🧠:Alright, so I need to figure out whether there are more rectangles with integer sides for a perimeter of 1996 or 1998. Hmm, okay. Let's start by recalling that the perimeter of a rectangle is given by P = 2*(length + width). Since we're dealing with integer sides, both the length and width must be positive integers. Also, the problem mentions that rectangles a×b and b×a are considered the same, so we have to account for that to avoid double-counting.First, let's express the perimeter formula in terms of variables. Let the sides be a and b, then 2*(a + b) = P. So for a given perimeter P, the sum of the sides a + b would be P/2. Therefore, for perimeter 1996, a + b = 1996/2 = 998, and for perimeter 1998, a + b = 1998/2 = 999. So the problem reduces to finding how many pairs (a, b) with a ≤ b satisfy a + b = 998 or 999. Since a and b are positive integers, a can range from 1 up to floor((P/2)/2) because beyond that, a would be greater than b, which we don't want due to considering them identical.Wait, let's clarify that. If a + b = S (where S is 998 or 999), then since a ≤ b, a can be from 1 to floor(S/2). Because if a were greater than floor(S/2), then a would be greater than b, which contradicts a ≤ b. So the number of rectangles is equal to the number of integer solutions (a, b) with a ≤ b and a + b = S. So for each perimeter, the number of rectangles is floor(S/2). Is that right?Wait, let's test with a small perimeter. Suppose perimeter is 4, then S = 2. Then a can be 1 and b =1. So floor(2/2)=1. Correct. Another example: perimeter 6, S=3. Then a can be 1, b=2. floor(3/2)=1. Wait, but that's only one rectangle. But 1x2 is the same as 2x1, so yes. Wait, but perimeter 6 allows for 1x2, which is the only rectangle. So floor(S/2) gives 1, which is correct. Another example: perimeter 8, S=4. Then a can be 1 and b=3, 2 and 2. So two rectangles. floor(4/2)=2. Correct. So that seems to hold. So the number of rectangles is floor(S/2). So for S even and odd?Wait, let's check with S even and odd. For S even: S=4, floor(4/2)=2. Which gives two rectangles: 1x3 and 2x2. Wait, no, 1x3 has perimeter 8? Wait, 2*(1+3)=8. Yes. So 1x3 and 2x2. So two rectangles. So floor(4/2)=2. Correct. For S=5 (perimeter 10), floor(5/2)=2. Then the rectangles would be 1x4 and 2x3. So two rectangles. Which is correct. So indeed, the number of rectangles is floor(S/2). So in general, the number of rectangles with integer sides for a perimeter P is floor(S/2), where S = P/2.Therefore, for the given problem, we need to compute floor(998/2) and floor(999/2). Let's compute them:For perimeter 1996: S = 998. floor(998/2) = 499.For perimeter 1998: S = 999. floor(999/2) = 499.5, but since we take the floor, it's 499.Wait, so both would have 499 rectangles? But that can't be right. Because when S is even or odd, there's a difference. Wait, let me re-examine the formula. For S even: S = 2k. Then the number of solutions is k - 1 +1? Wait, no.Wait, perhaps the formula is different. Let me think again.Suppose S = a + b, with a ≤ b, and a and b positive integers. Then a can range from 1 to floor(S/2). Because if a is 1, then b is S - 1. If a is 2, then b is S - 2, and so on, until a is floor(S/2). For example, if S is even, say S=4, then a can be 1 and 2. So floor(4/2) = 2. So 2 solutions: 1x3 and 2x2. If S is odd, say S=5, then floor(5/2) = 2, so a can be 1 and 2. Then b is 4 and 3, respectively. So 2 solutions: 1x4 and 2x3. So yes, in both cases, the number of solutions is floor(S/2). So even if S is even or odd, floor(S/2) gives the correct number of rectangles.Wait, but when S is even, S = 2k, floor(S/2) = k. So number of rectangles is k. For example, S=4, k=2. Then two rectangles. But when S is even, the middle point is when a = b. For S even, a = b = k is allowed (since 2k = S). So that is a square. For S odd, S=2k +1, then floor(S/2)=k. So a can be from 1 to k, and b is S - a. Since S is odd, all pairs will have a < b. Because a can be up to k, but 2k +1 -k = k +1, which is greater than k. So no square in that case.Therefore, for S even, the number of rectangles is S/2. Wait, S even: S=2k. Then floor(S/2)=k. So the number of rectangles is k. For example, S=4: 2 rectangles. S=6: 3 rectangles (1x5, 2x4, 3x3). Wait, 3x3 is a square, so that's correct. So yes, S even: floor(S/2) = S/2, which is integer. For S odd: floor(S/2)= (S -1)/2. So the number of rectangles is (S -1)/2.Therefore, in general, the number of rectangles is floor(S/2). So for S even, it's S/2, and for S odd, it's (S -1)/2.But let's compute for S=998 and S=999.For perimeter 1996: S=998, which is even. So number of rectangles = 998/2 = 499.For perimeter 1998: S=999, which is odd. So number of rectangles = (999 -1)/2 = 998/2 = 499.Wait, so both perimeters would have 499 rectangles? That seems odd. But according to the formula, they do. But maybe that's correct? Wait, let's test with similar smaller perimeters.Take perimeter 4 (S=2) vs perimeter 6 (S=3). For S=2: 1 rectangle (1x1). For S=3: 1 rectangle (1x2). So same number. For perimeter 8 (S=4) vs perimeter 10 (S=5). S=4: 2 rectangles (1x3, 2x2). S=5: 2 rectangles (1x4, 2x3). So same number again. So it seems that when you have perimeters differing by 2 (i.e., S differing by 1), the number of rectangles is the same if one S is even and the next is odd.Wait, so maybe when S increases by 1 (from even to odd), the number of rectangles remains the same. Because floor(S/2) for S even is S/2, and for S odd, it's (S-1)/2. So if S is even: 998/2=499. If S is 999: 999/2=499.5, floor is 499. So same number. So both perimeters have the same number of rectangles? But the problem is asking which has more. So the answer would be they have the same number? But maybe I made a mistake here.Wait, let's check another example. Suppose perimeter 12 (S=6) and 14 (S=7). For S=6: 3 rectangles (1x5, 2x4, 3x3). For S=7: 3 rectangles (1x6, 2x5, 3x4). So same number. Similarly, perimeter 16 (S=8) and 18 (S=9). For S=8: 4 rectangles (1x7, 2x6, 3x5, 4x4). For S=9: 4 rectangles (1x8, 2x7, 3x6, 4x5). Same number. So it seems that when you have perimeters P and P+2, where P is even, then the number of rectangles is the same.Therefore, according to this, both 1996 and 1998 perimeters would have the same number of rectangles, which is 499. But that contradicts the problem's implication that one has more than the other. Maybe the problem is trying to trick us, but according to the math, they are equal.Wait, but let me think again. Maybe when S is even, there is a square (a = b), but when S is odd, there is no square. So the number of rectangles for even S is (S/2 -1) +1 = S/2. For odd S: (S-1)/2. So actually, for even S, you have S/2 rectangles, which includes one square. For odd S, you have (S-1)/2 rectangles, all non-square. But in terms of count, they are equal when S is even and S+1 is odd.Wait, but in our previous examples, when S=6 (even), number of rectangles is 3, which includes 3x3. When S=7 (odd), number of rectangles is 3, which are all non-square. So the count is same. So even though even S has a square, the count is same as the next odd S. So in the problem's case, S=998 (even) has 499 rectangles, including a square (499x499?), wait no. Wait, S=998, a + b =998, so if a = b, then a = 499. So 499x499. Yes. So that's a square. For S=999 (odd), there's no square, so all rectangles are distinct a < b. But the number of rectangles is (999 -1)/2=499. So same number. So both perimeters have the same number of rectangles. Therefore, the answer is that they have the same number.But the problem is phrased: "Which has more rectangles with integer sides: those with a perimeter of 1996 or those with a perimeter of 1998?" So if they are the same, then the answer is neither, they are equal. But maybe the problem is expecting a different answer. Wait, perhaps I made a mistake in my reasoning.Wait, let's consider the formula again. The number of rectangles with integer sides and perimeter P is equal to the number of integer solutions (a, b) with a ≤ b and 2(a + b) = P. Therefore, a + b = P/2. Let S = P/2. The number of solutions is floor(S/2). Wait, but let's verify with another example. If P=8, S=4, floor(4/2)=2. The solutions are (1,3) and (2,2). So 2 rectangles. If P=10, S=5, floor(5/2)=2. The solutions are (1,4) and (2,3). So 2 rectangles. So same number. If P=6, S=3, floor(3/2)=1. Solution (1,2). P=4, S=2, floor(2/2)=1. Solution (1,1). So in these cases, when P increases by 2 (so S increases by 1), the number of rectangles remains the same if S was even and becomes odd. So for S even: floor(S/2)=S/2. For S odd: floor(S/2)=(S-1)/2. Therefore, if S increases by 1 from even to odd, floor(S/2) increases by 0.5, but floor'd down, so remains the same. Hence, the count remains the same.Therefore, for P=1996 (S=998 even) and P=1998 (S=999 odd), both have floor(S/2)=499. Therefore, they have the same number of rectangles.But wait, that contradicts the problem's implication that one is more than the other. Maybe the problem is designed to show that they are the same? But maybe I need to check my reasoning again.Alternatively, perhaps the formula isn't exactly floor(S/2). Let's re-examine.Suppose we have a perimeter P=2*(a + b). Therefore, a + b = P/2 = S. We need to find the number of pairs (a, b) where a and b are positive integers, a ≤ b. So a can range from 1 to floor(S/2), inclusive. For each a in 1 to floor(S/2), b = S - a. Since a ≤ b, then S - a ≥ a, which implies S ≥ 2a, hence a ≤ S/2. Therefore, the number of solutions is the number of integers a such that 1 ≤ a ≤ floor(S/2).So indeed, the number of solutions is floor(S/2). So for S=998 (even), floor(998/2)=499. For S=999 (odd), floor(999/2)=499. Therefore, both have 499 rectangles. Hence, they are equal.But maybe there's an error here. Let's take a concrete example with larger numbers.Suppose S=100 (perimeter 200). Then number of rectangles is floor(100/2)=50. So 50 rectangles. For S=101 (perimeter 202), floor(101/2)=50. So 50 rectangles. So again, same number. Thus, the formula holds. Therefore, it seems that for consecutive perimeters differing by 2 (i.e., S differing by 1), when moving from even to odd S, the number of rectangles remains the same. Therefore, the answer should be that they have the same number.But the problem asks, "Which has more rectangles with integer sides: those with a perimeter of 1996 or those with a perimeter of 1998?" According to the reasoning, they have the same number. But perhaps the problem is designed to test whether one considers that when S is even, there is a square, but maybe not? Wait, no, the count includes squares. For example, when S=4 (perimeter 8), the rectangle 2x2 is a square, and it's counted. So the count includes squares. But in the case of S even, you have one square and the rest non-squares. For S odd, all are non-squares. But the total count is the same. So the number of rectangles (including squares) is the same. Hence, the answer is they are equal.But the problem might be expecting a different answer. Let me check another approach.Alternatively, maybe the formula for the number of rectangles is different. Let's derive it step by step.Given a perimeter P, then a + b = P/2. Let S = P/2. We need to find the number of pairs (a, b) where a and b are positive integers, a ≤ b.So, for each a from 1 to floor(S/2), there is a corresponding b = S - a. Since a ≤ b, then b = S - a ≥ a => S - a ≥ a => S ≥ 2a => a ≤ S/2. Therefore, the number of solutions is the number of integers a in [1, floor(S/2)]. Hence, floor(S/2).Therefore, for S even: floor(S/2) = S/2. For S odd: floor(S/2) = (S - 1)/2. So when S increases by 1 from even to odd, the number of solutions remains the same. For example, S=998: floor(998/2)=499. S=999: floor(999/2)=499. Hence, same number of rectangles.Therefore, the conclusion is that both perimeters have the same number of rectangles. Therefore, neither has more; they are equal.But the problem asks, "Which has more...". Since they are equal, the answer is neither. But perhaps in the problem statement, there's a trick. Wait, maybe I misread the question. It says "rectangles with integer sides". So squares are included. But as per the formula, they are counted once. So even if one has a square and the other doesn't, the count remains the same. So answer is they are equal.But the original problem might have a typo or different intention. However, based on the mathematical derivation, both perimeters yield the same number of rectangles. Hence, the answer should be they have the same number.Wait, but let's check with actual numbers. Let's compute for S=998 and S=999.For S=998 (even):a ranges from 1 to 499. So 499 rectangles. Each a from 1 to 499 pairs with b=998 - a. The largest a is 499, which pairs with b=499 (since 998 -499=499). So that's the square.For S=999 (odd):a ranges from 1 to 499. Because floor(999/2)=499. Each a from 1 to 499 pairs with b=999 -a. The largest a is 499, which pairs with b=500 (since 999 -499=500). So there's no square here, but the number of pairs is still 499.Therefore, both S=998 and S=999 yield 499 rectangles. Hence, the same number. So the answer is that they are equal.But maybe I missed something. Let's think differently. Suppose for perimeter 1996, S=998. The number of rectangles is the number of integer solutions (a, b) with a ≤ b and a + b=998. This is equivalent to the number of ways to write 998 as the sum of two integers where the first is at most the second. This is the same as the number of integers a from 1 to 499.Similarly, for perimeter 1998, S=999. Number of rectangles is the number of a from 1 to 499 (since floor(999/2)=499). Each a gives a valid b=999 -a, which is greater than a since a ≤499 and 999 -a ≥500. Hence, all rectangles are non-square, but the count is the same.Therefore, the conclusion is that both have the same number of rectangles. So the answer is neither, they are equal.But the problem asks, "Which has more...". Maybe the answer is that they are equal. But perhaps the original problem had a different perimeter difference. For example, if the perimeters were 1996 and 2000, then S would be 998 and 1000. Then floor(998/2)=499, floor(1000/2)=500. So 500 rectangles for P=2000. But in our case, the difference is 2, leading to S differing by 1, and floor(S/2) same.Therefore, the correct answer is that they have the same number of rectangles.

✅:To determine which perimeter, 1996 or 1998, corresponds to more rectangles with integer sides (counting (a times b) and (b times a) as identical), we analyze the number of valid pairs ((a, b)) such that (a leq b) and (a + b = S), where (S = frac{text{perimeter}}{2}).1. For Perimeter 1996: - (S = frac{1996}{2} = 998) (even). - The valid integer pairs ((a, b)) satisfy (a leq b) and (a + b = 998). The smallest (a) is 1, and the largest is (leftlfloor frac{998}{2} rightrfloor = 499). - Number of rectangles: (499).2. For Perimeter 1998: - (S = frac{1998}{2} = 999) (odd). - Similarly, valid pairs ((a, b)) must satisfy (a leq b) and (a + b = 999). Here, (a) ranges from 1 to (leftlfloor frac{999}{2} rightrfloor = 499). - Number of rectangles: (499).Conclusion: Both perimeters yield the same number of rectangles. The key insight is that when (S) increases by 1 (from even to odd), the floor division (leftlfloor frac{S}{2} rightrfloor) remains unchanged. Thus, neither has more rectangles; they are equal.Answer: Both have the same number of rectangles.

🔑:To determine the number of rectangles with integer side lengths that have a given perimeter, we analyze both cases (1996 and 1998). Case 1: Perimeter (1996)1. Perimeter Formula: The perimeter (P) of a rectangle with side lengths (a) and (b) is given by: [ P = 2a + 2b ] For (P = 1996): [ 2a + 2b = 1996 ]2. Simplifying the Perimeter Equation: [ a + b = frac{1996}{2} = 998 ]3. Range of Integer Side Lengths: Let (a) be the smaller side, hence (1 le a le 998 -1 = 997). However, since (a le b), we only consider up to half of 998: [ 1 le a le 499 ] This implies (b = 998 - a), giving us 499 possible rectangles.Case 2: Perimeter (1998)1. Perimeter Formula: For (P = 1998): [ 2a + 2b = 1998 ]2. Simplifying the Perimeter Equation: [ a + b = frac{1998}{2} = 999 ]3. Range of Integer Side Lengths: Similarly, let (a) be the smaller side: [ 1 le a le 999 - 1 = 998 ] Again, since (a le b), we only consider up to half of 999: [ 1 le a le 499 ] This produces the same number of possible rectangles, 499.Conclusion: In both cases, the number of rectangles with integer sides is equal:[boxed{text{499}}]

🧠:Okay, so I need to prove that a composite number n always has a divisor greater than 1 but not greater than the square root of n. Hmm, let me start by recalling some basic number theory concepts. First, a composite number is a positive integer that has at least one positive divisor other than 1 and itself. So, if n is composite, there must be some integer d such that 1 < d < n and d divides n. The question is asking to show that there exists such a divisor d where d is not greater than sqrt(n). That is, there's a divisor d with 1 < d ≤ sqrt(n). Wait, but the original statement says "greater than 1 but not greater than sqrt(n)". So, does that mean d satisfies 1 < d ≤ sqrt(n)? Yes, that's how I should interpret it. So, I need to show that for any composite number n, there exists a divisor d where 1 < d ≤ sqrt(n). Let me think about how to approach this. If n is composite, it can be factored into two integers a and b, both greater than 1. So n = a * b. Now, if either a or b is less than or equal to sqrt(n), then we have our divisor. But what if both a and b are greater than sqrt(n)? Wait, but that's impossible, right? Because if both a and b are greater than sqrt(n), then their product would be greater than sqrt(n) * sqrt(n) = n. So that can't happen. Therefore, at least one of a or b must be less than or equal to sqrt(n). Ah, that seems like the key idea here. Let me formalize this. Suppose n is composite, so there exist integers a and b such that 1 < a, b < n and n = a * b. If both a > sqrt(n) and b > sqrt(n), then a * b > sqrt(n) * sqrt(n) = n, which contradicts the fact that a * b = n. Therefore, at least one of a or b must be ≤ sqrt(n). Hence, that divisor (either a or b) is greater than 1 (since a and b are both factors greater than 1) and ≤ sqrt(n). Therefore, such a divisor exists. Therefore, the statement is proven. Wait, but let me check if there are any edge cases. For example, what if n is a square number? Let's take n = 4. Then sqrt(4) = 2. The divisors are 2 and 2. So, 2 is equal to sqrt(4), which fits the condition. Another example, n = 9. sqrt(9) is 3, and 3 divides 9. So that works. What about n = 6? sqrt(6) is approximately 2.45. The divisors greater than 1 are 2, 3, 6. So 2 is ≤ 2.45, so that works. Similarly, n = 15. sqrt(15) ≈ 3.87. The divisors are 3, 5, 15. So 3 is ≤ 3.87. Wait, but in the case of n = 16, sqrt(16) is 4. The divisors are 2, 4, 8. So 2 and 4 are ≤ 4. So in all these examples, the smallest non-trivial divisor (greater than 1) is indeed ≤ sqrt(n). Wait, but is that always the case? Let's take n = 25. sqrt(25) is 5. The divisors are 5 and 25. So 5 is equal to sqrt(25). But what if n is a product of two primes, like 11*13=143. sqrt(143) is approximately 11.958. So the divisors are 11 and 13. So 11 is ≤ 11.958, which is sqrt(n). So here, 11 is the divisor less than sqrt(n). Wait, but 11 is less than sqrt(143)? Let's calculate sqrt(143). 11^2 is 121, 12^2 is 144. So sqrt(143) is approximately 11.958. So 11 is less than that. So yes, 11 is a divisor less than sqrt(n). Wait, but in this case, 11 is a prime divisor. So in general, if n is composite, the smallest prime divisor of n must be ≤ sqrt(n). Because if the smallest prime divisor is greater than sqrt(n), then the other divisor would have to be greater than sqrt(n) as well, leading to a contradiction. Let me think again. Let's suppose that n is composite, so it has some prime divisor p. Let p be the smallest prime divisor of n. Then if p > sqrt(n), then any other divisor would have to be at least p, so the other divisor would be n/p. But since p is the smallest prime divisor, n/p must be ≥ p. So n/p ≥ p > sqrt(n), which would imply that n/p * p > sqrt(n) * sqrt(n) = n, which is a contradiction because n/p * p = n. Therefore, p must be ≤ sqrt(n). Therefore, the smallest prime divisor p of n is ≤ sqrt(n). Hence, such a divisor exists. So, in summary, the reasoning is that if a composite number n had all its divisors greater than sqrt(n), then multiplying two such divisors would result in a number greater than n, which is impossible. Therefore, there must be at least one divisor ≤ sqrt(n). Alternatively, considering the smallest prime divisor, since n is composite, it must have at least one prime divisor, and the smallest one can't be greater than sqrt(n) because otherwise the other divisor would be less than the smallest prime divisor, which contradicts its minimality. Therefore, in either approach, the conclusion is that a composite number n must have a divisor greater than 1 and ≤ sqrt(n). I need to make sure that this proof is rigorous. Let me structure it step by step. 1. Let n be a composite number. Then n has at least one divisor d such that 1 < d < n. 2. Suppose for contradiction that all divisors d of n satisfying 1 < d < n are greater than sqrt(n). 3. Then, consider such a divisor d. Since d divides n, there exists another integer k = n/d, which is also a divisor of n. 4. If d > sqrt(n), then k = n/d < n/sqrt(n) = sqrt(n). 5. Therefore, k is a divisor of n such that k < sqrt(n). 6. But since k divides n and k < sqrt(n), and k must be at least 2 (since n is composite and k = n/d, and d < n implies k > 1). 7. Therefore, k is a divisor of n such that 1 < k ≤ sqrt(n), which contradicts our initial assumption that all divisors d with 1 < d < n are greater than sqrt(n). 8. Therefore, our assumption is false, and there must exist at least one divisor of n in the interval (1, sqrt(n)]. Hence, proven. Alternatively, using the smallest prime divisor approach:1. Let n be composite. Then n has a prime divisor p.2. Let p be the smallest prime divisor of n.3. Suppose p > sqrt(n). Then, since p divides n, we can write n = p * m for some integer m ≥ p (since p is the smallest divisor).4. Then, m ≥ p > sqrt(n) implies that m > sqrt(n).5. Therefore, p * m > sqrt(n) * sqrt(n) = n, which contradicts n = p * m.6. Therefore, our assumption that p > sqrt(n) is false, so p ≤ sqrt(n).7. Hence, p is a divisor of n such that 1 < p ≤ sqrt(n).Either way, the conclusion holds. So, in both approaches, by contradiction or by considering the smallest prime divisor, we arrive at the conclusion that such a divisor exists. Therefore, the proof is solid. I think that's all. Let me just recap: For any composite n, assuming all divisors are greater than sqrt(n) leads to a contradiction because their product would exceed n. Hence, at least one divisor must be ≤ sqrt(n), and since n is composite, that divisor is greater than 1. Therefore, the proof is complete.

✅:To prove that a composite number ( n ) always has a divisor greater than 1 but not greater than ( sqrt{n} ), we proceed as follows:Proof by Contradiction:1. Assumption: Suppose ( n ) is composite, so ( n = a cdot b ) where ( 1 < a, b < n ). Assume, for contradiction, that all divisors of ( n ) greater than 1 are strictly greater than ( sqrt{n} ).2. Analyze Divisors: If both ( a > sqrt{n} ) and ( b > sqrt{n} ), then their product satisfies: [ a cdot b > sqrt{n} cdot sqrt{n} = n, ] which contradicts ( a cdot b = n ).3. Conclusion: Therefore, at least one of ( a ) or ( b ) must satisfy ( leq sqrt{n} ). Since ( a ) and ( b ) are divisors of ( n ) greater than 1, this guarantees the existence of a divisor ( d ) such that ( 1 < d leq sqrt{n} ).Alternative Proof via Smallest Prime Divisor:1. Smallest Prime Divisor: Let ( p ) be the smallest prime divisor of ( n ). If ( p > sqrt{n} ), then ( n = p cdot m ), where ( m geq p ) (as ( p ) is the smallest divisor). This implies: [ p cdot m geq p cdot p > sqrt{n} cdot sqrt{n} = n, ] contradicting ( p cdot m = n ).2. Result: Hence, ( p leq sqrt{n} ), proving that ( p ) is a divisor of ( n ) satisfying ( 1 < p leq sqrt{n} ).Final Conclusion: A composite number ( n ) must have at least one divisor ( d ) where ( 1 < d leq sqrt{n} ). This follows from the impossibility of all divisors exceeding ( sqrt{n} ) and the necessity of a smallest prime divisor ( leq sqrt{n} ).Answer: (boxed{text{Every composite } n text{ has a divisor } d text{ with } 1 < d leq sqrt{n}.})

🔑:We need to prove that every composite number ( n ) always has a divisor greater than 1 but not greater than ( sqrt{n} ).1. Define Properties of Composite Numbers: A composite number ( n ) is defined as a natural number greater than 1 that is not prime. This means that ( n ) has positive divisors other than 1 and itself.2. Divisor Pair Concept: For every divisor ( a ) of ( n ), there exists a corresponding divisor ( frac{n}{a} ) such that: [ a cdot left(frac{n}{a}right) = n ]3. Consider Divisors Not Exceeding ( sqrt{n} ): To show that one of the divisors must not exceed ( sqrt{n} ), we need to examine the implications of their multiplicative relationship.4. Case Analysis Based on Divisor Magnitude: Let's assume that both ( a ) and ( frac{n}{a} ) are greater than ( sqrt{n} ). If both were greater than ( sqrt{n} ), then their product would be: [ a cdot frac{n}{a} = n > sqrt{n} cdot sqrt{n} = n ] This would lead to a contradiction because it would imply ( n > n ), which is impossible.5. Conclusion of Divisor Bound: Therefore, this contradiction implies that the assumption that both ( a ) and ( frac{n}{a} ) are greater than ( sqrt{n} ) is false. Hence, one of the divisors must be less than or equal to ( sqrt{n} ).6. Final Statement: Since ( n ) is composite and cannot be 1 or a prime number, it must have divisors other than 1 and itself, which leads us to conclude that:[boxed{text{A composite number } n text{ always has a divisor greater than 1 but not greater than } sqrt{n}.}]

🧠:Okay, so I have this geometry problem here. Let me try to understand it step by step. The problem is about a parallelogram ABCD, where AB is equal to 'a' and AD is equal to 'b'. There are two circles mentioned: the first one is centered at vertex A and passes through D, so its radius should be AD, which is 'b'. The second circle is centered at C and passes through D, so its radius would be CD. Wait, but in a parallelogram, opposite sides are equal, right? So AB equals CD, which is 'a', and AD equals BC, which is 'b'. Therefore, the radius of the second circle (centered at C) should be CD, which is 'a'? Wait, but CD is equal to AB, which is 'a', so yes, the radius of the second circle is 'a'. Then there's an arbitrary circle with center B that intersects the first circle (centered at A with radius b) at points M1 and N1, and intersects the second circle (centered at C with radius a) at points M2 and N2. The question is asking for the ratio of the lengths M1N1 to M2N2.Hmm. Let me visualize this. So we have a parallelogram ABCD. Let me sketch it mentally: points A, B, C, D with AB parallel to CD and AD parallel to BC. Since it's a parallelogram, vectors AB and DC are equal, and vectors AD and BC are equal. The coordinates might help here. Maybe I should assign coordinates to the points to make this easier?Let me place point A at the origin (0,0). Then, since AB is length 'a', and let's assume AB is along the x-axis for simplicity. So point B would be at (a, 0). Since AD is length 'b', and since it's a parallelogram, AD is one of the sides. If AD is length 'b' and angle at A is some angle, but without loss of generality, maybe we can assign coordinates. Let me think: if AD is from A(0,0) to D(0, b), but then ABCD would be a rectangle if AD is vertical. But in a general parallelogram, the sides AD and AB can be at any angle. Wait, but in a coordinate system, maybe we can let AD have coordinates (p, q) such that the length is 'b', and AB is (a, 0). Then point D would be at (p, q), and since ABCD is a parallelogram, point C would be at (a + p, q). Because in a parallelogram, the vector from B to C should be the same as the vector from A to D. So vector AD is (p, q), so vector BC should also be (p, q). Since point B is at (a, 0), adding (p, q) gives point C at (a + p, 0 + q) = (a + p, q). Then point D is at (p, q). Wait, but then AD is from (0,0) to (p, q), so its length is √(p² + q²) = b. Similarly, AB is from (0,0) to (a, 0), so its length is 'a'.Alternatively, maybe I can choose coordinates such that point A is (0,0), point B is (a, 0), point D is (0, b), and then point C is (a, b). But that's a rectangle, which is a specific type of parallelogram. But the problem doesn't specify it's a rectangle, so maybe that's too restrictive. So perhaps a better coordinate system is needed where the parallelogram isn't necessarily a rectangle.Alternatively, let's use vectors. Let me consider vectors for generality. Let me take point A as the origin. Vector AB is vector a (with magnitude 'a'), and vector AD is vector b (with magnitude 'b'). Then point B is at a, point D is at b, and point C is at a + b. The first circle is centered at A (origin) with radius AD, which is |b| = b. The second circle is centered at C (a + b) with radius CD. Since CD is equal to AB, which is |a| = a. So the second circle has radius a.Now, there's an arbitrary circle centered at B. Let's denote this circle as circle B with radius r (arbitrary). This circle intersects the first circle (center A, radius b) at points M1 and N1, and intersects the second circle (center C, radius a) at points M2 and N2. We need to find the ratio of the lengths M1N1 to M2N2.So, the problem is essentially asking: when you have two intersection points between two circles, the distance between those two points (the chord) can be found using the formula for the length of the chord given the distance between the centers and their radii. Since circle B is intersecting both circle A and circle C, we can compute the chord lengths M1N1 and M2N2 using the chord length formula, and then take their ratio.The chord length formula is: if two circles with radii r1 and r2 are separated by a distance d between their centers, then the length of their intersection chord is 2√[(r1² - ( (d² + r1² - r2²)/(2d) )² )]. Wait, let me recall the formula correctly.Yes, the chord length can be calculated using the formula:If two circles have centers separated by distance d, radii r1 and r2, then the length of their common chord is 2 * √[r1² - h²], where h is the distance from the center of the first circle to the chord. The value of h can be found using the formula h = (d² + r1² - r2²)/(2d). So substituting, the chord length is 2 * √[r1² - ( (d² + r1² - r2²)/(2d) )² ].Alternatively, the chord length can also be written as 2 * √[ (r1 + r2 + d)(r1 + r2 - d)(r1 - r2 + d)(-r1 + r2 + d) ) ] / (2d) ). Wait, maybe that's another formula. Let me verify.Alternatively, using the formula for the length of the common chord between two intersecting circles:Length = 2 * √[ (r1² - ( (d² + r1² - r2²)/(2d) )² ) ]Yes, that's correct. Let me derive it quickly. Suppose we have two circles, one at center O1 and the other at O2, separated by distance d. The line connecting O1O2 intersects the common chord at right angle at point M, which is the midpoint of the chord. The distance from O1 to M is h, then by Pythagoras in triangle O1MP, where P is one end of the chord:h² + ( (length/2) )² = r1².Similarly, in triangle O2MP:(d - h)² + (length/2)² = r2².Subtracting these two equations:(d - h)² - h² = r2² - r1².Expanding:d² - 2dh + h² - h² = r2² - r1².Simplify:d² - 2dh = r2² - r1².Solving for h:2dh = d² + r1² - r2².h = (d² + r1² - r2²)/(2d).Then, substituting h back into the first equation:(length/2)² = r1² - h².So length = 2√[r1² - h²] = 2√[r1² - ( (d² + r1² - r2²)/(2d) )² ].Yes, that seems correct.So, applying this formula to both intersections (M1N1 and M2N2), we can compute their lengths.First, for chord M1N1: this is the common chord between circle A (radius b) and circle B (radius r). The distance between centers A and B is AB = a. So here, d = AB = a, r1 = b (radius of circle A), r2 = r (radius of circle B).Wait, but actually, circle B is the arbitrary circle, so it has radius r, and it's intersecting circle A (radius b). So chord length M1N1 would be 2√[b² - ( (a² + b² - r²)/(2a) )² ].Similarly, for chord M2N2: this is the common chord between circle C (radius a) and circle B (radius r). The distance between centers B and C is BC. Since in the parallelogram, BC is equal to AD, which is length b. So the distance between centers B and C is BC = b. Therefore, d = BC = b, r1 = a (radius of circle C), r2 = r (radius of circle B). Therefore, chord length M2N2 is 2√[a² - ( (b² + a² - r²)/(2b) )² ].So now, the ratio M1N1 : M2N2 is [2√(b² - ((a² + b² - r²)/(2a))² )] : [2√(a² - ((a² + b² - r²)/(2b))² )].Wait, but here the radii of circle B (r) is arbitrary. However, the problem states that the circle centered at B is arbitrary, meaning that the ratio should not depend on r. Therefore, the ratio must be the same regardless of r, which suggests that when we compute the ratio, the terms involving r will cancel out. Let me check that.Let me denote k = a² + b² - r². Then,M1N1 = 2√[b² - (k/(2a))² ] = 2√[b² - k²/(4a²)].M2N2 = 2√[a² - (k/(2b))² ] = 2√[a² - k²/(4b²)].Therefore, the ratio is √[b² - k²/(4a²)] : √[a² - k²/(4b²)].But this still seems to depend on k, which includes r. However, the problem states that the circle is arbitrary, meaning that this ratio must be the same for any r where the circles intersect. Therefore, the ratio should not depend on r. This suggests that perhaps my approach is missing something, or maybe there's a geometric relationship that makes the ratio constant regardless of r.Alternatively, maybe the key lies in the properties of the parallelogram. Let me consider coordinates again, perhaps assigning coordinates to each point and using coordinate geometry.Let me set up a coordinate system. Let me place point A at (0, 0). Then, since AB = a, let me place point B at (a, 0). Since AD = b, but the direction of AD is not specified, in a general parallelogram, AD can be at some angle. Let me denote point D as (d_x, d_y), such that the distance from A to D is b: √(d_x² + d_y²) = b. Then, since ABCD is a parallelogram, point C must be at B + vector AD, which is (a + d_x, 0 + d_y).So coordinates:A: (0, 0)B: (a, 0)D: (d_x, d_y)C: (a + d_x, d_y)Now, circle centered at A (0,0) with radius AD = b: equation x² + y² = b².Circle centered at C (a + d_x, d_y) with radius CD. Wait, CD is equal to AB, which is a. So the radius is a. Therefore, the equation of the second circle is (x - (a + d_x))² + (y - d_y)² = a².Circle centered at B (a, 0) with arbitrary radius r: equation (x - a)² + y² = r².Now, intersections M1, N1 are the intersections between circle A and circle B. Similarly, intersections M2, N2 are intersections between circle C and circle B.We need to find the lengths of chords M1N1 and M2N2 and their ratio.To find the chord length between two circles, we can use the formula mentioned earlier. Alternatively, we can compute the distance between the intersection points by solving the equations of the circles.Let me first compute the chord length M1N1. The common chord between circle A and circle B.Circle A: x² + y² = b²Circle B: (x - a)² + y² = r²Subtracting the two equations to find the equation of the common chord:x² + y² - [(x - a)^2 + y^2] = b² - r²Expanding: x² + y² - (x² - 2a x + a² + y²) = b² - r²Simplify: x² + y² - x² + 2a x - a² - y² = b² - r²Which reduces to: 2a x - a² = b² - r²Therefore, 2a x = a² + b² - r²Thus, x = (a² + b² - r²)/(2a)So the common chord M1N1 lies on the vertical line x = (a² + b² - r²)/(2a). Therefore, to find the points M1 and N1, we substitute x into circle A's equation:[(a² + b² - r²)/(2a)]² + y² = b²Solving for y:y² = b² - [(a² + b² - r²)^2)/(4a²)]Thus, y = ±√[b² - ((a² + b² - r²)^2)/(4a²)]Therefore, the length of M1N1 is 2 * √[b² - ((a² + b² - r²)^2)/(4a²)] = 2 * √[ (4a²b² - (a² + b² - r²)^2 ) / (4a²) ) ] = 2 * [√(4a²b² - (a² + b² - r²)^2 ) / (2a) ) ] = √(4a²b² - (a² + b² - r²)^2 ) / aSimilarly, for the chord M2N2 between circle C and circle B.Circle C: (x - (a + d_x))² + (y - d_y)^2 = a²Circle B: (x - a)^2 + y² = r²Subtracting the equations:(x - (a + d_x))² + (y - d_y)^2 - [(x - a)^2 + y²] = a² - r²Expanding:[x² - 2(a + d_x)x + (a + d_x)^2] + [y² - 2 d_y y + d_y²] - [x² - 2a x + a²] - y² = a² - r²Simplify term by term:x² - 2(a + d_x)x + (a + d_x)^2 + y² - 2 d_y y + d_y² - x² + 2a x - a² - y² = a² - r²Cancelling x² and y² terms:-2(a + d_x)x + (a + d_x)^2 - 2 d_y y + d_y² + 2a x - a² = a² - r²Combine like terms:[-2(a + d_x)x + 2a x] + [(a + d_x)^2 - a² + d_y² - 2 d_y y] = a² - r²Compute coefficients for x:-2(a + d_x)x + 2a x = -2a x - 2 d_x x + 2a x = -2 d_x xThen the remaining terms:(a + d_x)^2 - a² + d_y² - 2 d_y y = [a² + 2a d_x + d_x² - a²] + d_y² - 2 d_y y = 2a d_x + d_x² + d_y² - 2 d_y yTherefore, the equation simplifies to:-2 d_x x + 2a d_x + d_x² + d_y² - 2 d_y y = a² - r²Wait, but in the parallelogram, vector AD is (d_x, d_y), and its length is b, so d_x² + d_y² = b². So substituting that:-2 d_x x + 2a d_x + b² - 2 d_y y = a² - r²Rearranging:-2 d_x x - 2 d_y y + 2a d_x + b² = a² - r²Bring constants to the right:-2 d_x x - 2 d_y y = a² - r² - 2a d_x - b²Divide both sides by -2:d_x x + d_y y = ( -a² + r² + 2a d_x + b² ) / 2This is the equation of the common chord M2N2.Alternatively, perhaps there's a better way to approach this. Let me recall that in a parallelogram, the coordinates of point C are (a + d_x, d_y), as before. The distance between centers B (a,0) and C (a + d_x, d_y) is √[(d_x)^2 + (d_y)^2] = √(b²) = b, since vector BC is equal to vector AD, which has length b.Therefore, the distance between centers B and C is b. The radius of circle C is a, and the radius of circle B is r. So using the chord length formula again, the length of M2N2 is 2√[a² - ( (b² + a² - r²)/(2b) )² ].Similarly, earlier for M1N1, length is 2√[b² - ( (a² + b² - r²)/(2a) )² ].Therefore, the ratio M1N1 : M2N2 is:[2√(b² - ((a² + b² - r²)^2)/(4a²)) ] : [2√(a² - ((a² + b² - r²)^2)/(4b²)) ]Simplify this ratio by dividing numerator and denominator by 2:√(b² - ((a² + b² - r²)^2)/(4a²)) / √(a² - ((a² + b² - r²)^2)/(4b²))Let me denote S = a² + b² - r². Then the ratio becomes:√[b² - (S²)/(4a²)] / √[a² - (S²)/(4b²)]= √[ (4a²b² - S²) / (4a²) ] / √[ (4a²b² - S²) / (4b²) ]Wait, let me compute the terms inside the square roots:For the numerator inside the sqrt:b² - (S²)/(4a²) = (4a²b² - S²) / (4a²)Similarly, denominator inside the sqrt:a² - (S²)/(4b²) = (4a²b² - S²) / (4b²)Therefore, the ratio becomes:√[ (4a²b² - S²)/(4a²) ] / √[ (4a²b² - S²)/(4b²) ]= [ √(4a²b² - S²) / (2a) ) ] / [ √(4a²b² - S²) / (2b) ) ]= [ (√(4a²b² - S²) ) / (2a) ) ] * [ (2b) / (√(4a²b² - S²) ) ) ]= (2b)/(2a) = b/aTherefore, the ratio M1N1 : M2N2 is b : a.So the ratio is b/a, meaning M1N1 : M2N2 = b : a.Wait, that's interesting. So despite S = a² + b² - r² being present in the expressions, when we take the ratio, the terms involving S cancel out, leaving us with a ratio of b/a. Therefore, the ratio is constant, regardless of the radius r of the arbitrary circle centered at B, as long as the intersections exist.Therefore, the answer should be boxed{dfrac{b}{a}}.But let me verify this once again to ensure there's no mistake. Let's go through the steps.1. The key was using the chord length formula for both intersections.2. Expressed both chord lengths in terms of S = a² + b² - r².3. Found that the ratio simplifies to b/a because all terms involving S canceled out.Therefore, the ratio is indeed b/a, so M1N1 : M2N2 = b : a.Another way to think about this is through the power of a point or geometric properties, but since the ratio simplifies algebraically, and the steps hold, this should be correct.Final AnswerThe ratio (M_1 N_1 : M_2 N_2) is boxed{dfrac{b}{a}}.

✅:Given a parallelogram (ABCD) with (AB = a) and (AD = b), we have two circles: the first circle is centered at (A) with radius (AD = b), and the second circle is centered at (C) with radius (CD = a). An arbitrary circle centered at (B) intersects the first circle at points (M_1, N_1) and the second circle at points (M_2, N_2). We need to find the ratio (M_1 N_1 : M_2 N_2).1. Chord Length Formula: - For the common chord between two circles with radii (r_1) and (r_2) and center distance (d), the chord length is given by: [ 2 sqrt{r_1^2 - left( frac{d^2 + r_1^2 - r_2^2}{2d} right)^2} ]2. Chord Length (M_1 N_1): - Circles: Center (A) (radius (b)) and center (B) (radius (r)). - Distance between centers (AB = a). - Chord length: [ M_1 N_1 = 2 sqrt{b^2 - left( frac{a^2 + b^2 - r^2}{2a} right)^2} ]3. Chord Length (M_2 N_2): - Circles: Center (C) (radius (a)) and center (B) (radius (r)). - Distance between centers (BC = b). - Chord length: [ M_2 N_2 = 2 sqrt{a^2 - left( frac{b^2 + a^2 - r^2}{2b} right)^2} ]4. Ratio Calculation: - Let (S = a^2 + b^2 - r^2). - Ratio: [ frac{M_1 N_1}{M_2 N_2} = frac{sqrt{b^2 - left( frac{S}{2a} right)^2}}{sqrt{a^2 - left( frac{S}{2b} right)^2}} = frac{sqrt{frac{4a^2b^2 - S^2}{4a^2}}}{sqrt{frac{4a^2b^2 - S^2}{4b^2}}} = frac{sqrt{frac{4a^2b^2 - S^2}{4a^2}}}{sqrt{frac{4a^2b^2 - S^2}{4b^2}}} = frac{frac{sqrt{4a^2b^2 - S^2}}{2a}}{frac{sqrt{4a^2b^2 - S^2}}{2b}} = frac{b}{a} ]Thus, the ratio (M_1 N_1 : M_2 N_2) is (boxed{dfrac{b}{a}}).

🔑:1. Geometric Configuration: We have a parallelogram (ABCD) with (AB = a) and (AD = b). There are two circles: - Circle 1 with center (A) and passing through (D). - Circle 2 with center (C) and passing through (D). 2. Intersections of Circles with Additional Circle: Another circle with center (B) intersects: - Circle 1 at points (M_1) and (N_1). - Circle 2 at points (M_2) and (N_2).3. Symmetry and Distances: - Points (M_1) and (N_1) are symmetric relative to line (AB). Hence, (M_1N_1) equals the double of the distance from (M_1) to line (AB). - Similarly, (M_2) and (N_2) are symmetric relative to line (BC). Thus, (M_2N_2) equals the double of the distance from (M_2) to line (BC).4. Triangles and Side Lengths: - Since (AB = a) and (AD = b), we have (BC = AD = b) and (CD = AB = a). - Consider point distances: - (CM_2 = CD = AB = a), - (AM_1 = AD = BC = b), - (BM_1 = BM_2).5. Triangles Equivalence: By the properties of the parallelogram: - (AB = CD = a), - (AD = BC = b), - (BM_1 = BM_2). This implies triangles (ABM_1) and (CBM_2) are congruent.6. Height Ratios: - The ratio of the line segments (M_1N_1 : M_2N_2) is proportional to the heights (altitudes) from (M_1) and (M_2) to lines (AB) and (BC) respectively. - These heights are inversely proportional to the corresponding sides (a) and (b).7. Conclusion: The ratio (M_1N_1 : M_2N_2) equals the ratio of the sides (b : a).[boxed{frac{M_1N_1}{M_2N_2} = frac{b}{a}}]

🧠:Okay, so I need to find all natural numbers n where the sum of the digits of 5^n equals 2^n. Hmm, let's start by understanding the problem. Natural numbers are positive integers, so n is 1, 2, 3, and so on. For each n, I have to compute 5 raised to that power, sum its digits, and check if that sum is equal to 2 raised to the same power. First, maybe I should test small values of n and see if any of them satisfy the condition. Let's start with n=1. When n=1, 5^1 is 5. The sum of its digits is 5. On the other side, 2^1 is 2. 5 ≠ 2, so n=1 is not a solution.Next, n=2. 5^2 is 25. The sum of digits is 2 + 5 = 7. 2^2 is 4. 7 ≠ 4, so n=2 isn't a solution either.n=3: 5^3 = 125. Sum of digits is 1 + 2 + 5 = 8. 2^3 is 8. Oh! 8 equals 8. So n=3 works. That's our first candidate.Let's check n=4. 5^4 is 625. Sum of digits: 6 + 2 + 5 = 13. 2^4 is 16. 13 ≠ 16. Not a solution.n=5: 5^5 = 3125. Sum: 3 + 1 + 2 + 5 = 11. 2^5 = 32. 11 ≠ 32. Nope.n=6: 5^6 = 15625. Sum: 1 + 5 + 6 + 2 + 5 = 19. 2^6 = 64. Doesn't match.n=7: 5^7 = 78125. Sum: 7 + 8 + 1 + 2 + 5 = 23. 2^7 = 128. Not close.n=8: 5^8 = 390625. Sum: 3 + 9 + 0 + 6 + 2 + 5 = 25. 2^8 = 256. Still no.n=9: 5^9 = 1953125. Sum: 1 + 9 + 5 + 3 + 1 + 2 + 5 = 26. 2^9 = 512. Not matching.n=10: 5^10 = 9765625. Sum: 9 + 7 + 6 + 5 + 6 + 2 + 5 = 40. 2^10 = 1024. Definitely not.So up to n=10, only n=3 works. Let's try a few more to see if another solution pops up.n=0: Wait, natural numbers usually start at 1, but sometimes 0 is included. Let me check. If n=0, 5^0=1, sum is 1. 2^0=1. So 1=1. But if the problem considers natural numbers starting at 1, then n=0 is excluded. The problem says natural numbers, which typically start at 1, so n=0 is probably not considered. But just in case, let's note that n=0 works, but the problem may not want it.Back to n=11: 5^11 is 48828125. Sum: 4 + 8 + 8 + 2 + 8 + 1 + 2 + 5 = 4+8=12, 12+8=20, 20+2=22, 22+8=30, 30+1=31, 31+2=33, 33+5=38. 2^11=2048. 38 vs 2048. No.n=12: 5^12 = 244140625. Sum: 2 + 4 + 4 + 1 + 4 + 0 + 6 + 2 + 5. Let's compute: 2+4=6, +4=10, +1=11, +4=15, +0=15, +6=21, +2=23, +5=28. 2^12=4096. 28 ≠ 4096.n=13: 5^13 = 1220703125. Sum: 1+2+2+0+7+0+3+1+2+5. 1+2=3, +2=5, +0=5, +7=12, +0=12, +3=15, +1=16, +2=18, +5=23. 2^13=8192. Nope.n=14: 5^14 = 6103515625. Sum: 6+1+0+3+5+1+5+6+2+5. 6+1=7, +0=7, +3=10, +5=15, +1=16, +5=21, +6=27, +2=29, +5=34. 2^14=16384. Not matching.n=15: 5^15 = 30517578125. Sum: 3+0+5+1+7+5+7+8+1+2+5. 3+0=3, +5=8, +1=9, +7=16, +5=21, +7=28, +8=36, +1=37, +2=39, +5=44. 2^15=32768. Still no.Hmm, seems like as n increases, 5^n's digit sum isn't growing nearly as fast as 2^n. Let's think about the growth rates. 2^n grows exponentially with base 2, while the digit sum of 5^n... Well, the digit sum of a number is at most 9 times the number of digits. The number of digits of 5^n is roughly log10(5^n) + 1 = n*log10(5) + 1 ≈ n*0.69897 + 1. So the maximum possible digit sum for 5^n is about 9*(0.69897n + 1) ≈ 6.2907n + 9. On the other hand, 2^n grows exponentially. So for large n, 2^n will be way larger than 6.2907n + 9. Therefore, there must be some point beyond which 2^n is always larger than the digit sum of 5^n, so no solutions exist beyond that. But we need to check if there are any solutions between n=3 and that theoretical cutoff. Since up to n=15, we didn't find any. Let's check a few more:n=16: 5^16 = 152587890625. Sum: 1+5+2+5+8+7+8+9+0+6+2+5. Let's compute step by step: 1+5=6, +2=8, +5=13, +8=21, +7=28, +8=36, +9=45, +0=45, +6=51, +2=53, +5=58. 2^16=65536. 58 < 65536. n=17: 5^17 = 762939453125. Sum: 7+6+2+9+3+9+4+5+3+1+2+5. 7+6=13, +2=15, +9=24, +3=27, +9=36, +4=40, +5=45, +3=48, +1=49, +2=51, +5=56. 2^17=131072. Still way larger.n=20: Let's jump to n=20. 5^20 is 95367431640625. Sum: 9+5+3+6+7+4+3+1+6+4+0+6+2+5. Let's add: 9+5=14, +3=17, +6=23, +7=30, +4=34, +3=37, +1=38, +6=44, +4=48, +0=48, +6=54, +2=56, +5=61. 2^20=1048576. 61 vs over a million. So clearly, as n increases, 2^n is growing much faster. Therefore, it's likely that n=3 is the only solution. But maybe we need to check some more mid-sized n? Let's try n=4 again: sum was 13 vs 16. Close, but not equal. n=5: 11 vs 32. n=6: 19 vs 64. n=7:23 vs 128. The digit sums are increasing, but not nearly as fast. Wait, perhaps there is another solution at a higher n where the digit sum coincidentally equals 2^n? Unlikely, but to be thorough, let's think about possible properties.Note that 5^n mod 9 is equal to (5 mod 9)^n = 5^n mod 9. The digit sum of a number is congruent to the number modulo 9. So the digit sum of 5^n ≡ 5^n mod 9. On the other side, 2^n mod 9. So for the equality digit sum(5^n) = 2^n to hold, it must be that 5^n ≡ 2^n mod 9. Let's compute 5^n mod 9 and 2^n mod 9. Compute 5^n mod 9:5^1 mod9=55^2=25 mod9=75^3=125 mod9= (125-13*9=125-117=8)5^4=5^3*5=8*5=40 mod9=45^5=4*5=20 mod9=25^6=2*5=10 mod9=15^7=1*5=5 mod9=5, then the cycle repeats: 5,7,8,4,2,1,...Similarly, 2^n mod9:2^1=22^2=42^3=82^4=16 mod9=72^5=14 mod9=52^6=10 mod9=12^7=2, cycle repeats: 2,4,8,7,5,1,...So cycles for 5^n mod9: period 6: [5,7,8,4,2,1]Cycles for 2^n mod9: period 6: [2,4,8,7,5,1]Therefore, 5^n ≡ 2^n mod9 implies that 5^n -2^n ≡0 mod9. So we need 5^n ≡2^n mod9.Looking at the cycles:For n modulo 6:If n ≡1 mod6: 5 vs2. 5≠2 mod9.n≡2 mod6: 5^2=7, 2^2=4. 7≠4.n≡3 mod6: 5^3=8, 2^3=8. So 8≡8 mod9. So when n≡3 mod6, 5^n ≡2^n mod9.n≡4 mod6: 5^4=4, 2^4=7. 4≠7.n≡5 mod6:5^5=2, 2^5=5. 2≠5.n≡0 mod6:5^6=1, 2^6=1. 1≡1. So when n≡0 or 3 mod6, 5^n≡2^n mod9.Therefore, possible solutions can only occur when n ≡3 mod6 or n≡0 mod6. But n=0 is not natural (if starting at 1). So possible n are 3,6,9,12,... etc. However, from our earlier tests, n=3 works, n=6 sum is19 vs64, n=9 sum26 vs512, n=12 sum28 vs4096. So even though these n satisfy the congruence condition, the actual values are not equal. So congruence is a necessary but not sufficient condition.Therefore, the only possible n are those congruent to 3 mod6 or 0 mod6. But we saw n=3 works, and n=6,9,12, etc., don't. So maybe n=3 is the only solution.But let's check n=6 again. 5^6=15625, sum digits 1+5+6+2+5=19. 2^6=64. 19≠64. So even though n=6 is 0 mod6, it doesn't work. Similarly, n=12: sum digits 28 vs 4096. So the congruence is necessary but not sufficient.Therefore, n=3 is the only solution where the congruence holds and the actual values match. To be thorough, maybe check n=18, which is 0 mod6.n=18: 5^18. Hmm, 5^10 is 9765625, 5^11=48828125, 5^12=244140625, 5^13=1220703125, 5^14=6103515625, 5^15=30517578125, 5^16=152587890625, 5^17=762939453125, 5^18=3814697265625. Sum the digits: 3+8+1+4+6+9+7+2+6+5+6+2+5. Let's compute step by step: 3+8=11, +1=12, +4=16, +6=22, +9=31, +7=38, +2=40, +6=46, +5=51, +6=57, +2=59, +5=64. 2^18=262144. 64≠262144. Still no.Even if the digit sum increases to 64 here, 2^18 is much larger. The digit sum can't catch up.Alternatively, maybe check n=3 again. 5^3=125, sum is8, 2^3=8. Works.n=3 is the only one. Let's confirm once more for n=3: yes. Also, since 2^n grows exponentially and the digit sum of 5^n grows roughly linearly (as per the number of digits), there's no way they can intersect again. Thus, the only solution is n=3.

✅:Step-by-Step Explanation and Answer:To find all natural numbers ( n ) where the sum of the digits of ( 5^n ) equals ( 2^n ):1. Initial Testing for Small ( n ): - For ( n = 1 ): ( 5^1 = 5 ), digit sum = 5; ( 2^1 = 2 ). Not equal. - For ( n = 2 ): ( 5^2 = 25 ), digit sum = 7; ( 2^2 = 4 ). Not equal. - For ( n = 3 ): ( 5^3 = 125 ), digit sum = ( 1 + 2 + 5 = 8 ); ( 2^3 = 8 ). Equal. Solution found: ( n = 3 ). - Testing up to ( n = 20 ) shows no further matches. The digit sums grow much slower than ( 2^n ).2. Analyzing Growth Rates: - ( 2^n ) grows exponentially. - The digit sum of ( 5^n ) grows linearly with ( n ) (bounded by ( 9 times text{number of digits} approx 6.29n + 9 )). - Beyond a certain ( n ), ( 2^n ) will always exceed the digit sum of ( 5^n ).3. Modular Congruence Check: - The digit sum modulo 9 must equal ( 5^n mod 9 ). - ( 5^n equiv 2^n mod 9 ) is necessary for equality. This holds when ( n equiv 0 ) or ( 3 mod 6 ). - Testing ( n equiv 0 mod 6 ) (e.g., ( n = 6, 12, 18 )) shows no solutions, confirming ( n = 3 ) is unique.Conclusion: The only natural number ( n ) satisfying the condition is ( boxed{3} ).

🔑:To find all natural numbers ( n ) for which the sum of the digits of ( 5^n ) is equal to ( 2^n ), we start with the following steps:1. Check initial values: Begin by verifying small values of ( n ): - For ( n = 1 ): [ 5^1 = 5 quad text{and} quad 2^1 = 2. ] The sum of the digits of ( 5 ) is ( 5 ), which is not equal to ( 2 ). - For ( n = 2 ): [ 5^2 = 25 quad text{and} quad 2^2 = 4. ] The sum of the digits of ( 25 ) is ( 2 + 5 = 7 ), which is not equal to ( 4 ). - For ( n = 3 ): [ 5^3 = 125 quad text{and} quad 2^3 = 8. ] The sum of the digits of ( 125 ) is ( 1 + 2 + 5 = 8 ), which is equal to ( 8 ). - For ( n = 4 ): [ 5^4 = 625 quad text{and} quad 2^4 = 16. ] The sum of the digits of ( 625 ) is ( 6 + 2 + 5 = 13 ), which is not equal to ( 16 ). - For ( n = 5 ): [ 5^5 = 3125 quad text{and} quad 2^5 = 32. ] The sum of the digits of ( 3125 ) is ( 3 + 1 + 2 + 5 = 11 ), which is not equal to ( 32 ).From these checks, we see that only ( n = 3 ) fits the condition so far.2. Prove for ( n geq 6 ): Now, consider ( n geq 6 ). We want to prove that the sum of the digits of ( 5^n ) is less than ( 2^n ). - Notice that the number ( 5^n ) has at most ( n ) digits. The sum of the digits of any number with ( n ) digits cannot exceed ( 9n ) because each digit is at most 9: [ text{Sum of digits of } 5^n leq 9n. ] - On the other hand, let's compare it with ( 2^n ). We claim: [ 2^n geq 9n. ] Check for ( n = 6 ): [ 2^6 = 64 quad text{and} quad 9 times 6 = 54. ] Clearly, ( 64 geq 54 ). - For ( n > 6 ), observe that as ( n ) increases by 1, ( 2^n ) increases by a factor of 2, which is much larger than the increase of ( 9n ) by just 9. More formally, the following inequality: [ 2^{n+1} = 2 cdot 2^n geq 2 cdot 9n = 18n. ] Even in less strict terms: [ 2^n text{ grows exponentially while } 9n text{ grows linearly, ensuring } 2^n geq 9n text{ for sufficient } n. ]Thus, for ( n geq 6 ):[text{Sum of the digits of } 5^n leq 9n < 2^n,]implying that for ( n geq 6 ), the sum of digits of ( 5^n neq 2^n ).Conclusion:The only natural number ( n ) that meets the condition is:[boxed{n = 3}]

🧠:Okay, so I need to prove that the sequence defined by a₁ = 1 and the recurrence relation a_{2ᵏ + j} = -aⱼ for k ≥ 0 and j = 1, 2, ..., 2ᵏ is not periodic. Hmm, let's start by understanding the sequence first. Maybe if I write out the first few terms, I can see a pattern.Starting with a₁ = 1. Now, let's apply the recurrence. For k = 0, 2⁰ = 1. So j can be 1. Then a_{1 + 1} = a₂ = -a₁ = -1. So the second term is -1.Next, for k = 1, 2¹ = 2. Then j can be 1 and 2. So:- a_{2 + 1} = a₃ = -a₁ = -1- a_{2 + 2} = a₄ = -a₂ = -(-1) = 1So now the sequence is [1, -1, -1, 1].For k = 2, 2² = 4. j runs from 1 to 4:- a_{4 + 1} = a₅ = -a₁ = -1- a_{4 + 2} = a₆ = -a₂ = 1- a_{4 + 3} = a₇ = -a₃ = 1- a_{4 + 4} = a₈ = -a₄ = -1Adding these, the sequence becomes [1, -1, -1, 1, -1, 1, 1, -1].Continuing for k = 3, 2³ = 8. j from 1 to 8:- a₉ = -a₁ = -1- a₁₀ = -a₂ = 1- a₁₁ = -a₃ = 1- a₁₂ = -a₄ = -1- a₁₃ = -a₅ = 1- a₁₄ = -a₆ = -1- a₁₅ = -a₇ = -1- a₁₆ = -a₈ = 1So the next eight terms are [-1, 1, 1, -1, 1, -1, -1, 1]. Putting it all together, the sequence starts as:1, -1, -1, 1, -1, 1, 1, -1, -1, 1, 1, -1, 1, -1, -1, 1, ...I notice that each time we double the length of the sequence by taking the previous block and negating each element. So the sequence is built recursively by appending the negation of the existing sequence. For example, the first term is [1]. Then [1, -1]. Then [1, -1, -1, 1], and so on. Each time, the next chunk is the negative of the previous chunk. So this is similar to the Thue-Morse sequence in some ways, but maybe different.But the problem is to prove that this sequence is not periodic. So periodic means that there exists some T > 0 such that for all n, a_{n + T} = a_n. So we need to show that no such T exists.One approach is to assume, for contradiction, that the sequence is periodic with period T. Then, we can derive some properties that would lead to a contradiction.Alternatively, we can analyze the structure of the sequence and show that its behavior is incompatible with periodicity. For instance, if the sequence has arbitrarily long runs of a particular property that can't exist in a periodic sequence.Looking at the sequence generated so far:n: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16...a_n:1,-1,-1,1,-1,1,1,-1,-1,1,1,-1,1,-1,-1,1...Looking at this, it doesn't seem to repeat any short period. For example, checking period 2: 1, -1 vs the next terms: -1, 1, which doesn't match. Period 4: 1, -1, -1, 1 vs next four terms: -1,1,1,-1. These are negatives of each other. Similarly, the next four terms are the negative of the previous four. So each block of length 2ᵏ is the negative of the block before it. So the structure is such that each subsequent block of length 2ᵏ is the negation of the previous.Therefore, if we suppose the sequence has period T, then T must be a multiple of some power of 2. Let me think. Because each time the sequence doubles in length when we apply the recurrence. So maybe the period, if it exists, must be 2ᵏ for some k. But if that's the case, then the next block of 2ᵏ terms would have to be equal to the previous block. However, according to the recurrence, each new block is the negative of the previous block. Therefore, if the period were 2ᵏ, then the next block (which is the negative) would have to equal the previous block, implying that each term is its own negative, which would mean all terms are zero. But our sequence has no zeros, which is a contradiction. Therefore, the period cannot be 2ᵏ.Alternatively, suppose the period is T, not necessarily a power of two. Let’s assume T is the minimal period. Then, for the periodicity, we must have a_{n + T} = a_n for all n. Let's see how the sequence is constructed. Each term a_{2ᵏ + j} = -a_j. So, if we take n = 2ᵏ + j, then a_n = -a_j. If the sequence is periodic with period T, then a_{2ᵏ + j} = a_{2ᵏ + j + T} = -a_j. But also, by periodicity, a_{2ᵏ + j + T} = a_{2ᵏ + j} = -a_j. Wait, maybe this isn't immediately helpful.Alternatively, perhaps we can look for a contradiction by assuming periodicity. Let's suppose there is a period T. Then, for all n, a_{n + T} = a_n. Let's choose n such that n = 2ᵏ for some k. Then, a_{2ᵏ + 1} = -a_1 = -1. But by periodicity, a_{2ᵏ + 1} = a_{1 + T} if T divides 2ᵏ. Wait, perhaps I need to relate T to the positions where the sign flips.Alternatively, consider that the sequence is constructed in blocks where each block is the negative of the previous. So the first block is [1], the second block is [-1], the third block is [-1, 1], the fourth block is [1, -1], wait, no. Wait, when k increases, the block size doubles. Wait, for k=0: block size 1 (a1), k=1: block size 2 (a2, a3), k=2: block size 4 (a4 to a7), etc. Wait, no. Wait, according to the recurrence, a_{2ᵏ + j} = -a_j for j=1,...,2ᵏ. So each time k increases by 1, we define the next 2ᵏ terms as the negative of the first 2ᵏ terms. So starting from a1=1, then for k=0, 2^0=1, so j=1: a2 = -a1 = -1. Then for k=1, 2^1=2, so j=1,2: a3 = -a1 = -1, a4 = -a2 = 1. Then for k=2, 2^2=4, j=1,2,3,4: a5 = -a1 = -1, a6 = -a2 = 1, a7 = -a3 = 1, a8 = -a4 = -1. Then for k=3, 2^3=8, j=1,...,8: a9 to a16 are negatives of a1 to a8. So each time, the next block of 2ᵏ terms is the negative of the first 2ᵏ terms. Therefore, the sequence can be seen as:Block 1 (length 1): [1]Block 2 (length 1): [-1]Block 3 (length 2): [-1, 1]Block 4 (length 4): [-1, 1, 1, -1]Wait, actually, no. Wait, each time k increases, we add a block of length 2ᵏ. So for k=0: length 1, k=1: length 2, k=2: length 4, k=3: length 8, etc. So each block doubles the length of the previous. Therefore, the entire sequence up to block m has length 2^m - 1. Wait, actually, the total length after k steps is 2^{k+1} -1. For example, after k=0: 1 term, k=1: 1 + 2 = 3 terms? Wait, no. Wait, when k=0, we have a1. Then for k=0, we compute a2. Then for k=1, compute a3 and a4. Then for k=2, compute a5 to a8. So each k corresponds to adding 2ᵏ terms. So the total length after k steps is 2^{k+1} -1. For example, k=0: 2^1 -1 =1 term, k=1: 2^2 -1=3 terms, but wait, after k=1, we have a1, a2, a3, a4. Wait, maybe my initial understanding is off.Wait, perhaps the way the recurrence is defined is that for each k ≥0, and j=1,…,2ᵏ, then a_{2ᵏ + j} = -a_j. So for each k, the indices from 2ᵏ +1 to 2ᵏ +2ᵏ =2^{k+1} are defined as the negative of the first 2ᵏ terms. Therefore, the sequence is built in chunks where each chunk of length 2ᵏ is the negative of the first chunk. Therefore, the entire sequence is:For k=0: a1=1For k=0: j=1 gives a2 = -a1 = -1For k=1: j=1,2 gives a3 = -a1 = -1, a4 = -a2 =1For k=2: j=1,2,3,4 gives a5=-a1=-1, a6=-a2=1, a7=-a3=1, a8=-a4=-1For k=3: j=1,…,8 gives a9 to a16 as negatives of a1 to a8.Therefore, the sequence is:1 (a1),-1 (a2),-1,1 (a3,a4),-1,1,1,-1 (a5-a8),-1,1,1,-1,1,-1,-1,1 (a9-a16),and so on.So each subsequent block of 2ᵏ terms is the negative of the first 2ᵏ terms. Therefore, the sequence is constructed recursively by taking the current sequence and appending its negative. So S₀ = [1], S₁ = [1, -1], S₂ = [1, -1, -1, 1], S₃ = [1, -1, -1, 1, -1, 1, 1, -1], etc. Each Sₖ is S_{k-1} followed by -S_{k-1}.This is similar to the Thue-Morse sequence, except that Thue-Morse is defined by appending the complement, but here it's the negative. So maybe similar properties hold, like it's not periodic.In the Thue-Morse case, it's known that the sequence is not periodic. The proof often involves showing that the sequence is overlap-free or has other properties that contradict periodicity. Maybe we can use a similar approach here.Suppose, for contradiction, that the sequence is periodic with period T. Then, the sequence repeats every T terms. Let's choose k such that 2ᵏ > T. Then, the block Sₖ has length 2ᵏ. Since 2ᵏ > T, the period T must divide 2ᵏ, but wait, not necessarily. The period could be any divisor of T, but since T is the minimal period, it's possible that T doesn't divide 2ᵏ. Hmm, maybe another approach.Alternatively, consider that the sequence contains the negation of its previous parts. So, for any m, the subsequence from 2ᵏ +1 to 2^{k+1} is the negative of the first 2ᵏ terms. If the sequence were periodic with period T, then the first T terms would repeat indefinitely. However, the term at position T +1 is equal to -a₁, but by periodicity, it should be equal to a₁. Therefore, -a₁ = a₁, which implies a₁ =0, but a₁=1. Contradiction. Wait, is this valid?Wait, if the period is T, then a_{T +1} = a₁ =1. But according to the recurrence, if T +1 is of the form 2ᵏ + j. Let's see when T +1 can be written as 2ᵏ +1. For example, if T is such that T +1 is a power of two. Suppose T +1 =2ᵏ, then a_{2ᵏ}= -a_{2ᵏ - 2^{k-1}} ? Wait, maybe not straightforward.Wait, perhaps if we take T as the period, then for n > T, a_n = a_{n - T}. But according to the recurrence, for n =2ᵏ + j, a_n = -a_j. So if 2ᵏ + j - T = m, then a_n = a_{m} = -a_j. But a_{m} = a_{m + T} = a_{m + 2T} = etc. Hmm, this might get too convoluted.Alternatively, suppose that the sequence is periodic with period T. Then, the entire sequence is determined by its first T terms. However, the way the sequence is constructed, each time we reach a block of length 2ᵏ, which is double the previous block, we negate it. So, if T is less than 2ᵏ for some k, then the terms beyond 2ᵏ would have to follow both the periodicity and the recurrence. Let's take k such that 2^{k} > T. Then, consider the term a_{2^{k} +1} = -a_1 = -1. But by periodicity, a_{2^{k} +1} = a_{1 + (2^{k} )} = a_{1 + m*T}, where m is some integer. But since T divides 2^{k} (because if the period is T, then 2^{k} must be a multiple of T, but wait, 2^{k} is a power of two, so unless T is also a power of two, this may not hold. However, even if T is a power of two, say T=2^m, then for k > m, 2^k is a multiple of T, so a_{2^k +1} = -a1 = -1. But by periodicity, a_{2^k +1} = a1 =1. Therefore, -1 =1, which is a contradiction. Therefore, the assumption of periodicity leads to a contradiction.Therefore, the sequence cannot be periodic.Wait, let me check that again. Suppose T is the period, and suppose T is a power of two, say T=2^m. Then, for k = m, 2^m +1 is the next index. But wait, 2^m +1 would be index T +1. By periodicity, a_{T +1} = a1 =1. But according to the recurrence, a_{2^m +1} = -a1 = -1. So 1 = -1, which is impossible. Hence, contradiction. Therefore, T cannot be a power of two.But what if T is not a power of two? Suppose T is some other integer. Then, choose k such that 2^k > T. Then, 2^k +1 > T, so a_{2^k +1} = -a1 = -1. By periodicity, a_{2^k +1} = a_{2^k +1 - n*T} for some n such that 1 ≤ 2^k +1 -n*T ≤ T. Wait, this might not be straightforward. Maybe need another approach.Alternatively, consider that the sequence contains arbitrarily long segments where each segment is the negative of the previous one. In a periodic sequence, after some point, the sequence repeats, so the behavior is consistent. However, in our case, each time we have a block that is the negative of the original, meaning that if the sequence were periodic, after some terms, it would have to be equal to its own negative, leading to all terms being zero, which they are not.More formally, suppose the sequence has period T. Then, for all n, a_n = a_{n+T}. Now, consider the term a_{2T +1}. According to the recurrence, since 2T is a power of two if T is a power of two, but if T isn't, maybe not. Wait, perhaps choosing T such that 2T is a power of two. Suppose T=2^{m-1}. Then 2T=2^m. Then, a_{2^m +1} = -a1 = -1. But also, a_{2^m +1} = a_{1 + 2^m} = a1 =1 (since the period is T=2^{m-1}, and 2^m is a multiple of T). Therefore, again, -1=1, contradiction.Alternatively, maybe for any T, we can find some term a_n such that a_n = -a_{n-T}, which would conflict with periodicity. For example, if n=2^m +1, then a_n=-a1. But if the period is T, then a_n = a_{n-T} = a_{2^m +1 - T}. If T <2^m, then 2^m +1 - T >2^{m} - T +1. But not sure.Wait, another angle. Suppose the sequence is periodic with period T. Then, the entire sequence is determined by the first T terms. Let's look at how the sequence is built. Each time we double the length by negating the previous block. So, after some steps, the sequence will have a block of length T which is the negative of the first T terms. But if the sequence is periodic with period T, then the block after the first T terms must be equal to the first T terms. However, according to the recurrence, it's the negative. Therefore, the first T terms must equal their own negative, implying that all terms are zero. But our sequence has 1 and -1, contradiction. Hence, the sequence cannot be periodic.Yes, this seems like a solid argument. Let me formalize it.Assume for contradiction that the sequence is periodic with period T. Then, the sequence satisfies a_{n + T} = a_n for all n ≥1. Consider the first T terms: a₁, a₂, ..., a_T. According to the recurrence relation, the next T terms (from a_{T +1} to a_{2T}) are defined as -a₁, -a₂, ..., -a_T. However, by periodicity, these next T terms must be equal to a₁, a₂, ..., a_T. Therefore, we have:For all j in 1, 2, ..., T: a_j = -a_jThis implies that for all j, 2a_j =0, so a_j=0. But this contradicts the fact that a₁=1 and all terms are either 1 or -1. Therefore, the assumption of periodicity leads to a contradiction. Hence, the sequence cannot be periodic.Therefore, the proof is complete.

✅:The sequence defined by (a_1 = 1) and (a_{2^k + j} = -a_j) for (k geq 0) and (j = 1, 2, dots, 2^k) is not periodic. Proof by contradiction:1. Assume the sequence is periodic with period (T), meaning (a_{n + T} = a_n) for all (n geq 1).2. Consider the first (T) terms: (a_1, a_2, dots, a_T).3. By the recurrence relation, the next (T) terms (positions (T + 1) to (2T)) are defined as: (a_{T + 1} = -a_1, , a_{T + 2} = -a_2, , dots, , a_{2T} = -a_T).4. However, due to periodicity, these terms must equal the first (T) terms: (a_{T + 1} = a_1, , a_{T + 2} = a_2, , dots, , a_{2T} = a_T).5. Combining steps 3 and 4, we get: (-a_j = a_j) for all (j = 1, 2, dots, T). This implies (2a_j = 0), so (a_j = 0) for all (j), which contradicts (a_1 = 1) and the fact that all terms are (pm 1).Thus, the assumption of periodicity leads to a contradiction. Therefore, the sequence is not periodic.(boxed{text{The sequence is not periodic.}})

🔑:The sequence (a_1, a_2, a_3, dots) is defined by (a_1 = 1) and the recursive relationship (a_{2^k + j} = -a_j) for (k = 0, 1, 2, dots) and (j = 1, 2, dots, 2^k). We need to prove that this sequence is not periodic.1. Initial Condition and Recursive Relationship: The initial condition is given as: [ a_1 = 1 ] The recursive relationship is: [ a_{2^k + j} = -a_j ] for ( k = 0,1,2, dots ) and ( j = 1,2, dots, 2^k ).2. Evaluating Sequence at Formulated Indices: Let's calculate a few terms of the sequence using the recursive relationship: - For ( k = 0 ): [ a_2 = a_{2^0 + 1} = -a_1 = -1 ] - For ( k = 1 ): [ a_3 = a_{2^1 + 1} = -a_1 = -1 ] [ a_4 = a_{2^1 + 2} = -a_2 = 1 ] - For ( k = 2 ): [ a_5 = a_{2^2 + 1} = -a_1 = -1 ] [ a_6 = a_{2^2 + 2} = -a_2 = 1 ] [ a_7 = a_{2^2 + 3} = -a_3 = 1 ] [ a_8 = a_{2^2 + 4} = -a_4 = -1 ]3. General Formula for (a_{2^k}): From the recursive relation and the computed terms, it can be observed that: [ a_{2^{k+1}} = -a_{2^k} ] Given (a_1 = 1), we determine: [ a_{2^k} = (-1)^k ]4. Contradiction for Periodicity: Assume that the sequence is periodic with period (T). Therefore, for all positive integers (m): [ a_{m+T} = a_m ] Specifically, for (m = 2^k) and (T leq 2^k): [ a_{2^k} = a_{2^k + T} = -a_T ] Since (a_{2^k} = (-1)^k), we have: [ (-1)^k = -a_T ] Since (a_T) must be either (1) or (-1), the above equation implies: [ a_T = (-1)^{k+1} ] Therefore: [ -a_T = (-1)^k ] If we consider both conditions: - ((a_T = 1 Rightarrow -a_T = -1)) - ((a_T = -1 Rightarrow -a_T = 1)) This results in a contradiction because it strictly equals to ((-1)^k) and ((-1)^{k+1}) for any positive integer (k). Conclusion:Thus, no such period (T) exists, and hence the sequence (a_n) is not periodic.[boxed{text{The sequence is not periodic.}}]

🧠:Okay, let me try to figure out this geometry problem. So we have triangle ABC with a point P, which is the foot of the perpendicular from C to AB. Then from P, we draw two more perpendiculars: p to AC and q to BC. These lines p and q intersect AC and BC at points T and S respectively. We're told that the sum of the angles at APT and BPS is 20 degrees, and we need to find the measure of angle ACB. First, I need to visualize this. Let me sketch a rough diagram in my mind. Triangle ABC, with AB as the base. Point P is somewhere between A and B on AB because it's the foot of the perpendicular from C. Then from P, we draw perpendiculars to AC and BC. So p is perpendicular to AC, meeting AC at T, and q is perpendicular to BC, meeting BC at S. Then angles at APT and BPS sum up to 20 degrees. Hmm, angles APT and BPS. Let's clarify which angles these are. Angle APT is at point P, between points A, P, T. Similarly, angle BPS is at point P, between points B, P, S. So those two angles at P add up to 20 degrees. Since P is the foot of the altitude from C, CP is perpendicular to AB. So CP is the altitude of triangle ABC. Then PT is perpendicular to AC, and PS is perpendicular to BC. So PT and PS are two more perpendiculars from P. I wonder if there's any relation between these perpendiculars and the original triangle. Maybe similar triangles? Let's see. First, let me note that PT is perpendicular to AC, so triangle APT is a right triangle at T. Similarly, triangle BPS is a right triangle at S. Wait, angle APT is in triangle APT. Since PT is perpendicular to AC, angle at T is 90 degrees, so angle APT is part of that triangle. Similarly, angle BPS is part of triangle BPS, which is also right-angled at S. But how do these angles relate to angle ACB? Maybe there's a way to relate these angles through cyclic quadrilaterals or some other properties. Alternatively, maybe coordinate geometry could help here. Assign coordinates to the points and compute the angles. Let's try that approach. Let me place AB on the x-axis for simplicity. Let’s let point A be at (0, 0), point B at (b, 0) for some positive b, and point C somewhere above AB. Then point P, the foot of the perpendicular from C to AB, will have coordinates (p, 0) where p is between 0 and b. The coordinates of C would be (p, h) where h is the height of the triangle. Then, from point P(p, 0), we need to draw perpendiculars to AC and BC. Let's find the equations of AC and BC first. Point A is (0, 0), and point C is (p, h), so the line AC has a slope of (h - 0)/(p - 0) = h/p. Therefore, the equation of AC is y = (h/p)x. Similarly, point B is (b, 0), so the line BC has a slope of (h - 0)/(p - b) = h/(p - b). The equation of BC is y = [h/(p - b)](x - b). Now, we need to find the equations of the perpendiculars from P(p, 0) to AC and BC. First, the perpendicular to AC from P. Since the slope of AC is h/p, the slope of the perpendicular is -p/h. The equation of this perpendicular line (which is line p) is y - 0 = (-p/h)(x - p). So y = (-p/h)(x - p). This line intersects AC at point T. To find T, we can solve the system of equations: y = (h/p)x (equation of AC)and y = (-p/h)(x - p) (equation of p)Setting them equal:(h/p)x = (-p/h)(x - p)Multiply both sides by ph to eliminate denominators:h^2 x = -p^2 (x - p)Expand the right side:h^2 x = -p^2 x + p^3Bring all terms to the left:h^2 x + p^2 x - p^3 = 0Factor x:x(h^2 + p^2) = p^3Therefore:x = p^3 / (h^2 + p^2)Then y = (h/p)x = (h/p)(p^3 / (h^2 + p^2)) = (h p^2)/(h^2 + p^2)So point T has coordinates (p^3 / (h^2 + p^2), h p^2 / (h^2 + p^2))Similarly, let's find point S, which is the intersection of perpendicular q from P to BC. The slope of BC is h/(p - b), so the slope of the perpendicular q is -(p - b)/h. The equation of q is y - 0 = [-(p - b)/h](x - p)So y = [-(p - b)/h](x - p)This line intersects BC at point S. The equation of BC is y = [h/(p - b)](x - b)Set them equal:[-(p - b)/h](x - p) = [h/(p - b)](x - b)Multiply both sides by h(p - b) to eliminate denominators:-(p - b)^2 (x - p) = h^2 (x - b)Bring all terms to the left:-(p - b)^2 (x - p) - h^2 (x - b) = 0Expand the first term:- (p - b)^2 x + (p - b)^2 p - h^2 x + h^2 b = 0Combine like terms:[- (p - b)^2 - h^2]x + (p - b)^2 p + h^2 b = 0Solve for x:x = [ (p - b)^2 p + h^2 b ] / [ (p - b)^2 + h^2 ]Similarly, y-coordinate can be found by plugging back into BC's equation:y = [h/(p - b)](x - b)So y = [h/(p - b)] [ ( (p - b)^2 p + h^2 b ) / ( (p - b)^2 + h^2 ) - b ]Simplify the numerator inside the brackets:( (p - b)^2 p + h^2 b ) - b( (p - b)^2 + h^2 )= (p - b)^2 p + h^2 b - b(p - b)^2 - b h^2= (p - b)^2 (p - b) + h^2 b - b h^2= (p - b)^3Therefore, y = [h/(p - b)] * [ (p - b)^3 / ( (p - b)^2 + h^2 ) ) ] = [ h (p - b)^3 ] / [ (p - b)( (p - b)^2 + h^2 ) ) ] = [ h (p - b)^2 ] / [ (p - b)^2 + h^2 ]So point S has coordinates ( [ (p - b)^2 p + h^2 b ] / [ (p - b)^2 + h^2 ], [ h (p - b)^2 ] / [ (p - b)^2 + h^2 ] )Okay, now we have coordinates for points T and S. Now, we need to find angles APT and BPS. Angle APT is the angle at point P between points A, P, T. Similarly, angle BPS is the angle at point P between points B, P, S. To compute these angles, we can use vectors. The angle between two vectors can be found using the dot product formula. First, let's find vectors PA and PT for angle APT. Point A is (0,0), point P is (p,0), and point T is (p^3/(h^2 + p^2), h p^2/(h^2 + p^2))Vector PA is A - P = (0 - p, 0 - 0) = (-p, 0)Vector PT is T - P = (p^3/(h^2 + p^2) - p, h p^2/(h^2 + p^2) - 0) = ( - p h^2/(h^2 + p^2), h p^2/(h^2 + p^2) )Similarly, angle BPS is the angle between vectors PB and PS. Point B is (b,0), so vector PB is B - P = (b - p, 0)Point S is ( [ (p - b)^2 p + h^2 b ] / [ (p - b)^2 + h^2 ], [ h (p - b)^2 ] / [ (p - b)^2 + h^2 ] )Vector PS is S - P = ( [ (p - b)^2 p + h^2 b ] / [ (p - b)^2 + h^2 ] - p, [ h (p - b)^2 ] / [ (p - b)^2 + h^2 ] - 0 )Compute the x-coordinate:[ (p - b)^2 p + h^2 b - p( (p - b)^2 + h^2 ) ] / [ (p - b)^2 + h^2 ]= [ (p - b)^2 p + h^2 b - p(p - b)^2 - p h^2 ] / [ (p - b)^2 + h^2 ]= [ h^2 b - p h^2 ] / [ (p - b)^2 + h^2 ]= h^2 (b - p) / [ (p - b)^2 + h^2 ]= - h^2 (p - b) / [ (p - b)^2 + h^2 ]The y-coordinate is [ h (p - b)^2 ] / [ (p - b)^2 + h^2 ]Therefore, vector PS is ( - h^2 (p - b)/D , h (p - b)^2 /D ) where D = (p - b)^2 + h^2Now, to find angles APT and BPS, we can compute the angles between vectors PA and PT, and between vectors PB and PS. First, let's compute angle APT. Vectors PA = (-p, 0) and PT = (- p h^2 / (h^2 + p^2), h p^2 / (h^2 + p^2))The dot product PA ⋅ PT = (-p)( - p h^2 / (h^2 + p^2) ) + 0 * (h p^2 / (h^2 + p^2)) = p^2 h^2 / (h^2 + p^2)The magnitude of PA is sqrt( (-p)^2 + 0^2 ) = pThe magnitude of PT is sqrt( [ - p h^2 / (h^2 + p^2) ]^2 + [ h p^2 / (h^2 + p^2) ]^2 )= sqrt( p^2 h^4 / (h^2 + p^2)^2 + h^2 p^4 / (h^2 + p^2)^2 )= sqrt( p^2 h^2 (h^2 + p^2) / (h^2 + p^2)^2 )= sqrt( p^2 h^2 / (h^2 + p^2) )= (p h)/sqrt(h^2 + p^2 )Therefore, the cosine of angle APT is (PA ⋅ PT)/( |PA| |PT| )= [ p^2 h^2 / (h^2 + p^2) ] / [ p * (p h)/sqrt(h^2 + p^2) ) ]= [ p^2 h^2 / (h^2 + p^2) ] / [ p^2 h / sqrt(h^2 + p^2) ) ]= [ h^2 / (h^2 + p^2) ] / [ h / sqrt(h^2 + p^2) ) ]= [ h^2 / (h^2 + p^2) ] * [ sqrt(h^2 + p^2) / h ]= h / sqrt(h^2 + p^2 )So angle APT = arccos( h / sqrt(h^2 + p^2 ) )Similarly, angle BPS. Let's compute the angle between vectors PB and PS. Vector PB is (b - p, 0)Vector PS is ( - h^2 (p - b)/D , h (p - b)^2 /D ) where D = (p - b)^2 + h^2Wait, note that (p - b) = -(b - p), so we can write vector PS as ( h^2 (b - p)/D , h (b - p)^2 /D )Therefore, PS = ( h^2 (b - p)/D , h (b - p)^2 /D )Vector PB is (b - p, 0). So let's compute the dot product:PB ⋅ PS = (b - p) * [ h^2 (b - p)/D ] + 0 * [ h (b - p)^2 /D ] = h^2 (b - p)^2 / DThe magnitude of PB is sqrt( (b - p)^2 + 0^2 ) = |b - p|The magnitude of PS is sqrt( [ h^2 (b - p)/D ]^2 + [ h (b - p)^2 /D ]^2 )= sqrt( h^4 (b - p)^2 / D^2 + h^2 (b - p)^4 / D^2 )= sqrt( h^2 (b - p)^2 [ h^2 + (b - p)^2 ] / D^2 )But D = (p - b)^2 + h^2 = (b - p)^2 + h^2, so D = h^2 + (b - p)^2Therefore, magnitude of PS = sqrt( h^2 (b - p)^2 D / D^2 ) = sqrt( h^2 (b - p)^2 / D ) = (h |b - p| ) / sqrt(D )Therefore, the cosine of angle BPS is (PB ⋅ PS)/( |PB| |PS| )= [ h^2 (b - p)^2 / D ] / [ |b - p| * (h |b - p| ) / sqrt(D ) ) ]= [ h^2 (b - p)^2 / D ] / [ h (b - p)^2 / sqrt(D ) )= [ h^2 / D ] / [ h / sqrt(D ) ]= [ h^2 / D ] * [ sqrt(D ) / h ]= h / sqrt(D )But D = (b - p)^2 + h^2, so angle BPS = arccos( h / sqrt( (b - p)^2 + h^2 ) )Wait, interesting. So angle APT = arccos( h / sqrt(h^2 + p^2 ) )and angle BPS = arccos( h / sqrt( (b - p)^2 + h^2 ) )The sum of these two angles is 20 degrees. So:arccos( h / sqrt(h^2 + p^2 ) ) + arccos( h / sqrt( (b - p)^2 + h^2 ) ) = 20 degreesHmm. Let me denote:Let’s set θ1 = arccos( h / sqrt(h^2 + p^2 ) )and θ2 = arccos( h / sqrt( (b - p)^2 + h^2 ) )So θ1 + θ2 = 20°But θ1 and θ2 are angles whose cosines are h over the hypotenuse of right triangles with legs h and p, or h and (b - p). Note that θ1 is the angle in a right triangle with adjacent side h and hypotenuse sqrt(h^2 + p^2), so θ1 is the angle opposite side p. Therefore, θ1 = arcsin( p / sqrt(h^2 + p^2 ) )Similarly, θ2 = arcsin( (b - p)/sqrt( (b - p)^2 + h^2 ) )But since θ1 + θ2 = 20°, which is a small angle. This might be tricky. Alternatively, perhaps there's a trigonometric identity that can relate these two arccos terms. Let me think. Let’s denote x = h / sqrt(h^2 + p^2 )and y = h / sqrt( h^2 + (b - p)^2 )So θ1 = arccos(x), θ2 = arccos(y)We know that θ1 + θ2 = 20°, so cos(θ1 + θ2) = cos(20°)Using the cosine addition formula:cos(θ1 + θ2) = cos θ1 cos θ2 - sin θ1 sin θ2 = x y - sqrt(1 - x^2) sqrt(1 - y^2 )Set this equal to cos(20°):x y - sqrt(1 - x^2) sqrt(1 - y^2 ) = cos(20° )But substituting x and y:x = h / sqrt(h^2 + p^2 )y = h / sqrt(h^2 + (b - p)^2 )Therefore, 1 - x^2 = p^2 / (h^2 + p^2 )sqrt(1 - x^2 ) = p / sqrt(h^2 + p^2 )Similarly, sqrt(1 - y^2 ) = (b - p)/sqrt(h^2 + (b - p)^2 )So:x y - sqrt(1 - x^2) sqrt(1 - y^2 ) = [ h / sqrt(h^2 + p^2 ) ][ h / sqrt(h^2 + (b - p)^2 ) ] - [ p / sqrt(h^2 + p^2 ) ][ (b - p)/sqrt(h^2 + (b - p)^2 ) ]= [ h^2 - p(b - p) ] / [ sqrt(h^2 + p^2 ) sqrt(h^2 + (b - p)^2 ) ]Set this equal to cos(20°):[ h^2 - p(b - p) ] / [ sqrt(h^2 + p^2 ) sqrt(h^2 + (b - p)^2 ) ] = cos(20° )Hmm, this seems complicated. Let me see if there's another approach. Maybe there's a geometric insight I'm missing. Alternatively, perhaps triangle ABC is a right-angled triangle? If angle ACB is 90°, but then the sum of angles APT and BPS would be... Not sure. Let's test with a specific case. Suppose angle ACB is 90°, then the altitude from C to AB would coincide with the orthocenter. Wait, no, in a right-angled triangle at C, the altitude from C to AB is just the same as the side CC, so P would coincide with C, but that's not possible since P is on AB. Wait, no, if triangle ABC is right-angled at C, then AB is the hypotenuse, and the altitude from C to AB is indeed the same as the height, which is CP. So in that case, P is the foot of the altitude. Then PT and PS would be perpendiculars from P to AC and BC. Wait, in this case, if ABC is right-angled at C, then AC and BC are perpendicular. Then drawing perpendiculars from P to AC and BC would result in PT and PS coinciding with parts of AC and BC? Wait, no. If AC is perpendicular to BC, then the perpendicular from P to AC would be a line parallel to BC, and the perpendicular from P to BC would be parallel to AC. Wait, maybe. Let me think. If ABC is right-angled at C, then coordinates could be A(0,0), B(b,0), C(0,c), but in this case, the foot of the altitude from C to AB would be at (0,0) if C is (0,c), but that's point A. Wait, no. Wait, in a right-angled triangle at C, the altitude from C to AB is the same as the vertex C, but since AB is the hypotenuse, the foot of the altitude from C to AB is a point P inside AB. Let me compute coordinates. Let’s take ABC with right angle at C, so coordinates A(0,0), B(b,0), C(0,c). Then AB is from (0,0) to (b,0), so the foot of the perpendicular from C(0,c) to AB is (0,0), which is point A. But in the problem statement, P is within AB, not at the endpoints. Therefore, ABC cannot be right-angled at C. So perhaps angle ACB is not 90°, but maybe 100°, 80°, etc. Alternatively, maybe angle ACB is 90° + 20° = 110°, but I need to verify. Alternatively, maybe there's some orthocenter properties here. Since P is the foot of the altitude, and PT and PS are perpendiculars to AC and BC. Then, perhaps T and S lie on the orthocenter or something. Not sure. Wait, in triangle ABC, the orthocenter is the intersection of the altitudes. Since CP is an altitude from C. If we draw another altitude from A, say, but not sure. Alternatively, quadrilateral CPTS might have some cyclic properties because PT and PS are perpendiculars. But PT is perpendicular to AC and PS is perpendicular to BC, so angles at T and S are right angles. So CPTS would have two right angles. Maybe not cyclic. Alternatively, considering that PT and PS are both perpendiculars from P to sides AC and BC, perhaps P is the orthocenter of some triangle? Not sure. Alternatively, maybe triangles APT and BPS are similar to ABC? Let's see. Triangle APT is right-angled at T, and triangle ABC is not necessarily right-angled. Hmm. Alternatively, look for some relation between angles. Let's think about angle APT and angle BPS. We know that angle APT + angle BPS = 20°, which is given. Maybe these angles are related to angles in triangle ABC. Alternatively, since PT is perpendicular to AC and PS is perpendicular to BC, maybe angle APT is equal to angle ACB or something like that. Wait, angle at P: angle APT is part of triangle APT, which is right-angled at T. Similarly, angle BPS is part of triangle BPS, right-angled at S. Wait, let's consider the quadrilateral PTSC. Since PT is perpendicular to AC and PS is perpendicular to BC, and CP is perpendicular to AB. Maybe there are some cyclic quadrilaterals here? Alternatively, maybe triangle TPS is similar to triangle ABC. Not sure. Alternatively, since PT and PS are both perpendiculars from P, maybe points T and S lie on the circle with diameter CP? Because in a circle with diameter CP, any point on the circle would form a right angle with C and P. But PT is perpendicular to AC and PS is perpendicular to BC, so maybe not. Alternatively, use trigonometric identities. Let me recall that in the earlier equation we had:[ h^2 - p(b - p) ] / [ sqrt(h^2 + p^2 ) sqrt(h^2 + (b - p)^2 ) ] = cos(20° )This seems complicated, but maybe there's a relation in triangle ABC that can simplify this. Let me note that in triangle ABC, the altitude CP has length h, and the coordinates we set were A(0,0), B(b,0), C(p,h). Therefore, AC has length sqrt(p^2 + h^2), and BC has length sqrt( (b - p)^2 + h^2 ). The angle at C, angle ACB, can be found using the Law of Cosines:cos(angle ACB) = (AC² + BC² - AB²) / (2 AC * BC )But AB is the base with length b. So:AC² = p² + h²BC² = (b - p)^2 + h²AB² = b²Therefore:cos(angle ACB) = [ (p² + h²) + ( (b - p)^2 + h² ) - b² ] / ( 2 sqrt(p² + h²) sqrt( (b - p)^2 + h² ) )Simplify the numerator:p² + h² + (b² - 2b p + p² ) + h² - b²= 2 p² + 2 h² - 2b pTherefore:cos(angle ACB) = (2 p² + 2 h² - 2b p ) / ( 2 sqrt(p² + h²) sqrt( (b - p)^2 + h² ) )Factor out 2:= [ 2 (p² + h² - b p ) ] / [ 2 sqrt(p² + h²) sqrt( (b - p)^2 + h² ) ]Cancel 2:= (p² + h² - b p ) / [ sqrt(p² + h²) sqrt( (b - p)^2 + h² ) ]But notice that in our previous equation derived from the angle sum:[ h^2 - p(b - p) ] / [ sqrt(h^2 + p^2 ) sqrt(h^2 + (b - p)^2 ) ] = cos(20° )Let me compare the two expressions. The numerator in the cosine of angle ACB is (p² + h² - b p )The numerator in the other equation is ( h² - p(b - p) ) = h² - p b + p² Which is the same as (p² + h² - b p )Therefore, the numerator in both expressions is the same. The denominators are also the same:sqrt(p² + h² ) sqrt( (b - p)^2 + h² )Therefore, cos(angle ACB) = [ p² + h² - b p ] / [ sqrt(p² + h² ) sqrt( (b - p)^2 + h² ) ] = cos(20° )Wait, but from our earlier equation:[ h² - p(b - p) ] / [ sqrt(p² + h² ) sqrt( (b - p)^2 + h² ) ] = cos(20° )But h² - p(b - p) = h² - pb + p² = p² + h² - pb, which is the same as the numerator in cos(angle ACB). Therefore:cos(angle ACB) = cos(20° )Therefore, angle ACB = 20°, but since cosine is positive here, and angles in a triangle are between 0° and 180°, so angle ACB is 20°, or possibly 340°, but that's not possible. Therefore, angle ACB must be 20°. Wait, but this seems too straightforward. Is this possible? But according to the steps, we derived that cos(angle ACB) = cos(20°), hence angle ACB is 20°. But in the problem statement, they mention that P lies within segment AB, so ABC is an acute triangle? If angle ACB were 20°, then the triangle would be acute if all angles are less than 90°, but 20° is possible. Wait, but in the equation, we had:From the angle sum:arccos(x) + arccos(y) = 20°, and we found that cos(arccos(x) + arccos(y)) = cos(20°) = cos(angle ACB)Therefore, angle ACB = 20°, which is the answer. But let me check with a specific example. Let's take a triangle where angle ACB is 20°, and see if the sum of angles APT and BPS is indeed 20°. Suppose angle ACB = 20°, and let's construct triangle ABC such that angle C is 20°, with AB as the base. Let's choose specific lengths for sides. Let’s say AB = 1 unit, and angle C = 20°, then using the Law of Sines, we can find other sides. But maybe this will get complicated. Alternatively, take coordinates again. Alternatively, consider that if angle ACB is 20°, then according to our derivation, the sum of angles APT and BPS is 20°, which matches the problem's given condition. Therefore, angle ACB is 20°. But let me verify once more. The key step was realizing that cos(angle ACB) equals cos(20°), hence angle ACB = 20°, since angles in a triangle are between 0 and 180°, and cosine is injective in that interval. Yes, that seems correct. Therefore, the measure of angle ACB is 20 degrees. Final AnswerThe measure of angle (angle ACB) is (boxed{20^circ}).

✅:Given a triangle (ABC) with the foot (P) of the perpendicular from (C) to (AB) lying within the segment (AB). From (P), perpendiculars (p) and (q) are drawn to (AC) and (BC) respectively, intersecting (AC) at (T) and (BC) at (S). We need to find the angle (angle ACB) given that (|angle APT| + |angle BPS| = 20^circ).1. Coordinate Setup: - Place (A) at ((0, 0)), (B) at ((b, 0)), and (C) at ((p, h)) where (P) is ((p, 0)). - The equations of lines (AC) and (BC) are derived with slopes (frac{h}{p}) and (frac{h}{p - b}) respectively.2. Perpendiculars from (P): - The perpendicular from (P) to (AC) (line (p)) has a slope of (-frac{p}{h}) and intersects (AC) at (T). - The perpendicular from (P) to (BC) (line (q)) has a slope of (-frac{p - b}{h}) and intersects (BC) at (S).3. Vectors and Dot Product: - Calculate vectors (PA), (PT), (PB), and (PS). - Use the dot product to find the angles (angle APT) and (angle BPS).4. Trigonometric Identities: - Express the angles using arccos and arcsin, leading to the equation involving the sum of angles. - Use the cosine addition formula to relate the angles to the given condition.5. Law of Cosines: - Apply the Law of Cosines in triangle (ABC) to relate the coordinates and angles. - Show that (cos(angle ACB) = cos(20^circ)), leading to (angle ACB = 20^circ).The key insight is recognizing that the given sum of angles (angle APT) and (angle BPS) translates directly to the angle (angle ACB) through trigonometric identities and geometric properties, leading to the conclusion that (angle ACB = 20^circ).[boxed{20^circ}]

🔑:1. Identify the angles from the given condition: Given that P is the foot of the perpendicular from point C to line AB, the angles alpha = | angle BAC | and beta = | angle ABC | are acute angles.2. Establish relationship from the right triangles: From the right triangles APT and BPS, we have the following relationships: [ alpha = 90^circ - | angle APT | quad text{and} quad beta = 90^circ - | angle BPS | ]3. Sum the relationships: Combining the two equations gives: [ alpha + beta = (90^circ - | angle APT |) + (90^circ - | angle BPS |) ] [ alpha + beta = 180^circ - (| angle APT | + | angle BPS |) ]4. Substitute given values: We know that: [ | angle APT | + | angle BPS | = 20^circ ] Therefore: [ alpha + beta = 180^circ - 20^circ = 160^circ ]5. Calculate the desired angle gamma = | angle ACB |: Since alpha, beta, and gamma are the internal angles of triangle ABC, we use the property that the sum of internal angles in a triangle equals 180^circ: [ alpha + beta + gamma = 180^circ ] Hence: [ gamma = 180^circ - (alpha + beta) = 180^circ - 160^circ = 20^circ ]Conclusion:[boxed{20^circ}]