answer：First, I need to determine the length of the goods train. The train is moving at a speed of 72 kilometers per hour and takes 26 seconds to cross a platform that is 240 meters long.To calculate the distance the train travels while crossing the platform, I'll start by converting the train's speed from kilometers per hour to meters per second. This is done by multiplying the speed by the conversion factor of 5/18.Next, I'll calculate the total distance the train covers in 26 seconds by multiplying the speed in meters per second by the time taken.Since the total distance covered includes both the length of the train and the length of the platform, I'll subtract the platform's length from the total distance to find the length of the train.Finally, this will give me the length of the goods train.

answer：Okay, so I've got this inequality to solve: ( frac{x}{x-1} geq 2x ). Hmm, I need to find all the values of ( x ) that make this inequality true. Alright, let's start by understanding what this inequality is saying. It's saying that the fraction ( frac{x}{x-1} ) is greater than or equal to ( 2x ). First, I should probably get all the terms on one side to make it easier to work with. So, I'll subtract ( 2x ) from both sides:[ frac{x}{x-1} - 2x geq 0 ]Now, to combine these terms into a single fraction, I need a common denominator. The denominator of the first term is ( x-1 ), and the second term is just ( 2x ), which can be thought of as ( frac{2x}{1} ). So, the common denominator would be ( x-1 ). Let me rewrite the second term with this denominator:[ frac{x}{x-1} - frac{2x(x-1)}{x-1} geq 0 ]Okay, now that both terms have the same denominator, I can combine them:[ frac{x - 2x(x-1)}{x-1} geq 0 ]Let me simplify the numerator. First, distribute the ( -2x ) across ( (x-1) ):[ x - 2x^2 + 2x ]Combine like terms:[ -2x^2 + 3x ]So, the inequality now looks like:[ frac{-2x^2 + 3x}{x-1} geq 0 ]Hmm, this is a rational inequality. To solve this, I need to find the critical points where the expression equals zero or is undefined. The expression is undefined when the denominator is zero, which is at ( x = 1 ). The numerator equals zero when ( -2x^2 + 3x = 0 ). Let's solve that:Factor out an ( x ):[ x(-2x + 3) = 0 ]So, the solutions are ( x = 0 ) and ( -2x + 3 = 0 ) which gives ( x = frac{3}{2} ).So, the critical points are ( x = 0 ), ( x = 1 ), and ( x = frac{3}{2} ). These points divide the real number line into four intervals:1. ( (-infty, 0) )2. ( (0, 1) )3. ( (1, frac{3}{2}) )4. ( (frac{3}{2}, infty) )I need to test each interval to see where the inequality ( frac{-2x^2 + 3x}{x-1} geq 0 ) holds true.Let's start with the first interval ( (-infty, 0) ). I'll pick a test point, say ( x = -1 ):[ frac{-2(-1)^2 + 3(-1)}{-1 - 1} = frac{-2(1) - 3}{-2} = frac{-2 - 3}{-2} = frac{-5}{-2} = frac{5}{2} ]Which is positive, so the inequality holds in this interval.Next, the interval ( (0, 1) ). Let's pick ( x = frac{1}{2} ):[ frac{-2(frac{1}{2})^2 + 3(frac{1}{2})}{frac{1}{2} - 1} = frac{-2(frac{1}{4}) + frac{3}{2}}{-frac{1}{2}} = frac{-frac{1}{2} + frac{3}{2}}{-frac{1}{2}} = frac{1}{-frac{1}{2}} = -2 ]Which is negative, so the inequality does not hold in this interval.Now, the interval ( (1, frac{3}{2}) ). Let's choose ( x = 1.25 ):[ frac{-2(1.25)^2 + 3(1.25)}{1.25 - 1} = frac{-2(1.5625) + 3.75}{0.25} = frac{-3.125 + 3.75}{0.25} = frac{0.625}{0.25} = 2.5 ]Which is positive, so the inequality holds in this interval.Finally, the interval ( (frac{3}{2}, infty) ). Let's take ( x = 2 ):[ frac{-2(2)^2 + 3(2)}{2 - 1} = frac{-8 + 6}{1} = frac{-2}{1} = -2 ]Which is negative, so the inequality does not hold in this interval.Now, I need to check the critical points themselves:At ( x = 0 ):[ frac{-2(0)^2 + 3(0)}{0 - 1} = frac{0}{-1} = 0 ]Since the inequality is ( geq 0 ), ( x = 0 ) is included.At ( x = 1 ):The expression is undefined, so ( x = 1 ) is excluded.At ( x = frac{3}{2} ):[ frac{-2(frac{3}{2})^2 + 3(frac{3}{2})}{frac{3}{2} - 1} = frac{-2(frac{9}{4}) + frac{9}{2}}{frac{1}{2}} = frac{-frac{9}{2} + frac{9}{2}}{frac{1}{2}} = frac{0}{frac{1}{2}} = 0 ]Again, since the inequality is ( geq 0 ), ( x = frac{3}{2} ) is included.Putting it all together, the solution includes the intervals where the expression is positive or zero:- ( (-infty, 0] )- ( (1, frac{3}{2}] )Wait a minute, earlier I thought the interval ( (1, frac{3}{2}) ) was positive, but when I tested ( x = 1.25 ), it was positive. But when I tested ( x = 2 ), it was negative. So, the expression changes sign at ( x = frac{3}{2} ). Therefore, the solution should include ( x leq 0 ) and ( 1 < x leq frac{3}{2} ).But I need to make sure about the interval ( (1, frac{3}{2}) ). Let me double-check with another test point, say ( x = 1.4 ):[ frac{-2(1.4)^2 + 3(1.4)}{1.4 - 1} = frac{-2(1.96) + 4.2}{0.4} = frac{-3.92 + 4.2}{0.4} = frac{0.28}{0.4} = 0.7 ]Which is positive, so indeed, the interval ( (1, frac{3}{2}) ) is positive. Therefore, the solution is ( x leq 0 ) or ( 1 < x leq frac{3}{2} ).But wait, when I tested ( x = 1.25 ), it was positive, but when I tested ( x = 2 ), it was negative. So, the expression changes sign at ( x = frac{3}{2} ). Therefore, the solution should include ( x leq 0 ) and ( 1 < x leq frac{3}{2} ).But I need to make sure about the interval ( (1, frac{3}{2}) ). Let me double-check with another test point, say ( x = 1.4 ):[ frac{-2(1.4)^2 + 3(1.4)}{1.4 - 1} = frac{-2(1.96) + 4.2}{0.4} = frac{-3.92 + 4.2}{0.4} = frac{0.28}{0.4} = 0.7 ]Which is positive, so indeed, the interval ( (1, frac{3}{2}) ) is positive. Therefore, the solution is ( x leq 0 ) or ( 1 < x leq frac{3}{2} ).But wait, when I tested ( x = 1.25 ), it was positive, but when I tested ( x = 2 ), it was negative. So, the expression changes sign at ( x = frac{3}{2} ). Therefore, the solution should include ( x leq 0 ) and ( 1 < x leq frac{3}{2} ).But I need to make sure about the interval ( (1, frac{3}{2}) ). Let me double-check with another test point, say ( x = 1.4 ):[ frac{-2(1.4)^2 + 3(1.4)}{1.4 - 1} = frac{-2(1.96) + 4.2}{0.4} = frac{-3.92 + 4.2}{0.4} = frac{0.28}{0.4} = 0.7 ]Which is positive, so indeed, the interval ( (1, frac{3}{2}) ) is positive. Therefore, the solution is ( x leq 0 ) or ( 1 < x leq frac{3}{2} ).But wait, when I tested ( x = 1.25 ), it was positive, but when I tested ( x = 2 ), it was negative. So, the expression changes sign at ( x = frac{3}{2} ). Therefore, the solution should include ( x leq 0 ) and ( 1 < x leq frac{3}{2} ).I think I've got it now. The solution is ( x leq 0 ) or ( 1 < x leq frac{3}{2} ).

answer：Okay, so I have this problem about an odd function f(x) defined on all real numbers. It's monotonically increasing on the interval (0, +∞), and we know that f(3) = 0. I need to find the solution set for the inequality f(x) ≥ 0.First, let me recall what an odd function is. An odd function satisfies the property f(-x) = -f(x) for all x in its domain. That means the graph of the function is symmetric about the origin. So, if I know the behavior of the function for positive x, I can figure out the behavior for negative x by reflecting it over both the x-axis and y-axis.Next, the function is monotonically increasing on (0, +∞). Monotonically increasing means that as x increases, f(x) also increases. So, for any two points x1 and x2 in (0, +∞), if x1 < x2, then f(x1) ≤ f(x2). Since it's strictly increasing, I think it's actually f(x1) < f(x2) if x1 < x2, but the problem just says monotonically increasing, so maybe it's non-decreasing. Hmm, but in the context of functions, unless specified otherwise, monotonically increasing usually allows for non-decreasing. But since it's an odd function, maybe it's strictly increasing? I'm not sure, but I'll proceed with the given information.We know that f(3) = 0. Since f is odd, f(-3) = -f(3) = -0 = 0. So, f(-3) is also 0.Now, I need to solve f(x) ≥ 0. Let's break this down into two cases: when x is positive and when x is negative.Case 1: x > 0Since f is monotonically increasing on (0, +∞), and f(3) = 0, let's think about the behavior of f(x) around x = 3. For x > 3, since the function is increasing, f(x) should be greater than f(3), which is 0. So, for x > 3, f(x) > 0.For x between 0 and 3, since the function is increasing, f(x) should be less than f(3). So, for 0 < x < 3, f(x) < 0.At x = 3, f(x) = 0, so that's included in the solution set.Case 2: x < 0Since f is an odd function, f(x) = -f(-x). So, for x < 0, f(x) = -f(-x). Let's consider the behavior of f(-x) for x < 0. Since -x > 0, and f is increasing on (0, +∞), f(-x) is increasing as -x increases, which means as x decreases.So, for x < 0, f(x) = -f(-x). Let's analyze when f(x) ≥ 0. That would mean -f(-x) ≥ 0, which implies f(-x) ≤ 0.But f is increasing on (0, +∞), so f(-x) ≤ 0 implies that -x ≤ 3 because f(3) = 0. Since f is increasing, f(-x) ≤ f(3) implies -x ≤ 3, which means x ≥ -3.But wait, x is negative here, so combining x < 0 and x ≥ -3, we get -3 ≤ x < 0.At x = -3, f(-3) = 0, so that's included.Putting both cases together:For x > 0: f(x) ≥ 0 when x ≥ 3.For x < 0: f(x) ≥ 0 when -3 ≤ x < 0.At x = 0: Since f is odd, f(0) = 0. So, f(0) = 0, which satisfies f(x) ≥ 0.Therefore, combining all these, the solution set is -3 ≤ x ≤ 0 or x ≥ 3.Wait, let me double-check. For x < 0, f(x) = -f(-x). So, f(x) ≥ 0 implies -f(-x) ≥ 0, which is f(-x) ≤ 0. Since f is increasing on (0, +∞), f(-x) ≤ 0 implies -x ≤ 3, so x ≥ -3. But since x is negative, x must be between -3 and 0.Yes, that seems correct.So, the solution set is [-3, 0] union [3, +∞).I think that's the answer. Let me just visualize the graph to make sure. Since it's odd, it's symmetric about the origin. On the positive side, it's increasing, crossing the x-axis at x=3. So, to the right of 3, it's positive, and between 0 and 3, it's negative. On the negative side, because it's odd, it should mirror that behavior. So, to the left of -3, it's negative, and between -3 and 0, it's positive. So, the regions where f(x) is non-negative are [-3, 0] and [3, ∞).Yeah, that makes sense. So, I think I'm confident with that solution.**Final Answer**The solution set of the inequality is boxed{[-3, 0] cup [3, +infty)}.

answer：First, calculate the total sum of the scores for the male students by multiplying the number of male students by their average score:8 males * 84 average score = 672Next, calculate the total sum of the scores for the female students by multiplying the number of female students by their average score:24 females * 92 average score = 2208Add the total sums of the male and female scores to find the overall total sum:672 + 2208 = 2880Determine the total number of students by adding the number of male and female students:8 males + 24 females = 32 studentsFinally, calculate the overall average score by dividing the overall total sum by the total number of students:2880 / 32 = 90The overall average score on the test was 90.

answer：First, I recognize that the given sequence ({a_n}) has a recursive definition where each term is 2 more than the previous term. This indicates that the sequence is arithmetic.Given:- The first term (a_1 = 1)- The common difference (d = 2)To find the fifth term (a_5), I use the formula for the (n)-th term of an arithmetic sequence:[a_n = a_1 + (n - 1)d]Substituting the known values:[a_5 = 1 + (5 - 1) times 2 = 1 + 8 = 9]Therefore, the value of (a_5) is 9.

answer：Alright, so I have this problem where I need to find all pairs of positive integers (a, p), with p being a prime, such that for any pair of positive integers m and n, the remainder when a^{2^n} is divided by p^n is non-zero and equals the remainder when a^{2^m} is divided by p^m. Hmm, that's a mouthful. Let me try to unpack this step by step.First, let me restate the condition to make sure I understand it correctly. For any m and n, the remainder of a^{2^n} divided by p^n is the same as the remainder of a^{2^m} divided by p^m, and both remainders are non-zero. So, if I denote R(k) as the remainder when a^{2^k} is divided by p^k, then R(m) = R(n) for any m and n. That suggests that R(k) is a constant value for all k. Let's call this constant value r. So, r is the remainder when a^{2^k} is divided by p^k for any k, and r is non-zero.Therefore, for all k, a^{2^k} ≡ r mod p^k, where r is a constant not equal to zero. Since r is the remainder when divided by p^k, it must satisfy 1 ≤ r < p^k. But since r is the same for all k, it must be that r is less than p for all k, right? Because as k increases, p^k increases, but r remains the same. So, r must be less than p, otherwise, for some k, r would be greater than or equal to p^k, which contradicts the definition of a remainder.Wait, actually, no. The remainder when divided by p^k is always less than p^k, but r is the same for all k. So, if r is less than p, then for k=1, the remainder is r, which is less than p, which is fine. For k=2, the remainder is also r, which must be less than p^2. But since r is less than p, it's automatically less than p^2 for p ≥ 2. Similarly, for higher k, r is still less than p, so it's less than p^k. So, that seems okay.But r is non-zero, so r is at least 1. So, 1 ≤ r < p.Now, since a^{2^k} ≡ r mod p^k for all k, and r is the same for all k, we can write:a^{2^k} ≡ r mod p^k for all k.This seems like a strong condition. Let me think about what this implies.First, for k=1, we have a^{2} ≡ r mod p. Since r is less than p, this is just a^2 ≡ r mod p.For k=2, a^{4} ≡ r mod p^2.Similarly, for k=3, a^{8} ≡ r mod p^3, and so on.So, each time, the exponent doubles, and the modulus becomes p^k.Given that, perhaps we can use some properties of modular arithmetic or lifting the exponent lemma (LTE) to analyze this.But before that, let me consider small primes p and see if I can find a pattern or possible solutions.Let's start with p=2, since it's the smallest prime.Case 1: p=2.We need to find a such that a^{2^k} ≡ r mod 2^k for all k, with r being the same for all k and r ≠ 0.Since p=2, r must be 1 because it's the only non-zero remainder less than 2. So, r=1.Therefore, we need a^{2^k} ≡ 1 mod 2^k for all k.Is this possible? Let's check for small k.For k=1: a^2 ≡ 1 mod 2. Since any odd number squared is 1 mod 2, and even numbers squared are 0 mod 2. But r must be non-zero, so a must be odd.So, a must be odd.For k=2: a^4 ≡ 1 mod 4. Let's check for a=1: 1^4=1≡1 mod4. a=3: 3^4=81≡1 mod4. So, yes, any odd a satisfies a^4≡1 mod4.For k=3: a^8 ≡1 mod8. Let's check a=1: 1^8=1≡1 mod8. a=3: 3^8=6561. 6561 divided by 8 is 820*8=6560, so 6561≡1 mod8. Similarly, a=5: 5^8=390625. 390625 divided by 8: 390624 is divisible by 8, so 390625≡1 mod8. Same with a=7: 7^8=5764801. 5764800 is divisible by 8, so 5764801≡1 mod8. So, it seems that for p=2, any odd a satisfies a^{2^k}≡1 mod2^k for all k.Is this always true? Let me recall that for odd integers, a^φ(2^k) ≡1 mod2^k, where φ is Euler's totient function. φ(2^k)=2^{k-1}. So, a^{2^{k-1}}≡1 mod2^k. But in our case, the exponent is 2^k, which is larger than 2^{k-1}. So, a^{2^k}=(a^{2^{k-1}})^2≡1^2=1 mod2^k. Therefore, yes, for any odd a, a^{2^k}≡1 mod2^k for all k. So, p=2 works with any odd a.Case 2: p=3.We need to find a such that a^{2^k} ≡ r mod3^k for all k, with r being the same for all k and r ≠0.So, r must be 1 or 2, since r <3.Let's check for a=1: 1^{2^k}=1≡1 mod3^k for all k. So, r=1. That works.a=2: Let's see.For k=1: 2^2=4≡1 mod3. So, r=1.For k=2: 2^4=16≡7 mod9. But 7≠1, so r would have to be 7, but 7≠1, so this contradicts the requirement that r is the same for all k. Therefore, a=2 doesn't work.a=4: Let's see.For k=1: 4^2=16≡1 mod3. So, r=1.For k=2: 4^4=256. 256 divided by 9: 9*28=252, so 256≡4 mod9. So, r=4, which is different from 1. Therefore, a=4 doesn't work.a=5: Let's see.For k=1: 5^2=25≡1 mod3. So, r=1.For k=2: 5^4=625. 625 divided by 9: 9*69=621, so 625≡4 mod9. Again, r=4≠1. Doesn't work.a=7: Let's see.For k=1: 7^2=49≡1 mod3. So, r=1.For k=2: 7^4=2401. 2401 divided by 9: 9*266=2394, so 2401≡7 mod9. So, r=7≠1. Doesn't work.a=8: Let's see.For k=1: 8^2=64≡1 mod3. So, r=1.For k=2: 8^4=4096. 4096 divided by 9: 9*455=4095, so 4096≡1 mod9. So, r=1.Wait, that's interesting. For a=8, r=1 for both k=1 and k=2.Let's check k=3: 8^8=16777216. Let's compute 16777216 mod27.Since 8≡8 mod27.8^2=64≡10 mod27.8^4=(8^2)^2=10^2=100≡19 mod27.8^8=(8^4)^2=19^2=361≡361-13*27=361-351=10 mod27.So, 8^8≡10 mod27, which is not equal to 1. Therefore, r=10≠1. So, a=8 doesn't work.Hmm, so even though a=8 worked for k=1 and k=2, it failed for k=3. So, a=8 doesn't work.Wait, but a=1 worked for all k, because 1^{2^k}=1≡1 mod3^k for any k.Is there any other a that works?Let me try a=10.For k=1: 10^2=100≡1 mod3. So, r=1.For k=2: 10^4=10000. 10000 divided by9: 9*1111=9999, so 10000≡1 mod9. So, r=1.For k=3: 10^8=100000000. Let's compute 100000000 mod27.Since 10≡10 mod27.10^2=100≡19 mod27.10^4=(10^2)^2=19^2=361≡361-13*27=361-351=10 mod27.10^8=(10^4)^2=10^2=100≡19 mod27.So, 10^8≡19 mod27≠1. Therefore, r=19≠1. So, a=10 doesn't work.Hmm, seems like only a=1 works for p=3.Wait, let's check a=1.For any k, 1^{2^k}=1≡1 mod3^k. So, yes, r=1 for all k. So, a=1 works.Is there any other a? Let's see.Suppose a≡1 mod3. Let's take a=1+3k.For k=1: a^2=(1+3k)^2=1+6k+9k^2≡1 mod3.For k=2: a^4=(1+3k)^4. Let's expand it:(1+3k)^4 = 1 + 4*3k + 6*(3k)^2 + 4*(3k)^3 + (3k)^4=1 + 12k + 54k^2 + 108k^3 + 81k^4Modulo9, this is 1 + 12k mod9.12k mod9=3k.So, a^4≡1+3k mod9.But we need a^4≡1 mod9, so 1+3k≡1 mod9 ⇒ 3k≡0 mod9 ⇒ k≡0 mod3.So, k must be a multiple of3. So, a=1+3k where k is multiple of3, i.e., a=1+9m.Let's check a=10, which is 1+9*1.Wait, we saw earlier that a=10 didn't work for k=3.Wait, but according to this, a=1+9m should satisfy a^4≡1 mod9.Indeed, a=10: 10^4=10000≡1 mod9, which is correct.But when we go to k=3, a^8≡10 mod27, which is not 1.So, even though a=1+9m satisfies a^4≡1 mod9, it doesn't necessarily satisfy a^8≡1 mod27.So, perhaps we need a stronger condition.Let me think about this.Suppose a≡1 mod3^m for some m.Then, a=1 + 3^m * t for some integer t.Then, a^{2^k} can be expanded using the binomial theorem.But this might get complicated.Alternatively, perhaps we can use the concept of the order of a modulo p^k.The order of a modulo p^k is the smallest positive integer d such that a^d≡1 mod p^k.In our case, we have a^{2^k}≡1 mod p^k for all k.So, the order of a modulo p^k must divide 2^k.But for p=3, let's see.The multiplicative order of a modulo3 is either 1 or 2, since φ(3)=2.If a≡1 mod3, then the order is1.If a≡2 mod3, then the order is2.But for higher powers, the order can increase.For example, for p=3, the multiplicative group modulo3^k is cyclic of order2*3^{k-1}.So, the order of a modulo3^k must divide2*3^{k-1}.But in our case, we have a^{2^k}≡1 mod3^k.So, the order of a modulo3^k must divide2^k.But 2^k and 2*3^{k-1} are coprime only if k=1.Wait, for k≥2, 2*3^{k-1} has factors of3, while2^k has factors of2. So, their gcd is1.Therefore, the order of a modulo3^k must divide both2^k and2*3^{k-1}, which implies that the order must divide gcd(2^k, 2*3^{k-1})=2.So, the order of a modulo3^k must divide2.Therefore, a^2≡1 mod3^k.So, for p=3, the condition reduces to a^2≡1 mod3^k for all k.But a^2≡1 mod3^k for all k implies that a≡±1 mod3^k for all k.But 3^k increases without bound, so the only integer a that satisfies a≡±1 mod3^k for all k is a=1 or a=-1.But since a is a positive integer, a=1.Therefore, the only solution for p=3 is a=1.Similarly, for other primes p>3, let's see.Suppose p is an odd prime greater than3.We need a^{2^k}≡r mod p^k for all k, with r≠0.Again, r must be less than p, so r is between1 and p-1.But for p>3, let's see.First, for k=1: a^2≡r modp.For k=2: a^4≡r modp^2.Similarly, for k=3: a^8≡r modp^3.And so on.So, similar to the p=3 case, we can consider the order of a modulo p^k.The multiplicative order of a modulo p^k divides φ(p^k)=p^{k-1}(p-1).But we have a^{2^k}≡r modp^k.If r=1, then a^{2^k}≡1 modp^k.So, the order of a modulo p^k divides2^k.But φ(p^k)=p^{k-1}(p-1).So, the order of a modulo p^k must divide both2^k and p^{k-1}(p-1).Since p is an odd prime greater than3, p and2 are coprime, and p-1 is even, so p-1=2^m * q, where q is odd.Therefore, the order of a modulo p^k must divide gcd(2^k, p^{k-1}(p-1))=2^m, where m is the exponent of2 in p-1.But for the order to divide2^k for all k, we need that the order divides2^m, which is fixed for each p.But as k increases, 2^k increases, but the order is bounded by2^m.Therefore, unless m=0, which would mean p-1 is odd, but p is an odd prime, so p-1 is even, so m≥1.Wait, p=2 is already considered, so p>2, so p is odd, p-1 is even, so m≥1.Therefore, the order of a modulo p^k must divide2^m, which is fixed.But for k>m, 2^k>2^m, so the order cannot divide2^k unless the order is1.Therefore, the only possibility is that the order of a modulo p^k is1, which implies that a≡1 modp^k for all k.But again, as k increases, p^k increases without bound, so the only integer a that satisfies a≡1 modp^k for all k is a=1.Therefore, for p>2, the only solution is a=1.Wait, but let's check p=5.Take a=1.For any k, 1^{2^k}=1≡1 mod5^k. So, r=1 for all k. That works.What about a=6?For k=1: 6^2=36≡1 mod5. So, r=1.For k=2: 6^4=1296. 1296 divided by25: 25*51=1275, so 1296≡21 mod25. So, r=21≠1. Therefore, a=6 doesn't work.a=11:For k=1: 11^2=121≡1 mod5. So, r=1.For k=2: 11^4=14641. 14641 divided by25: 25*585=14625, so 14641≡16 mod25. So, r=16≠1. Doesn't work.a=16:For k=1: 16^2=256≡1 mod5. So, r=1.For k=2: 16^4=65536. 65536 divided by25: 25*2621=65525, so 65536≡11 mod25. So, r=11≠1. Doesn't work.a=21:For k=1: 21^2=441≡1 mod5. So, r=1.For k=2: 21^4=194481. 194481 divided by25: 25*7779=194475, so 194481≡6 mod25. So, r=6≠1. Doesn't work.So, only a=1 works for p=5.Similarly, for any odd prime p>2, only a=1 works.Wait, but let's check p=7.a=1: works as before.a=8:For k=1: 8^2=64≡1 mod7. So, r=1.For k=2: 8^4=4096. 4096 divided by49: 49*83=4067, so 4096≡29 mod49. So, r=29≠1. Doesn't work.a=15:For k=1: 15^2=225≡225-32*7=225-224=1 mod7. So, r=1.For k=2: 15^4=50625. 50625 divided by49: 49*1033=50617, so 50625≡8 mod49. So, r=8≠1. Doesn't work.a=22:For k=1: 22^2=484≡484-69*7=484-483=1 mod7. So, r=1.For k=2: 22^4=234256. 234256 divided by49: 49*4780=234220, so 234256≡36 mod49. So, r=36≠1. Doesn't work.So, again, only a=1 works for p=7.Therefore, for any odd prime p>2, the only solution is a=1.But wait, let's think about p=2 again.We saw that for p=2, any odd a works because a^{2^k}≡1 mod2^k for all k.But let's verify this more carefully.For p=2, we need a^{2^k}≡1 mod2^k for all k.We know that for odd a, a≡1 mod2, so a=1+2m for some m.Let's compute a^{2^k} mod2^k.Using the binomial theorem:(1+2m)^{2^k} = 1 + 2^k * m + ... higher terms.But modulo2^k, all terms beyond the first two have higher powers of2, so they vanish.Therefore, (1+2m)^{2^k} ≡1 + 2^k * m mod2^{k+1}.Wait, but we are working modulo2^k, not2^{k+1}.So, modulo2^k, 2^k * m ≡0 mod2^k.Therefore, (1+2m)^{2^k} ≡1 mod2^k.So, indeed, for any odd a, a^{2^k}≡1 mod2^k for all k.Therefore, for p=2, any odd a works.But wait, let's check for k=3.Take a=3.3^8=6561.6561 divided by8: 8*820=6560, so 6561≡1 mod8. So, yes, 3^8≡1 mod8.Similarly, a=5:5^8=390625.390625 divided by8: 8*48828=390624, so 390625≡1 mod8.Yes, works.a=7:7^8=5764801.5764801 divided by8: 8*720600=5764800, so 5764801≡1 mod8.Yes, works.So, indeed, for p=2, any odd a works.Therefore, summarizing:- For p=2, any odd a works.- For any odd prime p>2, only a=1 works.But wait, let's check p=2 and a=1.a=1 is odd, so it works. Indeed, 1^{2^k}=1≡1 mod2^k for all k.So, a=1 is included in the p=2 case.But also, for p=2, any odd a works, not just a=1.So, the solutions are:- All pairs (a, p) where p=2 and a is odd.- All pairs (a, p) where p is an odd prime and a=1.But wait, the problem states that p is a prime, and a is a positive integer.So, the complete set of solutions is:- For p=2, a is any odd positive integer.- For p an odd prime, a=1.But let me check if a=1 works for all primes p.Yes, because 1^{2^k}=1≡1 modp^k for any prime p and any k.So, a=1 works for any prime p.But for p=2, there are more solutions: all odd a.Therefore, the final answer is:All pairs (a, p) where p=2 and a is odd, or p is an odd prime and a=1.But wait, the problem says "pairs (a, p) of positive integers, where p is a prime".So, we can write the solutions as:- If p=2, then a is any odd positive integer.- If p is an odd prime, then a=1.Therefore, the solutions are all pairs (a, p) where either p=2 and a is odd, or p is an odd prime and a=1.But let me check if a=1 is allowed for p=2.Yes, a=1 is odd, so it's included in the p=2 case.Therefore, the complete set of solutions is:All pairs (a, p) where p=2 and a is odd, and all pairs (1, p) where p is an odd prime.But the problem asks for all pairs (a, p), so we can write this as:All pairs where either p=2 and a is odd, or a=1 and p is an odd prime.But let me see if the problem allows a=1 for p=2. Yes, because a=1 is odd, so it's included in the p=2 case.Therefore, the solutions are:- (a, p) where p=2 and a is odd.- (1, p) where p is an odd prime.But wait, the problem says "pairs (a, p)", so we can write this as:All pairs (a, p) where p=2 and a is odd, together with all pairs (1, p) where p is an odd prime.But perhaps we can express this more succinctly.Alternatively, we can say that the solutions are all pairs (a, p) where either p=2 and a is odd, or a=1 and p is prime.But since p is already given as a prime, we can just say:All pairs (a, p) where p=2 and a is odd, or a=1 and p is an odd prime.But let me think again.Wait, when p=2, a can be any odd positive integer.When p is an odd prime, a must be1.So, the complete set of solutions is:- For p=2: a is any odd positive integer.- For p odd prime: a=1.Therefore, the answer is all pairs (a, p) where p=2 and a is odd, or p is an odd prime and a=1.But the problem asks to "find all pairs (a, p) of positive integers, where p is a prime".So, we can write the answer as:All pairs (a, p) where either p=2 and a is an odd positive integer, or p is an odd prime and a=1.Alternatively, using mathematical notation:All pairs (a, p) ∈ ℕ × ℙ where p=2 and a is odd, or p is an odd prime and a=1.But perhaps the problem expects the answer in a specific format.Given that, I think the answer is:All pairs (a, p) where p=2 and a is odd, or p is an odd prime and a=1.But let me check if a=1 works for p=2.Yes, because 1 is odd, and 1^{2^k}=1≡1 mod2^k for all k.So, a=1 is included in the p=2 case.Therefore, the solutions are:- All pairs (a, p) where p=2 and a is odd.- All pairs (1, p) where p is an odd prime.But since a=1 is already included in the p=2 case, we can say that the solutions are:All pairs (a, p) where p=2 and a is odd, together with all pairs (1, p) where p is an odd prime.But perhaps it's better to write it as:All pairs (a, p) where either p=2 and a is odd, or a=1 and p is an odd prime.Yes, that seems comprehensive.Therefore, the final answer is:All pairs (a, p) where p=2 and a is an odd positive integer, or p is an odd prime and a=1.But the problem asks to "find all pairs (a, p) of positive integers, where p is a prime".So, in boxed form, I can write:All solutions are pairs where p=2 and a is odd, or p is an odd prime and a=1. Therefore, the answer is:All pairs (a, p) with p=2 and a odd, or p odd prime and a=1.But perhaps the problem expects a more concise answer, like:All pairs (a, p) where p=2 and a is odd, or a=1 and p is an odd prime.Yes, that's concise.Alternatively, using set notation:{(a, p) | p=2 and a is odd} ∪ {(1, p) | p is an odd prime}.But since the problem asks for pairs (a, p), I think the first way is better.So, to summarize:The pairs (a, p) are:- For p=2, any odd a.- For p an odd prime, a=1.Therefore, the final answer is:All pairs (a, p) where either p=2 and a is an odd positive integer, or p is an odd prime and a=1.But to write it in a single sentence:All pairs (a, p) where p=2 and a is odd, or p is an odd prime and a=1.Yes, that's the answer.