Okay, let's try to figure out this problem. The question is asking: For how many positive integers n > 1 is it true that 2²⁴ is a perfect n-th power? Hmm, okay. Let me break this down step by step.

First, I need to recall what a perfect n-th power means. If a number is a perfect n-th power, that means it can be expressed as some integer raised to the power of n. So, if 2²⁴ is a perfect n-th power, there must exist some integer k such that kⁿ = 2²⁴. 

Wait, but 2²⁴ is already a power of 2. So maybe k has to be a power of 2 as well? Let me check. If k is 2 raised to some exponent, say 2^m, then k^n would be (2^m)^n = 2^(m*n). So, if we set that equal to 2²⁴, then m*n = 24. Therefore, the exponents m and n must be positive integers such that their product is 24. 

So, the problem reduces to finding the number of positive integers n > 1 for which there exists a positive integer m such that m*n = 24. But since m has to be a positive integer, n must be a divisor of 24. Therefore, the number of such n is equal to the number of divisors of 24, excluding 1.

Wait, let me confirm this. Let's think again. If 2²⁴ is a perfect n-th power, then n must divide 24, right? Because 2²⁴ = (2^(24/n))ⁿ. So, 24/n must be an integer, which means that n must be a divisor of 24. 

Therefore, the number of positive integers n > 1 is equal to the number of divisors of 24 minus 1 (since we exclude n=1). So, first, I need to find all the divisors of 24, then subtract 1 to exclude n=1.

Let me list the divisors of 24. The prime factorization of 24 is 2³ * 3¹. The number of divisors is (3+1)(1+1) = 4*2=8. So there are 8 divisors. Let's list them: 1, 2, 3, 4, 6, 8, 12, 24. 

So, excluding 1, we have 7 divisors. Therefore, there are 7 positive integers n > 1 such that 2²⁴ is a perfect n-th power. 

Wait, but let me verify this with an example. Let's take n=2. Then 2²⁴ = (2¹²)², which is a perfect square. That's correct. Similarly, n=3: 2²⁴ = (2⁸)³, which is a perfect cube. n=4: 2²⁴ = (2⁶)⁴. Also a perfect fourth power. n=6: 2²⁴ = (2⁴)⁶. n=8: (2³)⁸. n=12: (2²)¹². n=24: (2¹)²⁴. All of these are perfect powers. 

So, it seems that the logic holds. The number of divisors of 24 is 8, subtract 1 for n>1, giving 7. Therefore, the answer should be 7. 

But just to make sure I didn't make a mistake in counting the divisors. Let's list them again: 1, 2, 3, 4, 6, 8, 12, 24. That's 8 divisors. Yes, correct. Excluding 1, we have 7. 

Alternatively, the number of divisors of a number can be found by adding 1 to each exponent in the prime factorization and multiplying. For 24, which is 2³ * 3¹, the number of divisors is (3+1)(1+1) = 4*2=8. So that's correct. 

Therefore, I think the answer is 7. 

But let me check if there's any other way this could be interpreted. For example, could there be a case where even though n divides 24, 2²⁴ is not a perfect n-th power? But since 2 is prime, and 2²⁴ is a prime power, then any perfect n-th power must have exponents that are multiples of n. So, if n divides 24, then 24/n is an integer, so 2²⁴ = (2^(24/n))ⁿ. Therefore, it's a perfect n-th power. 

So there's no issue here. Therefore, the answer is indeed 7. 

Another way to think about it: For a number to be a perfect n-th power, all exponents in its prime factorization must be multiples of n. Since 2²⁴ has only one prime factor, 2, with exponent 24, then n must divide 24. So, the number of possible n's is the number of divisors of 24, excluding 1. 

Yes, that's consistent with what I thought earlier. 

Therefore, after all this checking, I'm confident the answer is 7.
Therefore, the number of positive integers \( n > 1 \) for which \( 2^{24} \) is a perfect \( n \)-th power is \( \boxed{7} \).
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