12x^2 - 3 ) is between ( 12*0.25 -3 = 3 -3 = 0 ) and ( 12*6.25 -3 = 75 -3 = 72 ). Wait, but when ( x ) is between -2.5 and -0.5, ( x^2 ) is between ( (-0.5)^2 = 0.25 ) and ( (-2.5)^2 = 6.25 ). So ( 12x^2 -3 ) is between 12*0.25 -3 = 0 and 12*6.25 -3 = 72. So the derivative is non-negative in this interval, but actually, at ( x = -0.5 ), the derivative is ( 12*(0.25) -3 = 3 - 3 = 0 ). So near ( x = -0.5 ), the derivative approaches zero. So the function is increasing on the interval ( (-2.5, -0.5) ), but with derivative starting at 72 when x is near -2.5 and approaching 0 as x approaches -0.5. Therefore, the function is increasing throughout the interval ( (-2.5, -0.5) ). Therefore, the supremum is approaching 1.0 as ( x ) approaches -0.5 from the left, but since -0.5 is not included in the interval, there's no maximum. Therefore, when ( a = -2.5 ), the interval does not include the critical point, and the function is increasing, so no maximum. Hence, ( a ) must be greater than -2.5 (i.e., -5/2) to ensure that the interval ( (a, a+2) ) includes -1/2.Wait, but if ( a = -2.5 ), then ( a +2 = -0.5 ), so the interval is (-2.5, -0.5). The critical point is at -0.5, which is not included. So when ( a = -5/2 ), the interval is (-5/2, -5/2 + 2) = (-5/2, -1/2). Wait, that's the interval. So when ( a = -5/2 ), the interval is (-2.5, -0.5). But the critical point at -0.5 is not included. Therefore, the maximum is not attained. Therefore, ( a ) must be greater than -5/2 so that the upper bound ( a + 2 ) is greater than -0.5. Similarly, ( a ) must be less than -0.5 so that the lower bound is less than -0.5. Therefore, the interval (a, a+2) must contain -0.5. Therefore, the range of ( a ) is all real numbers such that ( a < -0.5 ) and ( a + 2 > -0.5 ). Which simplifies to ( a < -0.5 ) and ( a > -2.5 ). Therefore, combining these, ( -2.5 < a < -0.5 ), which is ( (-frac{5}{2}, -frac{1}{2}) ).Therefore, I think that's the answer. Let me just check another value inside the range. Let's take ( a = -2 ). Then the interval is (-2, 0). Does this interval contain the critical point at -0.5? Yes. So in that interval, the function has a critical point at -0.5 which is a local maximum. Therefore, the function attains a maximum at x = -0.5. Therefore, yes, that works. Similarly, if ( a = -1 ), the interval is (-1, 1). The critical point at -0.5 is inside, so the maximum is there. The other critical point at 0.5 is a local minimum, but since the interval includes both, the maximum is still at -0.5.Wait, but what if the interval includes both critical points? For example, if ( a = -1.5 ), the interval is (-1.5, 0.5). This interval includes both critical points: -0.5 and 0.5. Wait, but 0.5 is not included because the interval is up to 0.5, not including. So at x = 0.5, the interval is open. So in (-1.5, 0.5), the critical points are -0.5 and 0.5. But 0.5 is not included. Therefore, the function in (-1.5, 0.5) has a local maximum at -0.5 and a local minimum approaching 0.5. But since 0.5 is not included, the local minimum at 0.5 is not in the interval. However, the critical point at -0.5 is inside, so the maximum is there. Therefore, even if the interval includes both critical points, as long as the local maximum is inside, the function will have a maximum there. The presence of the local minimum in the interval doesn't affect the existence of the maximum. Therefore, the key is that the local maximum at -0.5 is inside the interval. If the interval includes -0.5, then the function has a maximum there. If not, then it doesn't. Therefore, the range of ( a ) is ( -5/2 < a < -1/2 ).Therefore, the answer is ( (-frac{5}{2}, -frac{1}{2}) ).Final AnswerThe range of the real number ( a ) is boxed{left( -dfrac{5}{2}, -dfrac{1}{2} right)}.</think>To determine the range of the real number ( a ) such that the function ( f(x) = 4x^3 - 3x ) has a maximum value on the interval ( (a, a+2) ), we first find the critical points of the function. The critical points are found by setting the derivative equal to zero:[f'(x) = 12x^2 - 3 = 0 implies 4x^2 - 1 = 0 implies x^2 = frac{1}{4} implies x = pm frac{1}{2}]Next, we analyze the nature of these critical points using the second derivative:[f''(x) = 24x]- At ( x = frac{1}{2} ), ( f''left(frac{1}{2}right) = 12 > 0 ), indicating a local minimum.- At ( x = -frac{1}{2} ), ( f''left(-frac{1}{2}right) = -12 < 0 ), indicating a local maximum.For the function to have a maximum value on the interval ( (a, a+2) ), the interval must contain the local maximum at ( x = -frac{1}{2} ). Therefore, we need:[a < -frac{1}{2} < a + 2]Solving these inequalities:1. ( a < -frac{1}{2} )2. ( -frac{1}{2} < a + 2 implies a > -frac{5}{2} )Combining these results, we get:[-frac{5}{2} < a < -frac{1}{2}]Thus, the range of the real number ( a ) is boxed{left( -dfrac{5}{2}, -dfrac{1}{2} right)}.]

🔑:Solution: Given f′(x)=12x^{2}-3, let f′(x) 0 to find x dfrac {1}{2}. From this, we know the function is increasing on (-∞,- dfrac {1}{2}), decreasing on (- dfrac {1}{2}, dfrac {1}{2}), and increasing on ( dfrac {1}{2},+∞). Therefore, the function has a local maximum at x=- dfrac {1}{2}. It is inferred that this local maximum must be the maximum value on the interval (a,a+2) therefore we have a - dfrac {1}{2}…②. Solving ② gives: a > - dfrac {5}{2}, and since f(- dfrac {1}{2}) > f(a+2), we get 1 > 4(a+2)^{3}-3(a+2), …③. Solving ③ gives: a < 3. In summary, we have a∈(- dfrac {5}{2},- dfrac {1}{2}) Therefore, the answer is: boxed{(- dfrac {5}{2},- dfrac {1}{2})}. To find the derivative of the function f(x)=4x^{3}-3x and study the location of its maximum value, since the function has a maximum value on the interval (a,a+2), the x-coordinate of the maximum value point is an element of the set (a,a+2). From this, we can derive an equation regarding the parameter a and solve it to find the range of the real number a. This problem examines the use of derivatives to study the maximum and minimum values of a function. Using derivatives to study the extremum of a function is an important application of derivatives in mathematics. It is important to grasp the steps of solving such problems: differentiate, determine monotonicity, and conclude the extremum.

🧠:[<think>Okay, so I need to solve this problem about a sequence where the sum of the first n terms is given by a specific formula, and then find the difference q - p when there are three terms forming an arithmetic sequence. Let me try to break this down step by step.First, let's parse the problem statement again. We have a sequence {a_n} where the first term a_1 is 3. The sum of the first n terms, S_n, is given by the formula S_n = 2a_n + (3/2)((-1)^n - 1). Then, it says that in this sequence, there are three terms a_1, a_p, a_q (with p and q positive integers, 1 < p < q) that form an arithmetic sequence. We need to find q - p.Alright, so the key steps here are probably:1. Find a general formula for a_n using the given S_n.2. Determine the conditions under which a_1, a_p, a_q form an arithmetic sequence.3. Solve for p and q, then compute q - p.Starting with step 1: finding a_n. Remember that S_n is the sum of the first n terms, so S_n = a_1 + a_2 + ... + a_n. Also, S_{n} = 2a_n + (3/2)((-1)^n - 1). Given that a_1 = 3.But we can also express a_n in terms of S_n and S_{n-1}. Specifically, a_n = S_n - S_{n-1} for n ≥ 2. That's a standard approach when dealing with sums of sequences.So let's compute a_n for n ≥ 2.Given S_n = 2a_n + (3/2)((-1)^n - 1)Similarly, S_{n-1} = 2a_{n-1} + (3/2)((-1)^{n-1} - 1)Therefore, a_n = S_n - S_{n-1} = [2a_n + (3/2)((-1)^n - 1)] - [2a_{n-1} + (3/2)((-1)^{n-1} - 1)]Let me write that out:a_n = 2a_n + (3/2)((-1)^n - 1) - 2a_{n-1} - (3/2)((-1)^{n-1} - 1)Hmm, let's simplify this. Let's collect like terms.First, bring the 2a_n to the left side:a_n - 2a_n = -2a_{n-1} + (3/2)[(-1)^n - 1 - (-1)^{n-1} + 1]Simplify left side:- a_n = -2a_{n-1} + (3/2)[(-1)^n - (-1)^{n-1}]Simplify the right side. Let's compute the term inside the brackets:(-1)^n - (-1)^{n-1} = (-1)^{n-1}(-1 - 1) = (-1)^{n-1}(-2) = 2*(-1)^nWait, let me check that:(-1)^n - (-1)^{n - 1} = (-1)^{n} + (-1)^n = 2*(-1)^n?Wait, actually:Let's factor out (-1)^{n - 1}:= (-1)^{n - 1} [ (-1) - 1 ] = (-1)^{n - 1} (-2) = 2*(-1)^nBecause (-1)^{n} = (-1)^{n - 1} * (-1), so (-1)^{n - 1}*(-2) = -2*(-1)^{n - 1} = 2*(-1)^{n} (since - (-1)^{n - 1} = (-1)^n)Yes, that's correct.So the term inside the brackets simplifies to 2*(-1)^n. Then:- a_n = -2a_{n-1} + (3/2)(2*(-1)^n) = -2a_{n-1} + 3*(-1)^nSo bringing all terms to the left side:- a_n + 2a_{n-1} = 3*(-1)^nMultiply both sides by -1:a_n - 2a_{n-1} = -3*(-1)^n = 3*(-1)^{n + 1}So we have the recurrence relation:a_n - 2a_{n-1} = 3*(-1)^{n + 1} for n ≥ 2With the initial condition a_1 = 3.So this is a linear nonhomogeneous recurrence relation. To solve this, we can use the method of solving linear recurrences. The general solution will be the solution to the homogeneous equation plus a particular solution.The homogeneous equation is a_n - 2a_{n-1} = 0. The characteristic equation is r - 2 = 0, so r = 2. Therefore, the homogeneous solution is A*2^n.Now, we need a particular solution. The nonhomogeneous term is 3*(-1)^{n + 1} = -3*(-1)^n. Let's write the nonhomogeneous term as C*(-1)^n, where C = -3.So we can assume a particular solution of the form D*(-1)^n. Let's substitute into the recurrence:a_n - 2a_{n-1} = C*(-1)^nAssume a_p(n) = D*(-1)^n. Then:D*(-1)^n - 2D*(-1)^{n - 1} = C*(-1)^nSimplify:D*(-1)^n + 2D*(-1)^{n} = C*(-1)^nFactor out (-1)^n:(D + 2D)*(-1)^n = C*(-1)^n => 3D = C => D = C/3Given that C = -3, so D = -3/3 = -1.Therefore, the particular solution is -1*(-1)^n = (-1)^{n + 1}.Therefore, the general solution is:a_n = A*2^n + (-1)^{n + 1}Now, apply the initial condition a_1 = 3.When n = 1:3 = A*2^1 + (-1)^{1 + 1} => 3 = 2A + (-1)^2 => 3 = 2A + 1 => 2A = 2 => A = 1Therefore, the general formula for a_n is:a_n = 2^n + (-1)^{n + 1}Let me check this formula with the given S_n to see if it's consistent.Wait, maybe we should check if this a_n gives the correct S_n. Let's compute S_n using the formula S_n = sum_{k=1}^n a_k, which should equal 2a_n + (3/2)((-1)^n - 1).Given a_n = 2^n + (-1)^{n + 1}So sum_{k=1}^n a_k = sum_{k=1}^n [2^k + (-1)^{k + 1}] = sum_{k=1}^n 2^k + sum_{k=1}^n (-1)^{k + 1}Sum of 2^k from k=1 to n is 2^{n + 1} - 2 (since it's a geometric series with ratio 2).Sum of (-1)^{k + 1} from k=1 to n is an alternating series: 1 - 1 + 1 - 1 + ... + (-1)^{n + 1}This sum is equal to:If n is even: 0If n is odd: 1Alternatively, we can write it as (1 - (-1)^n)/2. Wait:Let me recall that sum_{k=1}^n (-1)^{k + 1} = [1 - (-1)^n]/2. Let me verify:For n = 1: [1 - (-1)^1]/2 = (1 + 1)/2 = 1, which matches.For n = 2: [1 - (-1)^2]/2 = (1 - 1)/2 = 0, which matches.Yes, so sum_{k=1}^n (-1)^{k + 1} = [1 - (-1)^n]/2Therefore, S_n = (2^{n + 1} - 2) + [1 - (-1)^n]/2But according to the problem statement, S_n should equal 2a_n + (3/2)((-1)^n - 1)Let me compute 2a_n + (3/2)((-1)^n - 1):Given a_n = 2^n + (-1)^{n + 1}, so 2a_n = 2^{n + 1} + 2*(-1)^{n + 1}Then, 2a_n + (3/2)((-1)^n - 1) = 2^{n + 1} + 2*(-1)^{n + 1} + (3/2)((-1)^n - 1)Let's simplify this:First, 2*(-1)^{n + 1} = -2*(-1)^nSo:= 2^{n + 1} - 2*(-1)^n + (3/2)(-1)^n - 3/2Combine the terms with (-1)^n:= 2^{n + 1} + (-1)^n*(-2 + 3/2) - 3/2= 2^{n + 1} + (-1)^n*(-1/2) - 3/2= 2^{n + 1} - (1/2)(-1)^n - 3/2Now, compare with S_n from the sum:S_n = (2^{n + 1} - 2) + [1 - (-1)^n]/2= 2^{n + 1} - 2 + 1/2 - (-1)^n/2= 2^{n + 1} - 3/2 - (-1)^n/2Which matches the expression above: 2^{n + 1} - (1/2)(-1)^n - 3/2Yes, that's correct. So the formula for a_n is verified.So the explicit formula for a_n is a_n = 2^n + (-1)^{n + 1}So now, moving to step 2: find p and q such that a_1, a_p, a_q form an arithmetic sequence.Given that a_1 = 3.An arithmetic sequence requires that the difference between consecutive terms is constant. So, for a_1, a_p, a_q to be in arithmetic sequence, we need:a_p - a_1 = a_q - a_pWhich simplifies to:2a_p = a_1 + a_qSo 2a_p = 3 + a_qWe need to find integers p and q (with 1 < p < q) such that this holds.Given that a_n = 2^n + (-1)^{n + 1}Therefore, a_p = 2^p + (-1)^{p + 1}Similarly, a_q = 2^q + (-1)^{q + 1}Therefore, substituting into the equation:2[2^p + (-1)^{p + 1}] = 3 + [2^q + (-1)^{q + 1}]Let's simplify this equation.Left side: 2*2^p + 2*(-1)^{p + 1} = 2^{p + 1} + 2*(-1)^{p + 1}Right side: 3 + 2^q + (-1)^{q + 1}So the equation becomes:2^{p + 1} + 2*(-1)^{p + 1} = 3 + 2^q + (-1)^{q + 1}Let me rearrange terms:2^{p + 1} - 2^q + 2*(-1)^{p + 1} - (-1)^{q + 1} = 3Hmm, this seems complex. Let's consider possible simplifications.First, note that 2^{p + 1} - 2^q = 2^{p + 1} - 2^q = 2^q(2^{p + 1 - q} - 1). Hmm, but unless we know something about p and q, this might not help.Alternatively, since 2^{p + 1} and 2^q are powers of 2, and 3 is a constant. Maybe we can analyze possible values of p and q.But given that p and q are integers greater than 1, and p < q, perhaps we can consider small values of p and q. But since the problem states that such p and q exist, and asks for q - p, maybe there is a unique solution or a pattern.Alternatively, maybe the equation can be manipulated modulo 2 or modulo some number to find constraints on p and q.Alternatively, since the terms involve (-1)^{p + 1} and (-1)^{q + 1}, which alternate between -1 and 1 depending on whether p is even or odd.Let me consider the parity of p and q. Let's denote:Let’s denote:If p is even, then (-1)^{p + 1} = (-1)^{odd} = -1If p is odd, then (-1)^{p + 1} = (-1)^{even} = 1Similarly for q.Let me handle different cases based on the parity of p and q.Case 1: p is even, q is even.Case 2: p is even, q is odd.Case 3: p is odd, q is even.Case 4: p is odd, q is odd.Let's check each case.Case 1: p even, q even.Then:(-1)^{p + 1} = (-1)^{even + 1} = (-1)^{odd} = -1Similarly, (-1)^{q + 1} = -1So the equation becomes:2^{p + 1} - 2^q + 2*(-1) - (-1) = 3Simplify:2^{p + 1} - 2^q - 2 + 1 = 32^{p + 1} - 2^q - 1 = 32^{p + 1} - 2^q = 4Factor out 2^{p + 1}:2^{p + 1}(1 - 2^{q - (p + 1)}) = 4Let me write this as:2^{p + 1} - 2^q = 4We can factor 2^{p + 1} - 2^q = 2^{p + 1}(1 - 2^{q - p - 1}) = 4So 1 - 2^{q - p - 1} must be 4 / 2^{p + 1}Since 2^{p + 1} divides 4, possible values of p + 1:p + 1 ≤ 2 (since 2^{p + 1} divides 4, which is 2^2), so p + 1 can be 1 or 2. But p is even and ≥2 (since p >1), so p ≥2, so p +1 ≥3. But 2^{p +1} divides 4 only if p +1 ≤2, which is impossible here. Therefore, no solution in this case.Case 2: p even, q odd.Then:(-1)^{p + 1} = -1 (as p is even)(-1)^{q + 1} = (-1)^{odd + 1} = (-1)^{even} = 1So equation:2^{p +1} -2^q + 2*(-1) -1 = 3Simplify:2^{p +1} -2^q -2 -1 =32^{p +1} -2^q -3 =32^{p +1} -2^q =6So 2^{p +1} -2^q=6Factor:2^{p +1} - 2^q =6Again, we can write this as 2^{p+1} - 2^q =6Perhaps factor 2^q:But if q > p +1, then 2^{p +1} -2^q is negative. So maybe q < p +1?But since q > p, the only possibility is q = p +1. Let's check:If q = p +1:Then 2^{p +1} -2^{p +1} =0 ≠6. Not possible.If q = p -1: But q > p, which is impossible.Therefore, no solution here.Alternatively, rearrange the equation:2^{p +1} =2^q +6Looking for integer p, q with q > p ≥2 (since p is even and >1)Let me test possible small q:Let’s take q=3 (odd). Then 2^q=8. So 2^{p +1}=8 +6=14. But 14 is not a power of 2.q=4: even, but we are in case 2 where q is odd. So next q=5: 2^5=32. Then 2^{p +1}=32 +6=38. Not power of 2.q=1: q must be at least p +1 > p ≥2. So q=1 invalid.This case seems impossible.Case 3: p odd, q even.Then:(-1)^{p +1} =1 (since p is odd)(-1)^{q +1}= -1 (since q is even)Equation:2^{p +1} -2^q +2*1 -(-1)=3Simplify:2^{p +1} -2^q +2 +1 =32^{p +1} -2^q +3 =3So 2^{p +1} -2^q =0Therefore, 2^{p +1} =2^q ⇒ p +1 = qSince q > p, and p is odd, q = p +1. So q - p =1.But we need to check if this holds with the parity. Since p is odd, q = p +1 is even, which matches the case (q even). So this is possible.So if p is odd and q = p +1, then 2^{p +1} =2^q, which is valid, and equation holds. So this suggests that when p is odd and q = p +1, the equation holds.But let's verify this with actual terms. Let me take p odd, say p=3, q=4.Compute a_1=3, a_p=a_3, a_q=a_4.Compute a_3=2^3 + (-1)^{4}=8 +1=9a_4=2^4 + (-1)^5=16 -1=15Check if 3,9,15 form an arithmetic sequence.Difference between a_p and a_1: 9 -3=6Difference between a_q and a_p:15 -9=6Yes, so 3,9,15 is an arithmetic sequence with common difference 6. Therefore, p=3, q=4, q -p=1.But the problem states that p and q are positive integers with 1 < p < q. Here p=3, q=4, which satisfies the condition, and q - p=1.But wait, let me check another example to be sure.Take p=5 (odd), q=6 (even). Compute a_5=2^5 + (-1)^6=32 +1=33a_6=2^6 + (-1)^7=64 -1=63Then a1=3, a5=33, a6=63.Check if 3,33,63 is arithmetic.33 -3=30, 63 -33=30. Yes, common difference 30. So this is also arithmetic. So in this case, q - p=1 as well.Therefore, according to this case, for any odd p and q=p+1, the three terms a1, a_p, a_q form an arithmetic sequence, leading to q - p=1.But wait, the problem states that "contains three terms a1, a_p, a_q" so perhaps there are multiple possibilities, but according to this, every odd p and q=p+1 would work. But the problem might be expecting a specific answer.But the answer is q - p=1. However, let's check if this is the only possibility. Let me check if there are other cases.Case 4: p odd, q odd.Then:(-1)^{p +1}=1 (p is odd)(-1)^{q +1}=1 (q is odd)So the equation:2^{p +1} -2^q +2*1 -1=3Simplify:2^{p +1} -2^q +2 -1=32^{p +1} -2^q +1=3Therefore:2^{p +1} -2^q=2So 2^{p +1} -2^q=2Factor:2^{p +1} -2^q=2 ⇒2^q(2^{p +1 - q} -1)=2Since the left side is 2^q*(something) =2, and 2 is 2^1.Thus, possibilities:Either:2^q=2 and (2^{p +1 - q} -1)=1Or 2^q=1 and (2^{p +1 - q} -1)=2, but 2^q=1 implies q=0, invalid.Thus, first possibility:2^q=2 ⇒ q=1But q >p >1, so q=1 is invalid.Alternatively, maybe other factors. Wait, but 2^{p +1} -2^q=2. Let's rewrite this as 2^q(2^{p +1 - q} -1)=2.The factors of 2 are 2 and 1. Since 2^q is a power of 2, and (2^{p +1 - q} -1) must be an integer. So possible:Either:2^q=2 and (2^{p +1 - q} -1)=1Which gives q=1, but q> p>1 invalid.Or 2^q=1 and (2^{p +1 - q} -1)=2. But 2^q=1 gives q=0, invalid.Therefore, no solutions in this case.Thus, the only possible case is case 3, where p is odd, q is even, and q = p +1, leading to q -p=1.But wait, the problem states "three terms a1, a_p, a_q (p, q ∈ Z_+, 1 < p < q)" so p must be at least 2. Wait, in our previous examples, p=3 and q=4, which is acceptable. But if p is odd and greater than 1, like p=3,5,7,..., then q=p+1 is even and greater than p.But in the problem, the answer is to find q - p. If this is always 1, then the answer is 1. But let me confirm with the first example.Take p=3, q=4: a1=3, a3=9, a4=15. 3,9,15: common difference 6. Yes, arithmetic.Similarly, p=5, q=6: a1=3, a5=33, a6=63. Common difference 30. Also arithmetic.But let's check p=1: but p must be greater than 1, so p=3 is the first possible.But the problem says "three terms a1, a_p, a_q", so a1 is fixed. Therefore, if for every odd p ≥3, q=p+1 gives an arithmetic sequence, then there are infinitely many solutions, but the problem says "contains three terms", so likely the minimal possible q - p is 1, which is the answer.But the problem might be designed such that q - p=1, so the answer is 1.But wait, let me check the problem statement again: it says "then q - p=_______". So we need to compute q - p given that such p and q exist. Since according to our analysis, the difference is always 1, regardless of p (as long as p is odd and q=p+1), then the answer is 1.But wait, hold on. Let me check another term to be sure.Take p=7, q=8:a7=2^7 + (-1)^8=128 +1=129a8=2^8 + (-1)^9=256 -1=255Check if 3,129,255 form an arithmetic sequence. 129-3=126, 255-129=126. Yes, common difference 126. So yes, arithmetic. q - p=1.So this seems consistent. Therefore, the answer is 1.But wait, let me check if there are other possibilities where q - p is different. Suppose there exists another pair p, q where q - p is different.Suppose p=2, even. Let's check if there is a q >2 such that a1=3, a2, a_q form an arithmetic sequence.Compute a2=2^2 + (-1)^3=4 -1=3So a1=3, a2=3, then for arithmetic sequence, the next term should also be 3. But a_q=3.But a_q=2^q + (-1)^{q +1}. Let's see if this can be 3.2^q + (-1)^{q +1}=3.Case q even:2^q + (-1)^{q +1}=2^q -1=3 ⇒2^q=4 ⇒q=2. But q>p=2, so invalid.Case q odd:2^q +1=3 ⇒2^q=2 ⇒q=1. But q>p=2 invalid.So no solution here.If p=4 (even), check:a4=15. Then 3, a4=15, a_q. To form arithmetic sequence, need 15 -3 = a_q -15 ⇒a_q=27.Find q such that a_q=27.a_q=2^q + (-1)^{q +1}=27.Case q even: 2^q -1=27 ⇒2^q=28 ⇒q=log2(28) not integer.Case q odd: 2^q +1=27 ⇒2^q=26 ⇒q not integer.No solution.Alternatively, maybe p=5 (odd), q=7 (odd). Let's check.a5=33, a7=129.Check if 3,33,129 form an arithmetic sequence. 33 -3=30, 129 -33=96. Not equal. Not arithmetic.Alternatively, p=3, q=5. a3=9, a5=33.3,9,33. Differences 6 and 24. Not equal.So if we take p=3 and q=5, that's not arithmetic. So seems like the only way to get arithmetic sequence is when q=p+1 where p is odd. Then the difference is 1.Alternatively, let's check if there's a case where q - p=2.Suppose p=3, q=5. As above, not arithmetic.p=5, q=7: 3,33,129. Not arithmetic.Alternatively, maybe another combination.Wait, maybe take p=2, q=5. a1=3, a2=3, a5=33. 3,3,33. Differences 0 and 30. Not arithmetic.Alternatively, p=4, q=6. a4=15, a6=63. Then 3,15,63. Differences 12 and 48. Not equal.Alternatively, p=3, q=6. 3,9,63. Differences 6 and 54. Not equal.Hence, it seems the only possible way is q=p+1 with p odd, leading to q-p=1.Therefore, the answer should be 1.But let me check once again with the recurrence relation.From earlier, we had the explicit formula for a_n=2^n + (-1)^{n+1}So, for a general term:a_n = 2^n + (-1)^{n+1}Therefore, let's compute the difference between a_{p} and a1:a_p - a1 = 2^p + (-1)^{p+1} -3Similarly, a_q - a_p =2^q + (-1)^{q+1} -2^p - (-1)^{p+1}For these to be equal:2^p + (-1)^{p+1} -3 =2^q + (-1)^{q+1} -2^p - (-1)^{p+1}Simplify:2^p -3 + (-1)^{p+1} =2^q -2^p + (-1)^{q+1} - (-1)^{p+1}Bring all terms to left side:2^p -3 + (-1)^{p+1} -2^q +2^p - (-1)^{q+1} + (-1)^{p+1}=0Combine like terms:(2^p +2^p) -2^q -3 + [ (-1)^{p+1} + (-1)^{p+1} ] - (-1)^{q+1} =0Which simplifies:2^{p+1} -2^q -3 + 2*(-1)^{p+1} - (-1)^{q+1}=0Which is the same equation as before:2^{p +1} -2^q +2*(-1)^{p +1} - (-1)^{q +1}=3So same equation. So our previous analysis holds.Therefore, the only solution is when p is odd, q=p+1, leading to q-p=1.Thus, the answer is 1.But the problem is in Chinese, and sometimes there might be translation issues, but the key steps seem to lead to 1.Wait, but in the example I did earlier, p=3, q=4, the difference is 1, but maybe in the problem statement's context, maybe there is another answer. Alternatively, perhaps I made a mistake in the arithmetic sequence condition.Wait, the problem states that a1, a_p, a_q form an arithmetic sequence. So the indices are 1, p, q. In an arithmetic sequence, the difference between consecutive terms is constant. Therefore, the condition is that a_p - a1 = a_q - a_p, which is 2a_p =a1 +a_q. This is correct.But let's check if in the case p=3, q=4, this holds.a_p =a3=9, a_q=a4=15.2a_p=18, a1 +a_q=3+15=18. Yes, 18=18. Correct.Similarly, p=5, q=6: 2a5=66, a1 +a6=3+63=66. Correct.Therefore, yes, the condition holds.Therefore, since p must be odd, q=p+1, then q-p=1.Thus, the answer is 1. Therefore, the answer should be boxed{1}Final Answerboxed{1}</think>Given the sequence ({a_n}) with the sum of the first (n) terms as (S_n), where (a_1 = 3) and (S_n = 2a_n + frac{3}{2}left((-1)^n - 1right)), we need to find (q - p) when (a_1, a_p, a_q) form an arithmetic sequence.1. Finding the general formula for (a_n): - The sum (S_n) is given by (S_n = 2a_n + frac{3}{2}left((-1)^n - 1right)). - Using the relation (a_n = S_n - S_{n-1}) for (n geq 2), we derive the recurrence relation: [ a_n - 2a_{n-1} = 3(-1)^{n+1} ] - Solving this recurrence relation with the initial condition (a_1 = 3), we find: [ a_n = 2^n + (-1)^{n+1} ]2. Checking the arithmetic sequence condition: - For (a_1, a_p, a_q) to form an arithmetic sequence, we need (2a_p = a_1 + a_q). - Substituting (a_p = 2^p + (-1)^{p+1}) and (a_q = 2^q + (-1)^{q+1}) into the equation, we get: [ 2(2^p + (-1)^{p+1}) = 3 + (2^q + (-1)^{q+1}) ] - Simplifying and analyzing the equation, we find that (p) must be odd and (q = p + 1) to satisfy the condition. This results in (q - p = 1).3. Verification: - Checking with specific values (e.g., (p = 3) and (q = 4)), we confirm that (a_1 = 3), (a_3 = 9), and (a_4 = 15) form an arithmetic sequence with a common difference of 6.Thus, the difference (q - p) is (boxed{1}).,<think>Okay, let's try to solve this problem. Hmm, we have a sequence {a_n} where the sum of the first n terms is S_n. The first term is given as a₁ = 3. The formula for S_n is given as 2a_n + (3/2)[(-1)^n - 1]. We need to find q - p where a₁, a_p, a_q form an arithmetic sequence with 1 < p < q.First, let me recall that in an arithmetic sequence, the difference between consecutive terms is constant. So, if a₁, a_p, a_q are in arithmetic sequence, then the difference between a_p and a₁ should be the same as the difference between a_q and a_p. That means 2a_p = a₁ + a_q. Since a₁ is 3, this becomes 2a_p = 3 + a_q. So, we need to find p and q such that this equation holds. Then, the answer would be q - p.But to find such p and q, we need to first find a general formula for a_n. Since S_n is the sum of the first n terms, S_n = a₁ + a₂ + ... + a_n. Also, S_n is given by 2a_n + (3/2)[(-1)^n - 1]. Let me write this down:S_n = 2a_n + (3/2)[(-1)^n - 1]We also know that S_n = S_{n-1} + a_n for n ≥ 2. So, maybe we can use this recursive relation to find a formula for a_n.Let's try to express S_n and S_{n-1} using the given formula and then find a relation between a_n and a_{n-1}.Given S_n = 2a_n + (3/2)[(-1)^n - 1]Similarly, S_{n-1} = 2a_{n-1} + (3/2)[(-1)^{n-1} - 1]But S_n = S_{n-1} + a_n, so:2a_n + (3/2)[(-1)^n - 1] = 2a_{n-1} + (3/2)[(-1)^{n-1} - 1] + a_nLet me simplify this equation.Left side: 2a_n + (3/2)[(-1)^n - 1]Right side: 2a_{n-1} + (3/2)[(-1)^{n-1} - 1] + a_nSubtract a_n from both sides:Left side: a_n + (3/2)[(-1)^n - 1]Right side: 2a_{n-1} + (3/2)[(-1)^{n-1} - 1]Now, let's rearrange terms:a_n = 2a_{n-1} + (3/2)[(-1)^{n-1} - 1] - (3/2)[(-1)^n - 1]Simplify the terms involving (-1)^n.First, compute the difference:(3/2)[(-1)^{n-1} - 1 - (-1)^n + 1] = (3/2)[(-1)^{n-1} - (-1)^n]Note that (-1)^{n-1} = -(-1)^n. So, substituting:= (3/2)[ -(-1)^n - (-1)^n ] = (3/2)[ -2(-1)^n ] = -3(-1)^nTherefore, the equation becomes:a_n = 2a_{n-1} - 3(-1)^nSo, we have the recurrence relation:a_n = 2a_{n-1} - 3(-1)^n, with a₁ = 3.Hmm, this seems like a nonhomogeneous linear recurrence relation. Let me see if we can solve this.The homogeneous part is a_n - 2a_{n-1} = 0, which has the solution a_n^h = C(2)^{n-1}, where C is a constant.Now, we need a particular solution for the nonhomogeneous part. The nonhomogeneous term is -3(-1)^n. Let's assume a particular solution of the form A(-1)^n.Substitute into the recurrence:A(-1)^n = 2A(-1)^{n-1} - 3(-1)^nDivide both sides by (-1)^{n-1}:A(-1) = 2A - 3(-1)Simplify:- A = 2A + 3Bring terms together:- A - 2A = 3 => -3A = 3 => A = -1Therefore, the particular solution is -(-1)^n = (-1)^{n+1}So, the general solution is:a_n = a_n^h + a_n^p = C(2)^{n-1} + (-1)^{n+1}Now, apply the initial condition a₁ = 3.When n = 1:a₁ = C(2)^{0} + (-1)^{2} = C(1) + 1 = C + 1 = 3Therefore, C = 3 - 1 = 2.Thus, the general formula for a_n is:a_n = 2(2)^{n-1} + (-1)^{n+1} = 2^n + (-1)^{n+1}Let me check this formula with the given initial term.For n = 1: 2^1 + (-1)^{2} = 2 + 1 = 3. Correct.Now, let's check if this formula satisfies the recurrence relation.Suppose n = 2: a_2 = 2^2 + (-1)^3 = 4 - 1 = 3.Using the recurrence: a_2 = 2a₁ - 3(-1)^2 = 2*3 - 3*1 = 6 - 3 = 3. Correct.Similarly, n=3: a_3 = 2^3 + (-1)^4 = 8 + 1 = 9.Using recurrence: a_3 = 2a_2 -3(-1)^3 = 2*3 -3*(-1) = 6 + 3 = 9. Correct.Good, so the formula seems correct.Now, we have a general term a_n = 2^n + (-1)^{n+1}Also, let's check the sum S_n. According to the problem, S_n = 2a_n + (3/2)[(-1)^n -1]Using our formula for a_n:2a_n = 2[2^n + (-1)^{n+1}] = 2^{n+1} + 2(-1)^{n+1}Now, compute the other term: (3/2)[(-1)^n -1] = (3/2)(-1)^n - 3/2Adding these together:2^{n+1} + 2(-1)^{n+1} + (3/2)(-1)^n - 3/2Simplify the terms with (-1)^n:2(-1)^{n+1} = -2(-1)^n, so:= 2^{n+1} - 2(-1)^n + (3/2)(-1)^n - 3/2Combine like terms:= 2^{n+1} - (2 - 3/2)(-1)^n - 3/2= 2^{n+1} - (1/2)(-1)^n - 3/2Alternatively, factor:= 2^{n+1} - (1/2)(-1)^n - 3/2But S_n should also be the sum of the first n terms. Let's compute S_n using the formula for a_n and check if it matches.Sum_{k=1}^n a_k = Sum_{k=1}^n [2^k + (-1)^{k+1}] = Sum_{k=1}^n 2^k + Sum_{k=1}^n (-1)^{k+1}Compute each sum:Sum of 2^k from k=1 to n is 2(2^n -1)/(2 -1) = 2^{n+1} - 2Sum of (-1)^{k+1} from k=1 to n is alternating series. Let's compute:If n is even: Sum_{k=1}^{n} (-1)^{k+1} = (1 -1 +1 -1 + ... +1 -1) = 0If n is odd: Sum_{k=1}^{n} (-1)^{k+1} = (1 -1 +1 -1 + ... +1) = 1Therefore, Sum_{k=1}^n (-1)^{k+1} = [1 + (-1)^{n+1}]/2Wait, maybe better to use the formula for the sum of a geometric series:Sum_{k=1}^n (-1)^{k+1} = Sum_{k=1}^n (-1)^{k+1} = [1 - (-1)^{n+1}]/(1 - (-1)) = [1 - (-1)^{n+1}]/2Yes, because the common ratio is -1. So:Sum_{k=1}^n (-1)^{k+1} = [1 - (-1)^{n+1}]/2 = [1 + (-1)^n]/2Wait, let me check:Sum_{k=1}^n r^{k-1} = (1 - r^n)/(1 - r)But here, starting at k=1, first term is (-1)^{1+1} = 1, common ratio is -1, so:Sum_{k=1}^n (-1)^{k+1} = [1 - (-1)^n]/[1 - (-1)] = [1 - (-1)^n]/2Wait, no:Wait, Sum_{k=1}^n (-1)^{k+1} can be written as Sum_{m=0}^{n-1} (-1)^{m} = [1 - (-1)^n]/[1 - (-1)] by the geometric series formula. Because when m = 0, (-1)^0 =1, which corresponds to k=1: (-1)^{1+1}=1. So, yes:Sum_{k=1}^n (-1)^{k+1} = Sum_{m=0}^{n-1} (-1)^m = [1 - (-1)^n]/[1 - (-1)] = [1 - (-1)^n]/2Therefore, Sum_{k=1}^n (-1)^{k+1} = [1 - (-1)^n]/2Therefore, total sum S_n = (2^{n+1} - 2) + [1 - (-1)^n]/2Simplify:= 2^{n+1} - 2 + 1/2 - (-1)^n / 2= 2^{n+1} - 3/2 - (-1)^n / 2But according to the problem statement, S_n is given by 2a_n + (3/2)[(-1)^n -1]Compute that:2a_n + (3/2)[(-1)^n -1] = 2[2^n + (-1)^{n+1}] + (3/2)(-1)^n - 3/2= 2^{n+1} + 2(-1)^{n+1} + (3/2)(-1)^n - 3/2Note that 2(-1)^{n+1} = -2(-1)^nSo,= 2^{n+1} - 2(-1)^n + (3/2)(-1)^n - 3/2Combine the (-1)^n terms:= 2^{n+1} - [2 - 3/2](-1)^n - 3/2= 2^{n+1} - (1/2)(-1)^n - 3/2Which matches the expression we had earlier for S_n. So that's correct. Therefore, the formula for a_n is correct.So, a_n = 2^n + (-1)^{n+1}Now, we need to find p and q such that a₁, a_p, a_q form an arithmetic sequence. Given that a₁ = 3, so:The arithmetic sequence condition is 2a_p = a₁ + a_qSubstitute a₁ = 3, a_p = 2^p + (-1)^{p+1}, a_q = 2^q + (-1)^{q+1}Thus:2[2^p + (-1)^{p+1}] = 3 + 2^q + (-1)^{q+1}Simplify:2^{p+1} + 2(-1)^{p+1} = 3 + 2^q + (-1)^{q+1}Let's rearrange:2^{p+1} - 2^q + 2(-1)^{p+1} - (-1)^{q+1} = 3Hmm, this seems a bit complicated. Maybe we can write 2(-1)^{p+1} as -2(-1)^p, and (-1)^{q+1} as -(-1)^q.Wait, let me see:2(-1)^{p+1} = 2*(-1)^{p}*(-1) = -2(-1)^pSimilarly, (-1)^{q+1} = -(-1)^qTherefore, substituting back:2^{p+1} - 2^q -2(-1)^p - (-(-1)^q) = 3Wait, the term is 2(-1)^{p+1} - (-1)^{q+1} = -2(-1)^p - (-1)^{q+1}But then:= -2(-1)^p - (-1)^{q+1}But (-1)^{q+1} = -(-1)^q, so:= -2(-1)^p - (-(-1)^q) = -2(-1)^p + (-1)^qSo, the equation becomes:2^{p+1} - 2^q -2(-1)^p + (-1)^q = 3Hmm, this equation is still quite complicated. Maybe we can look for some patterns or possible values of p and q. Since 2^{p+1} and 2^q are powers of 2, their difference might be manageable.Given that 1 < p < q, p and q are integers greater than 1.Let me try small values of p and q to see if any satisfy the equation.Start with p = 2, then q must be greater than 2.Compute for p = 2:a_p = a_2 = 2^2 + (-1)^{3} = 4 - 1 = 3So, if a_p = 3, then 2a_p = 6. So, 6 = 3 + a_q => a_q = 3. So, need to find q > 2 such that a_q = 3.Check a_3: 2^3 + (-1)^4 = 8 + 1 = 9 ≠ 3a_4: 2^4 + (-1)^5 = 16 - 1 = 15 ≠ 3a_5: 32 + 1 = 33 ≠3, etc. So, no such q exists. So p=2 is not possible.Next, p=3:a_p = a_3 = 2^3 + (-1)^4 = 8 +1=9So, 2a_p = 18. Therefore, 18 = 3 + a_q => a_q = 15. Check if any a_q =15.a_4: 16 -1=15. Yes! a_4=15. Therefore, q=4.Therefore, p=3, q=4. Then q - p =1. But wait, the question says p and q are integers with 1 < p < q, which is satisfied here. But let me confirm if a₁, a_3, a_4 form an arithmetic sequence.a₁=3, a_3=9, a_4=15. The differences are 6 and 6, so yes, it's an arithmetic sequence with common difference 6. So, 3,9,15. Thus, q-p=1.But wait, the problem says "three terms a₁、 a_p、 a_q" that form an arithmetic sequence. So, if the answer is 1, but maybe there are other possibilities. Let me check other p and q.Wait, but maybe I made a mistake here. Let's verify:If p=3, q=4, then q-p=1. But the problem might have other solutions. Let's check p=4:a_p = a_4=15So, 2a_p=30. Then 30 = 3 + a_q => a_q=27.Check if any a_q=27. Let's compute a_5=32 +1=33, a_6=64 -1=63, a_7=128 +1=129, etc. 27 is not a power of 2 plus or minus 1. So, no solution here.p=5:a_p = 2^5 + (-1)^6 =32 +1=332a_p=66=3 +a_q =>a_q=63. Check a_6=64 -1=63. Yes, a_6=63. So q=6.So p=5, q=6. Then q-p=1 again. Similarly, 3,33,63. The differences are 30 and 30. So arithmetic sequence. Thus, q-p=1.Wait, so maybe the difference is always 1? But according to the problem statement, it's supposed to have three terms a1, ap, aq with p and q integers, 1 <p <q. Then, the answer is q - p.But in both cases p=3,q=4 and p=5,q=6, the difference is 1. Is this the only possible solution? Let me check p=4, q=5:But a_p=15, then 2a_p=30=3 + a_q =>a_q=27. But a_5=33≠27. So no.Wait, maybe there are more possibilities. Let's check p=2:As before, p=2 gives a_p=3, so need a_q=3. But a_q for q>2 is not 3. So, no.p=6:a_p=63. Then 2a_p=126=3 +a_q =>a_q=123. Check a_7=128 +1=129, a_8=256 -1=255. 123 isn't in the sequence. So no.p=1 is not allowed since p>1.Wait, so so far, p=3,q=4 and p=5,q=6. Then, in both cases, q-p=1.But the problem states that "form an arithmetic sequence". So, are there multiple solutions, but they all result in q-p=1? Then the answer is 1.But let's check if there are other possibilities.Let me check p=4:a_p=15. Then 2*15=30=3 + a_q => a_q=27. Is there a term a_n=27? Let's compute a_n=2^n + (-1)^{n+1}Check n=5: 32 +1=33≠27n=4: 16 -1=15≠27n=6: 64 -1=63≠27n=3:8 +1=9≠27So no. So no solution here.Similarly, p=5 gives a_p=33. Then 2*33=66=3 +a_q =>a_q=63. Which is a_6=63. So q=6. q-p=1.Similarly, if we take p=6, then a_p=63. 2*63=126=3 +a_q =>a_q=123. But a_7=129, a_8=255, etc. So no.Similarly, p=7: a_p=129. 2*129=258=3 +a_q =>a_q=255. Which is a_8=255. So then p=7,q=8. q-p=1.So, the pattern seems to be that for p=2k+1, q=2k+2, with k≥1, q-p=1. Therefore, the difference is always 1.But wait, let me check:a_p =2^p + (-1)^{p+1}For odd p: If p is odd, then (-1)^{p+1}=(-1)^{even}=1. So a_p=2^p +1.Similarly, for q=p+1 (even q), a_q=2^{q} + (-1)^{q+1}=2^{p+1} + (-1)^{p+2}=2^{p+1} -1.So, let's check if these satisfy 2a_p =3 +a_q.2*(2^p +1) =3 +2^{p+1} -1Left side:2^{p+1} +2Right side:2^{p+1} +2So equality holds. Therefore, for any odd p, q=p+1, this equation holds.Therefore, there are infinitely many solutions where p is odd, q=p+1. Hence, q-p=1.But the problem states that {a_n} contains three terms a₁, a_p, a_q. So since the sequence is infinite, there are infinitely many such pairs, but the question is asking for q-p. Since all such pairs have q-p=1, the answer is 1.But the problem is given in Chinese, maybe there's something lost in translation, but according to the problem statement, it's likely that the answer is 1.But wait, let me check once again. For p=3 and q=4: a₁=3, a_3=9, a_4=15. 3,9,15: common difference 6.Similarly, p=5, q=6: a₁=3, a_5=33, a_6=63. Differences:30 and 30. Yes.Similarly, p=7,q=8: 3, 129, 255. Differences 126, 126.But the problem says "contains three terms", so there are multiple possibilities, but all have q-p=1. So the answer should be 1.But wait, the problem might have a different solution. Wait, let's check again.Wait, the problem says that the three terms form an arithmetic sequence, but in the problem statement, it's possible that the three terms are not consecutive terms of the arithmetic sequence. But in our case, the terms a₁, a_p, a_q form an arithmetic sequence, so the indices are 1, p, q. So the positions 1, p, q must be in arithmetic progression? No, the terms themselves, regardless of their positions, should form an arithmetic sequence. So, the terms a₁, a_p, a_q must satisfy 2a_p = a₁ + a_q. That's all.But according to our analysis, this is true for any odd p and q=p+1. Hence, q-p=1.But in the problem statement, maybe it's required that p and q are such that the terms are in positions forming an arithmetic sequence, but no, the problem says the three terms form an arithmetic sequence, regardless of their positions.So, according to our deduction, the answer should be 1.But let's check again with p=3 and q=4:a₁=3, a_p=9, a_q=15. 3,9,15 is indeed an arithmetic sequence with common difference 6.Similarly, p=5 and q=6: 3,33,63 with common difference 30.So, in each case, q-p=1.Therefore, the answer is 1.But wait, the problem is given in a math competition style, and often such problems have unique answers, so maybe it's 2 or another number. But according to our analysis, the answer is 1. Hmm.Wait, perhaps I made a mistake in the general solution.Wait, let me re-examine the recurrence relation.We had the recurrence a_n = 2a_{n-1} -3(-1)^n.We solved it by finding the homogeneous solution and a particular solution.The homogeneous solution is C*2^{n-1}, correct.For the particular solution, we assumed A(-1)^n. Let's verify:If particular solution is A(-1)^n, then substituting into the equation:A(-1)^n = 2A(-1)^{n-1} -3(-1)^nDivide both sides by (-1)^{n-1}:A(-1) = 2A -3(-1)Which gives:- A = 2A +3-3A =3 => A= -1. So that's correct. So particular solution is -(-1)^n = (-1)^{n+1}. So the general solution is C*2^{n-1} + (-1)^{n+1}Using a₁=3:3= C*2^{0} + (-1)^2 = C +1 => C=2. So a_n=2*2^{n-1} + (-1)^{n+1}=2^n + (-1)^{n+1}Yes, this is correct.Therefore, the formula is correct.So, the problem reduces to solving 2a_p =3 + a_q, which leads to q=p+1 for odd p, hence q-p=1.Therefore, the answer is 1.But let's check the problem statement again. It says "three terms a₁, a_p, a_q (p, q ∈ ℤ₊, 1 < p < q) that form an arithmetic sequence". So, p must be greater than 1, and q greater than p.So, according to our solution, the difference q - p is 1. Therefore, the answer is 1.But perhaps the problem is expecting a different answer? Maybe I need to check if there are other solutions where q - p is different.Suppose there exists p and q where q - p is not 1. Let's assume such a case.For example, suppose p=3 and q=5. Then, check if 2a_3 = a₁ + a_5.Compute a_3=9, a_5=33. So, 2*9=18, and 3 +33=36. Not equal. So no.p=3, q=6: 2*9=18 vs 3 +63=66. Not equal.p=4, q=6: 2*15=30 vs 3 +63=66. Not equal.p=2, q=3: 2*3=6 vs 3 +9=12. Not equal.p=5, q=7: 2*33=66 vs 3 +129=132. Not equal.Alternatively, maybe there is a case where q-p=2.But in all the tried cases, the only solutions are q-p=1.Therefore, the answer is 1.Wait, but the problem says "then q - p =". Maybe in Chinese math problems, sometimes answers are different. Wait, perhaps the user expects a different answer. Let me check the calculation again.Wait, when p is even, for example p=2, a_p=3, so 2a_p=6=3 + a_q =>a_q=3. But a_q=3 only when q=1 or q=2, but q>p=2, so no solution. So p=2 not possible.Similarly, for p=4, a_p=15, then 2*15=30=3 +a_q =>a_q=27. But there is no a_q=27. So no solution.For p=6, a_p=63, then 2*63=126=3 +a_q=123, but a_q=123 is not in the sequence.Therefore, the only solutions are when p is odd and q=p+1, which gives q-p=1.Hence, the answer is 1. Therefore, I think the correct answer is boxed{1}.Final Answerboxed{1}</think>Given the sequence {a_n} with the sum of the first n terms as S_n, and the conditions a_1 = 3 and S_n = 2a_n + frac{3}{2}left((-1)^n - 1right), we need to find q - p if a_1, a_p, a_q form an arithmetic sequence with 1 < p < q.First, we derive the recurrence relation for a_n using the given sum formula S_n:[S_n = 2a_n + frac{3}{2}left((-1)^n - 1right)][S_{n-1} = 2a_{n-1} + frac{3}{2}left((-1)^{n-1} - 1right)]Since S_n = S_{n-1} + a_n, we equate and simplify:[2a_n + frac{3}{2}left((-1)^n - 1right) = 2a_{n-1} + frac{3}{2}left((-1)^{n-1} - 1right) + a_n][a_n = 2a_{n-1} - 3(-1)^n]Solving this recurrence relation with a_1 = 3, we find the general solution:[a_n = 2^n + (-1)^{n+1}]Next, we check the condition for a_1, a_p, a_q to form an arithmetic sequence:[2a_p = a_1 + a_q]Substituting a_1 = 3, a_p = 2^p + (-1)^{p+1}, and a_q = 2^q + (-1)^{q+1}:[2(2^p + (-1)^{p+1}) = 3 + 2^q + (-1)^{q+1}]Testing small values of p and q, we find that for odd p, q = p + 1 satisfies the equation. For example:- p = 3, q = 4: a_3 = 9, a_4 = 15 gives 2 cdot 9 = 3 + 15- p = 5, q = 6: a_5 = 33, a_6 = 63 gives 2 cdot 33 = 3 + 63Thus, the difference q - p = 1 for all valid pairs (p, q). Therefore, the answer is:[boxed{1}]]

🔑:4. 1.begin{array}{l}text { Given } S_{n}=2 a_{n}+frac{3}{2}left((-1)^{n}-1right) Rightarrow S_{n-1}=2 a_{n-1}+frac{3}{2}left((-1)^{n-1}-1right)(n geqslant 2) .end{array}Subtracting the two equations gives a_{n}=2 a_{n-1}-3(-1)^{n}(n geqslant 2).Let b_{n}=frac{a_{n}}{(-1)^{n}}. Thenbegin{array}{l}b_{n}=-2 b_{n-1}-3 Rightarrow b_{n}+1=-2left(b_{n-1}+1right) quad=(-2)^{n-1}left(b_{1}+1right)=(-2)^{n} Rightarrow b_{n}=(-2)^{n}-1 Rightarrow a_{n}=2^{n}-(-1)^{n} .end{array}Assume there exists a positive integer c such that a_{1} 、 a_{n} 、 a_{n+c} form an arithmetic sequence. Thenbegin{array}{l}2left(2^{n}-(-1)^{n}right)=2^{n+c}-(-1)^{n+c}+3 Rightarrow 2^{n+1}left(2^{c-1}-1right)+(-1)^{n}left(2-(-1)^{c}right)+3=0 .end{array}Clearly, the equation holds if and only if c=1 and n is a positive odd number.Thus, p=n, q=n+1 Rightarrow q-p=1.