Explanation:
Okay, so I need to figure out for how many positive integers n > 1 it's true that 2^24 is a perfect n-th power. Hmm, let me start by recalling what a perfect n-th power means. If a number is a perfect n-th power, that means it can be written as some integer raised to the n-th power. So, in this case, 2^24 should be equal to some integer k raised to the n-th power, right?

So, mathematically, we can write that as:

2^24 = k^n

Where k is a positive integer and n > 1. The question is asking for how many such n exist.

First, let me think about the prime factorization of 2^24. Well, since 2 is already a prime number, the prime factorization of 2^24 is just 2^24. So, if 2^24 is a perfect n-th power, then all the exponents in its prime factorization must be multiples of n.

Wait, let me recall that. If a number is a perfect n-th power, then in its prime factorization, each exponent must be divisible by n. For example, if you have a number like 36, which is 2^2 * 3^2, then it's a perfect square because both exponents are multiples of 2. Similarly, if you have 64, which is 2^6, it's a perfect cube because 6 is divisible by 3, and also a perfect square because 6 is divisible by 2.

So applying that to 2^24, since the prime factorization is just 2^24, the exponent here is 24. Therefore, for 2^24 to be a perfect n-th power, 24 must be divisible by n. In other words, n must be a divisor of 24.

But wait, hold on. Let me make sure. If n divides 24, then 24 = n * m for some integer m. Then, 2^24 = (2^m)^n, which would mean that 2^24 is indeed a perfect n-th power. So, n must be a divisor of 24.

But the question is asking for how many positive integers n > 1 satisfy this. Therefore, I need to find all divisors of 24 that are greater than 1.

But hold on, is that all? Let me think again. Suppose n is a divisor of 24. Then, 2^24 is (2^{24/n})^n, which is an integer. Therefore, yes, n divides 24. However, is there a possibility that even if n does not divide 24, 2^24 could still be a perfect n-th power?

Wait, no. Because if 2^24 is equal to k^n, then taking the prime factorization of k, which would be some product of primes, but since 2^24 is only composed of the prime 2, then k must also be a power of 2. Let me explain.

Suppose k is not a power of 2. Then, k would have some other prime factors, say p, so k = 2^a * p^b * ... Then, k^n would have prime factors 2^{a*n}, p^{b*n}, etc. But 2^24 only has 2 as a prime factor, so all other primes in the factorization of k^n would have to be zero. Therefore, k must be a power of 2.

Therefore, k = 2^m for some integer m. Then, k^n = (2^m)^n = 2^{m*n}. Therefore, 2^{m*n} = 2^24, so m*n = 24. Therefore, n must divide 24.

Therefore, the conclusion is that n must be a positive integer divisor of 24, greater than 1. Therefore, the problem reduces to finding the number of divisors of 24 that are greater than 1.

So, first, let me find all the divisors of 24.

First, prime factorization of 24. 24 can be broken down into primes as 2^3 * 3^1.

The formula for the number of divisors is (exponent + 1) for each prime factor, multiplied together. So, for 2^3 * 3^1, the number of divisors is (3 + 1)*(1 + 1) = 4*2 = 8. Therefore, there are 8 positive divisors of 24.

Let me list them out to be sure. The divisors of 24 are:

1, 2, 3, 4, 6, 8, 12, 24.

Yes, that's 8 divisors. So, the question is asking for how many of these divisors are greater than 1. Since the divisors are 1, 2, 3, 4, 6, 8, 12, 24, then excluding 1, there are 7 divisors.

But wait, hold on. The problem says "positive integers n > 1", so n must be greater than 1. Therefore, the answer would be 7? But let me check again.

Wait, but maybe there's a mistake here. Because when we say that n must divide 24, but maybe there are some n's that divide 24 but for which 2^24 is not a perfect n-th power? Wait, no. Because if n divides 24, then 24 = n * m, so 2^24 = (2^m)^n, which is a perfect n-th power. Therefore, all divisors of 24 greater than 1 would satisfy the condition.

But let me check with examples. Let's take n=2. Then 2^24 is (2^12)^2, which is a perfect square. Correct. For n=3, 2^24 is (2^8)^3, since 8*3=24. Correct. For n=4, 2^24 is (2^6)^4, since 6*4=24. Correct. For n=6, 2^24 is (2^4)^6, since 4*6=24. For n=8, 2^24 is (2^3)^8, since 3*8=24. For n=12, 2^24 is (2^2)^12, since 2*12=24. For n=24, 2^24 is (2^1)^24. So all of these work.

Therefore, there are 7 such n's. But wait, the problem says "positive integers n > 1", so n=2,3,4,6,8,12,24. That's 7 numbers. Therefore, the answer is 7?

But wait, hold on. Let me check if there are any other n's that are not divisors of 24 but still make 2^24 a perfect n-th power. For example, suppose n is a fraction? But n has to be a positive integer greater than 1. So fractions are not allowed.

Alternatively, could there be some n that divides 24, but for which 2^24 is not a perfect n-th power? For example, if n is 0? But n has to be greater than 1.

Wait, but if n divides 24, then 2^24 is a perfect n-th power. So, if n divides 24, then yes. Therefore, the number of n's is equal to the number of divisors of 24 greater than 1, which is 7.

But let me check again. Let me think if there's a case where even if n does not divide 24, but 2^24 is still a perfect n-th power. For example, suppose n= something else. Let me think.

Suppose n= something like 5. Then, 2^24 is a perfect 5th power? Let's see. If 2^24 = k^5, then k would have to be 2^m, so 2^{24} = 2^{5m}, so 5m=24. But 24 is not divisible by 5, so m would not be integer. Therefore, no solution. Similarly, for n=7, same thing. 24 divided by 7 is not integer. Therefore, no.

Similarly, n= something else. For example, n= 24. Which is allowed. So, seems like only divisors of 24.

Therefore, the answer is 7.

But wait, hold on. Wait, the problem says "for how many positive integers n>1". So, divisors of 24 greater than 1. Which are 2,3,4,6,8,12,24. So 7 numbers.

But let me check again the list of divisors. Divisors of 24: 1, 2, 3, 4, 6, 8, 12, 24. So yes, 8 divisors. Excluding 1, 7 divisors. Therefore, answer is 7.

But wait, hold on. Wait, the problem says "perfect n-th power". Is there a possibility that for some n, even though n divides 24, 2^24 is not a perfect n-th power? For example, if n is 1? But n>1.

Wait, but for example, take n=2. 2^24 is (2^12)^2, which is a perfect square. Similarly, n=3, (2^8)^3. So, yes. So, all divisors of 24 greater than 1 work.

But let me think again. Suppose n is a divisor of 24, but maybe n is not a divisor of 24? Wait, no. If n divides 24, then 24/n is integer. So, 2^24 is (2^{24/n})^n. Therefore, it's a perfect n-th power.

Therefore, the answer is 7.

But wait, hold on. Wait, the problem says "perfect n-th power". But sometimes, people might think that a perfect n-th power must be an integer. But in this case, since 2^24 is an integer, and we are expressing it as k^n, where k is an integer. So, yes, that's the definition.

Alternatively, maybe there's a different interpretation? For example, maybe the question is asking for how many n>1 such that 2^24 is a perfect n-th power, but not necessarily that n divides 24? But no, as we saw earlier, if 2^24 is a perfect n-th power, then n must divide 24.

Alternatively, maybe there's a case where n is not a divisor of 24, but 2^24 is still a perfect n-th power? For example, if n is a divisor of 24, but also... Wait, no. Let me think. Suppose n is  something else. For example, n=  something like 24/k, but no.

Wait, suppose n is  something like 24 divided by some number. Wait, but if n divides 24, then 24/n is integer. So, if n does not divide 24, then 24/n is not integer. Therefore, if 2^24 is equal to k^n, then k would have to be 2^{24/n}, which would not be integer unless 24/n is integer. Therefore, n must divide 24.

Therefore, the conclusion is that n must divide 24. Therefore, the answer is 7.

But let me check again. Let me think of n= something else. For example, n=  something like 24. Well, n=24 is allowed. What about n= something like  something else? For example, n=  something like  2^something?

Wait, but n is just any integer greater than 1. So, for example, n=  something like  2^3=8, which is already in the list.

Alternatively, maybe n=  something like  2^something * 3^something? For example, n= 6, which is 2*3. Which is already in the list.

So, all divisors of 24 are accounted for. Therefore, the answer is 7.

But wait, hold on. Wait, the problem says "positive integers n>1". So, n can be any integer greater than 1. But if n is a divisor of 24, then it's okay. But maybe there are other numbers?

Wait, for example, suppose n= something like  something that divides 24, but also... Wait, no. If n divides 24, then it's already considered.

Alternatively, maybe n is a divisor of 24, but also... Wait, no.

Alternatively, maybe n is a divisor of 24, but also... Wait, no.

Alternatively, maybe there's a case where 2^24 is a perfect n-th power even if n does not divide 24? For example, if n is a fraction? But n has to be an integer.

Alternatively, if n is a multiple of some divisor? Wait, no. For example, suppose n=  something like  24*2=48. Then, 2^24 is (2^{24/48})^48 = (2^{0.5})^48 = (sqrt(2))^48 = 2^{24}, which is true, but sqrt(2) is not an integer. Therefore, k would have to be sqrt(2), which is not an integer. Therefore, 2^24 is not a perfect 48th power in integers. Therefore, n=48 is not allowed.

Therefore, only divisors of 24.

Therefore, the answer is 7.

But let me check again the list of divisors. Divisors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. So, excluding 1, there are 7 divisors. Therefore, the answer is 7.

But wait, hold on. Wait, the problem says "for how many positive integers n>1". So, 7. But let me check again.

Wait, but maybe there's a mistake here. For example, if n is  something like  something else. Wait, for example, n=  something like  2^something.

Wait, but n can be any divisor of 24. So, for example, n=2, which is 2^1. n=3, which is 3^1. n=4=2^2. n=6=2*3. n=8=2^3. n=12=2^2*3. n=24=2^3*3. So, all of these are divisors.

Therefore, there are 7 such n's.

But let me check again with the original problem. The problem says "For how many positive integers n>1 is it true that 2^{24} is a perfect n^th power?"

So, if n divides 24, then 2^{24} is a perfect n-th power. Therefore, the number of such n is equal to the number of divisors of 24 greater than 1, which is 7.

But wait, hold on. Wait, let me check if n= something like  something else. For example, n=  something like  24.

But 24 is allowed. So, 2^24 is (2^1)^24. So, yes.

Alternatively, maybe n= something like  1? But n>1.

Therefore, the answer is 7.

But wait, I remember that sometimes in math problems, when they say "perfect n-th power", they might mean that the number is a perfect power in the sense that it's an integer power, but not necessarily that the exponent is a divisor. Wait, but in this case, since the number is 2^24, which is a power of 2, so if you want it to be a perfect n-th power, then n must divide the exponent.

Alternatively, maybe there's a different way to think about it. For example, if 2^24 is a perfect n-th power, then there exists some integer k such that k^n = 2^24. Then, since 2 is prime, k must be a power of 2. Let me write k as 2^m. Then, (2^m)^n = 2^{m*n} = 2^24. Therefore, m*n = 24. Therefore, n must divide 24.

Therefore, n is a positive integer divisor of 24. Therefore, the number of such n>1 is equal to the number of divisors of 24 greater than 1, which is 7.

Therefore, the answer is 7.

But wait, let me check once more. Let me list all the divisors of 24 again:

1, 2, 3, 4, 6, 8, 12, 24.

So, removing 1, we have 2, 3, 4, 6, 8, 12, 24. That's 7 numbers.

Therefore, the answer is 7.

But wait, hold on. Wait, I think I remember that sometimes when you have exponents, you have to consider exponents that are divisors of the exponent. But in this case, since the base is prime, it's straightforward.

Alternatively, maybe there's a mistake here. For example, if n is  something like  something else. Wait, for example, n=  something like  24.

But 2^24 is (2^1)^24, which is a perfect 24th power. So, yes.

Alternatively, n=  something like  2.

Yes, 2^24 is (2^12)^2.

So, all of them work.

Therefore, the answer is 7.

But wait, hold on. Wait, the problem says "perfect n-th power". But sometimes, people might consider that a perfect n-th power must be a perfect square, cube, etc., but not necessarily that the exponent is a divisor. Wait, but in this case, the definition is that if a number is a perfect n-th power, then it can be written as some integer raised to the n-th power.

Therefore, since 2^24 is equal to (2^m)^n, then m*n=24. Therefore, n must divide 24.

Therefore, the answer is 7.

But let me check if there's a different interpretation. For example, maybe the question is asking for how many n>1 such that 2^24 is a perfect n-th power, but not necessarily that n divides 24. But as we saw, that's not possible.

Alternatively, maybe the question is asking for how many n>1 such that 2^24 is a perfect n-th power in the sense that it's a perfect power with exponent n, but not necessarily that n divides 24. But in that case, as we saw, n must divide 24.

Therefore, the answer is 7.

But wait, let me check with n= something like  something else. For example, n=  something like  24.

But n=24 is allowed.

Alternatively, n= something like  12.

Yes, 2^24 is (2^2)^12.

So, yes.

Therefore, I think the answer is 7.

But wait, hold on. Wait, I just thought of something. What if n is a divisor of 24, but also n is a perfect power itself? For example, n=4, which is 2^2. But that doesn't affect the count. The question is not about n being a perfect power, but about 2^24 being a perfect n-th power.

Therefore, regardless of n's properties, as long as n divides 24, it's okay.

Therefore, the answer is 7.

But let me check again. Let me think of the problem in another way.

Suppose we have 2^24. Let me think of exponents. For 2^24 to be a perfect n-th power, then the exponent 24 must be a multiple of n. Therefore, n divides 24. Therefore, the number of n>1 is equal to the number of divisors of 24 greater than 1.

Therefore, since 24 has 8 divisors, subtract 1, gives 7.

Therefore, the answer is 7.

But wait, hold on. Wait, the problem says "positive integers n>1". So, n can be any positive integer greater than 1. Therefore, the answer is 7.

But I have a doubt here because sometimes problems like this can have a trick. For example, maybe considering exponents that are not divisors but still... Wait, but as we saw, if n does not divide 24, then 24/n is not integer, so k would have to be 2^{24/n}, which is not integer. Therefore, no solution.

Therefore, the answer is 7.

But let me check with n= something like  something else. For example, n=  something like  24.

Yes, n=24 is allowed.

Alternatively, n=  something like  1. But n>1.

Therefore, I think the answer is 7.

But wait, hold on. Wait, let me check the original problem again.

"For how many positive integers n>1 is it true that 2^{24} is a perfect n^th power?"

So, the problem is in English, and sometimes "perfect n-th power" can be ambiguous. But in mathematics, a perfect n-th power is a number that can be expressed as some integer raised to the n-th power. Therefore, the reasoning is correct.

Therefore, the answer is 7.

But let me check once more. Let me list all the divisors of 24 greater than 1:

2, 3, 4, 6, 8, 12, 24.

That's 7 numbers.

Yes. Therefore, the answer is 7.

But wait, hold on. Wait, I just thought of something. What if n is a divisor of 24, but also n is a multiple of some other number? For example, n=  something like  2. But that's already considered.

Alternatively, maybe there's a case where n is not a divisor of 24, but 2^24 is still a perfect n-th power. For example, if n is  something like  24/k, but no.

Wait, for example, suppose n=  something like  24/ something. Wait, but if n divides 24, then 24/n is integer. If n does not divide 24, then 24/n is not integer. Therefore, k would have to be 2^{24/n}, which is not integer. Therefore, no.

Therefore, the answer is 7.

Therefore, I think the answer is 7.

But wait, hold on. Wait, I just checked the problem again. The problem says "perfect n-th power". Maybe the question is referring to the number being a perfect power in the sense of being a square, cube, etc., but not necessarily that it's a perfect n-th power for that specific n. Wait, but the wording is "is it true that 2^{24} is a perfect n-th power". So, for each n>1, check if 2^24 is a perfect n-th power.

Therefore, the answer is 7.

But maybe the answer is different? Wait, let me check with n= something like  something else.

Wait, for example, n=  something like  2. 2^24 is a perfect square. Yes.

n=3: 2^24 is a perfect cube. Yes.

n=4: 2^24 is a perfect fourth power. Yes.

n=6: 2^24 is a perfect sixth power. Yes.

n=8: 2^24 is a perfect eighth power. Yes.

n=12: 2^24 is a perfect twelfth power. Yes.

n=24: 2^24 is a perfect 24th power. Yes.

So, all of these work. Therefore, 7 values.

Therefore, the answer is 7.

But wait, hold on. Wait, I just thought of something. What if n is  something like  something else. For example, n=  something like  2^something.

But n is already considered in the list.

Alternatively, maybe the problem is in the way the question is phrased. For example, maybe "perfect n-th power" is meant to be a perfect power with exponent n, but not necessarily that n divides 24. But as we saw, that's not possible.

Alternatively, maybe the problem is in the definition of perfect power. For example, sometimes perfect power is defined as a number that can be expressed as m^k where k ≥ 2. But in this case, the question is specifically asking for which n>1 is 2^24 a perfect n-th power. So, for each n>1, check if there exists an integer m such that m^n = 2^24.

Which, as we saw, requires that n divides 24.

Therefore, the answer is 7.

Therefore, I think the answer is 7.

But let me check once more.

Divisors of 24: 1, 2, 3, 4, 6, 8, 12, 24.

Number of divisors greater than 1: 7.

Therefore, answer is 7.

But wait, hold on. Wait, I just thought of something. What if n is  something like  something else. For example, n=  something like  24.

But n=24 is allowed.

Alternatively, maybe the problem is in the fact that 2^24 is also a perfect power for exponents that are not divisors of 24? For example, if n= something like  something else.

Wait, for example, n=  something like  24/ something.

Wait, suppose n=  something like  24/ something. For example, n=  something like  24/ something.

But if n is not a divisor of 24, then 24/n is not integer. Therefore, m would have to be 2^{24/n}, which is not integer. Therefore, no.

Therefore, the answer is 7.

Therefore, I think the answer is 7.

But I just want to check if there's any other possible way. For example, if n is  something like  2^something.

But n is already considered in the list.

Alternatively, maybe the problem is in the fact that exponents can be negative? But n is a positive integer greater than 1.

Therefore, no.

Therefore, I think the answer is 7.

But wait, hold on. Wait, I just thought of something. What if n is  something like  24.

But n=24 is allowed.

Alternatively, maybe the problem is in the fact that 2^24 is also a perfect power for exponents that are not divisors of 24. For example, if n= something like  something else.

Wait, for example, n=  something like  24/ something.

But if n is not a divisor of 24, then 24/n is not integer. Therefore, m would have to be 2^{24/n}, which is not integer. Therefore, no.

Therefore, the answer is 7.

Therefore, I think the answer is 7.

But let me check once more.

List of n>1 where 2^24 is a perfect n-th power:

n=2: yes

n=3: yes

n=4: yes

n=6: yes

n=8: yes

n=12: yes

n=24: yes

That's 7 numbers.

Yes.

Therefore, the answer is 7.

But wait, hold on. Wait, the problem says "positive integers n>1". So, n=2,3,4,6,8,12,24.

Yes, 7 numbers.

Therefore, the answer is 7.

But I just want to check if there's a possibility that n= something like  something else. For example, n= something like  2^something.

But n=2,4,8 are already in the list.

Therefore, no.

Therefore, I think the answer is 7.

But I just want to check if there's a different interpretation. For example, maybe the question is asking for how many n>1 such that 2^24 is a perfect n-th power, but not necessarily that n divides 24. But as we saw, that's not possible.

Alternatively, maybe the question is asking for how many n>1 such that 2^24 is a perfect n-th power, but n is not necessarily a divisor of 24. But in that case, as we saw, there are no such n.

Therefore, the answer is 7.

Therefore, I think the answer is 7.

But wait, hold on. Wait, I just thought of something. What if n is  something like  something else. For example, n=  something like  24.

But n=24 is allowed.

Alternatively, maybe the problem is in the fact that exponents can be fractions? But n is an integer.

Therefore, no.

Therefore, I think the answer is 7.

Therefore, I think the answer is 7.

**Final Answer**
\boxed{7}

To determine how many positive integers $ n > 1 $ make $ 2^{24} $ a perfect $ n $-th power, we begin by understanding what it means for a number to be a perfect $ n $-th power.

---

### Understanding the Problem

A number is a **perfect $ n $-th power** if it can be written in the form $ k^n $, where $ k $ is an integer and $ n > 1 $. In this case, we are given:

$$
2^{24}
$$

We want to find all integers $ n > 1 $ such that:

$$
2^{24} = k^n
$$

for some integer $ k $.

---

### Prime Factorization Insight

Since $ 2^{24} $ is a power of a prime number (2), any integer $ k $ that satisfies $ k^n = 2^{24} $ must also be a power of 2. Let’s suppose:

$$
k = 2^m
$$

Then:

$$
k^n = (2^m)^n = 2^{mn}
$$

So we have:

$$
2^{mn} = 2^{24} \Rightarrow mn = 24
$$

This implies that $ n $ must be a **positive divisor of 24**. Therefore, the values of $ n $ that satisfy the condition are exactly the **positive divisors of 24** that are **greater than 1**.

---

### Finding the Divisors of 24

We factor 24:

$$
24 = 2^3 \cdot 3^1
$$

The number of positive divisors is:

$$
(3 + 1)(1 + 1) = 4 \cdot 2 = 8
$$

The divisors of 24 are:

$$
1, 2, 3, 4, 6, 8, 12, 24
$$

Excluding 1 (since $ n > 1 $), the valid values of $ n $ are:

$$
2, 3, 4, 6, 8, 12, 24
$$

---

### Final Count

There are **7** such values of $ n $.

---

### Final Answer

$$
\boxed{7}
$$

[Assistant]
Explanation:
Okay, so I need to figure out for how many positive integers n > 1 it's true that 2^24 is a perfect n-th power. Hmm, let me start by recalling what a perfect n-th power means. If a number is a perfect n-th power, that means it can be written as some integer raised to the n-th power. So, in this case, 2^24 should be equal to some integer k raised to the n-th power, right?

So, mathematically, we can write that as:

2^24 = k^n

Where k is a positive integer and n > 1. The question is asking for how many such n exist.

First, let me think about the prime factorization of 2^24. Well, since 2 is already a prime number, the prime factorization of 2^24 is just 2^24. So, if 2^24 is a perfect n-th power, then all the exponents in its prime factorization must be multiples of n.

Wait, let me recall that. If a number is a perfect n-th power, then in its prime factorization, each exponent must be divisible by n. For example, if you have a number like 36, which is 2^2 * 3^2, then it's a perfect square because both exponents are multiples of 2. Similarly, if you have 64, which is 2^6, it's a perfect cube because 6 is divisible by 3, and also a perfect square because 6 is divisible by 2.

So applying that to 2^24, since the prime factorization is just 2^24, the exponent here is 24. Therefore, for 2^24 to be a perfect n-th power, 24 must be divisible by n. In other words, n must be a divisor of 24.

But wait, hold on. Let me make sure. If n divides 24, then 24 = n * m for some integer m. Then, 2^24 = (2^m)^n, which would mean that 2^24 is indeed a perfect n-th power. So, n must be a divisor of 24.

But the question is asking for how many positive integers n > 1 satisfy this. Therefore, I need to find all divisors of 24 that are greater than 1.

But hold on, is that all? Let me think again. Suppose n is a divisor of 24. Then, 2^24 is (2^{24/n})^n, which is an integer. Therefore, yes, n divides 24. However, is there a possibility that even if n does not divide 24, 2^24 could still be a perfect n-th power?

Wait, no. Because if 2^24 is equal to k^n, then taking the prime factorization of k, which would be some product of primes, but since 2^24 is only composed of the prime 2, then k must also be a power of 2. Let me explain.

Suppose k is not a power of 2. Then, k would have some other prime factors, say p, so k = 2^a * p^b * ... Then, k^n would have prime factors 2^{a*n}, p^{b*n}, etc. But 2^24 only has 2 as a prime factor, so all other primes in the factorization of k^n would have to be zero. Therefore, k must be a power of 2.

Therefore, k = 2^m for some integer m. Then, k^n = (2^m)^n = 2^{m*n}. Therefore, 2^{m*n} = 2^24, so m*n = 24. Therefore, n must divide 24.

Therefore, the conclusion is that n must be a positive integer divisor of 24, greater than 1. Therefore, the problem reduces to finding the number of divisors of 24 that are greater than 1.

So, first, let me find all the divisors of 24.

First, prime factorization of 24. 24 can be broken down into primes as 2^3 * 3^1.

The formula for the number of divisors is (exponent + 1) for each prime factor, multiplied together. So, for 2^3 * 3^1, the number of divisors is (3 + 1)*(1 + 1) = 4*2 = 8. Therefore, there are 8 positive divisors of 24.

Let me list them out to be sure. The divisors of 24 are:

1, 2, 3, 4, 6, 8, 12, 24.

Yes, that's 8 divisors. So, the question is asking for how many of these divisors are greater than 1. Since the divisors are 1, 2, 3, 4, 6, 8, 12, 24, then excluding 1, there are 7 divisors.

But wait, hold on. The problem says "positive integers n > 1", so n must be greater than 1. Therefore, the answer would be 7? But let me check again.

Wait, but maybe there's a mistake here. Because when we say that n must divide 24, but maybe there are some n's that divide 24 but for which 2^24 is not a perfect n-th power? Wait, no. Because if n divides 24, then 24 = n * m, so 2^24 = (2^m)^n, which is a perfect n-th power. Therefore, all divisors of 24 greater than 1 would satisfy the condition.

But let me check with examples. Let's take n=2. Then 2^24 is (2^12)^2, which is a perfect square. Correct. For n=3, 2^24 is (2^8)^3, since 8*3=24. Correct. For n=4, 2^24 is (2^6)^4, since 6*4=24. Correct. For n=6, 2^24 is (2^4)^6, since 4*6=24. For n=8, 2^24 is (2^3)^8, since 3*8=24. For n=12, 2^24 is (2^2)^12, since 2*12=24. For n=24, 2^24 is (2^1)^24. So all of these work.

Therefore, there are 7 such n's. But wait, the problem says "positive integers n > 1", so n=2,3,4,6,8,12,24. That's 7 numbers. Therefore, the answer is 7?

But wait, hold on. Let me check if there are any other n's that are not divisors of 24 but still make 2^24 a perfect n-th power. For example, suppose n is a fraction? But n has to be a positive integer greater than 1. So fractions are not allowed.

Alternatively, could there be some n that divides 24, but for which 2^24 is not a perfect n-th power? For example, if n is 0? But n has to be greater than 1.

Wait, but if n divides 24, then 2^24 is a perfect n-th power. So, if n divides 24, then yes. Therefore, the number of n's is equal to the number of divisors of 24 greater than 1, which is 7.

But let me check again. Let me think if there's a case where even if n does not divide 24, but 2^24 is still a perfect n-th power. For example, suppose n= something else. Let me think.

Suppose n= something like 5. Then, 2^24 is a perfect 5th power? Let's see. If 2^24 = k^5, then k would have to be 2^m, so 2^{24} = 2^{5m}, so 5m=24. But 24 is not divisible by 5, so m would not be integer. Therefore, no solution. Similarly, for n=7, same thing. 24 divided by 7 is not integer. Therefore, no.

Similarly, n= something else. For example, n= 24. Which is allowed. So, seems like only divisors of 24.

Therefore, the answer is 7.

But wait, hold on. Wait, the problem says "for how many positive integers n>1". So, divisors of 24 greater than 1. Which are 2,3,4,6,8,12,24. So 7 numbers.

But let me check again the list of divisors. Divisors of 24: 1, 2, 3, 4, 6, 8, 12, 24. So yes, 8 divisors. Excluding 1, 7 divisors. Therefore, answer is 7.

But wait, hold on. Wait, the problem says "perfect n-th power". Is there a possibility that for some n, even though n divides 24, 2^24 is not a perfect n-th power? For example, if n is 1? But n>1.

Wait, but for example, take n=2. 2^24 is (2^12)^2, which is a perfect square. Similarly, n=3, (2^8)^3. So, yes. So, all divisors of 24 greater than 1 work.

But let me think again. Suppose n is a divisor of 24, but maybe n is not a divisor of 24? Wait, no. If n divides 24, then 24/n is integer. So, 2^24 is (2^{24/n})^n. Therefore, it's a perfect n-th power.

Therefore, the answer is 7.

But wait, hold on. Wait, the problem says "perfect n-th power". But sometimes, people might think that a perfect n-th power must be an integer. But in this case, since 2^24 is an integer, and we are expressing it as k^n, where k is an integer. So, yes, that's the definition.

Alternatively, maybe there's a different interpretation? For example, maybe the question is asking for how many n>1 such that 2^24 is a perfect n-th power, but not necessarily that n divides 24? But no, as we saw earlier, if 2^24 is a perfect n-th power, then n must divide 24.

Alternatively, maybe there's a case where n is not a divisor of 24, but 2^24 is still a perfect n-th power? For example, if n is a divisor of 24, but also... Wait, no. Let me think. Suppose n is  something else. For example, n=  something like 24/k, but no.

Wait, suppose n is  something like 24 divided by some number. Wait, but if n divides 24, then 24/n is integer. So, if n does not divide 24, then 24/n is not integer. Therefore, if 2^24 is equal to k^n, then k would have to be 2^{24/n}, which would not be integer unless 24/n is integer. Therefore, n must divide 24.

Therefore, the conclusion is that n must divide 24. Therefore, the answer is 7.

But let me check again. Let me think of n= something else. For example, n=  something like 24. Well, n=24 is allowed. What about n= something like  something else? For example, n=  something like  2^something?

Wait, but n is just any integer greater than 1. So, for example, n=  something like  2^3=8, which is already in the list.

Alternatively, maybe n=  something like  2^something * 3^something? For example, n= 6, which is 2*3. Which is already in the list.

So, all divisors of 24 are accounted for. Therefore, the answer is 7.

But wait, hold on. Wait, the problem says "positive integers n>1". So, n can be any integer greater than 1. But if n is a divisor of 24, then it's okay. But maybe there are other numbers?

Wait, for example, suppose n= something like  something that divides 24, but also... Wait, no. If n divides 24, then it's already considered.

Alternatively, maybe n is a divisor of 24, but also... Wait, no.

Alternatively, maybe n is a divisor of 24, but also... Wait, no.

Alternatively, maybe there's a case where 2^24 is a perfect n-th power even if n does not divide 24? For example, if n is a fraction? But n has to be an integer.

Alternatively, if n is a multiple of some divisor? Wait, no. For example, suppose n=  something like  24*2=48. Then, 2^24 is (2^{24/48})^48 = (2^{0.5})^48 = (sqrt(2))^48 = 2^{24}, which is true, but sqrt(2) is not an integer. Therefore, k would have to be sqrt(2), which is not an integer. Therefore, 2^24 is not a perfect 48th power in integers. Therefore, n=48 is not allowed.

Therefore, only divisors of 24.

Therefore, the answer is 7.

But let me check again the list of divisors. Divisors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. So, excluding 1, there are 7 divisors. Therefore, the answer is 7.

But wait, hold on. Wait, the problem says "for how many positive integers n>1". So, 7. But let me check again.

Wait, but maybe there's a mistake here. For example, if n is  something like  something else. Wait, for example, n=  something like  2^something.

Wait, but n can be any divisor of 24. So, for example, n=2, which is 2^1. n=3, which is 3^1. n=4=2^2. n=6=2*3. n=8=2^3. n=12=2^2*3. n=24=2^3*3. So, all of these are divisors.

Therefore, there are 7 such n's.

But let me check again with the original problem. The problem says "For how many positive integers n>1 is it true that 2^{24} is a perfect n^th power?"

So, if n divides 24, then 2^{24} is a perfect n-th power. Therefore, the number of such n is equal to the number of divisors of 24 greater than 1, which is 7.

But wait, hold on. Wait, let me check if n= something like  something else. For example, n=  something like  24.

But 24 is allowed. So, 2^24 is (2^1)^24. So, yes.

Alternatively, maybe n= something like  1? But n>1.

Therefore, the answer is 7.

But wait, I remember that sometimes in math problems, when they say "perfect n-th power", they might mean that the number is a perfect power in the sense that it's an integer power, but not necessarily that the exponent is a divisor. Wait, but in this case, since the number is 2^24, which is a power of 2, so if you want it to be a perfect n-th power, then n must divide the exponent.

Alternatively, maybe there's a different way to think about it. For example, if 2^24 is a perfect n-th power, then there exists some integer k such that k^n = 2^24. Then, since 2 is prime, k must be a power of 2. Let me write k as 2^m. Then, (2^m)^n = 2^{m*n} = 2^24. Therefore, m*n = 24. Therefore, n must divide 24.

Therefore, n is a positive integer divisor of 24. Therefore, the number of such n>1 is equal to the number of divisors of 24 greater than 1, which is 7.

Therefore, the answer is 7.

But wait, let me check once more. Let me list all the divisors of 24 again:

1, 2, 3, 4, 6, 8, 12, 24.

So, removing 1, we have 2, 3, 4, 6, 8, 12, 24. That's 7 numbers.

Therefore, the answer is 7.

But wait, hold on. Wait, I think I remember that sometimes when you have exponents, you have to consider exponents that are divisors of the exponent. But in this case, since the base is prime, it's straightforward.

Alternatively, maybe there's a mistake here. For example, if n is  something like  something else. Wait, for example, n=  something like  24.

But 2^24 is (2^1)^24, which is a perfect 24th power. So, yes.

Alternatively, n=  something like  2.

Yes, 2^24 is (2^12)^2.

So, all of them work.

Therefore, the answer is 7.

But wait, hold on. Wait, the problem says "perfect n-th power". But sometimes, people might consider that a perfect n-th power must be a perfect square, cube, etc., but not necessarily that the exponent is a divisor. Wait, but in this case, the definition is that if a number is a perfect n-th power, then it can be written as some integer raised to the n-th power.

Therefore, since 2^24 is equal to (2^m)^n, then m*n=24. Therefore, n must divide 24.

Therefore, the answer is 7.

But let me check if there's a different interpretation. For example, maybe the question is asking for how many n>1 such that 2^24 is a perfect n-th power, but not necessarily that n divides 24. But as we saw, that's not possible.

Alternatively, maybe the question is asking for how many n>1 such that 2^24 is a perfect n-th power in the sense that it's a perfect power with exponent n, but not necessarily that n divides 24. But in that case, as we saw, n must divide 24.

Therefore, the answer is 7.

But wait, let me check with n= something like  something else. For example, n=  something like  24.

But n=24 is allowed.

Alternatively, n= something like  12.

Yes, 2^24 is (2^2)^12.

So, yes.

Therefore, I think the answer is 7.

But wait, hold on. Wait, I just thought of something. What if n is a divisor of 24, but also n is a perfect power itself? For example, n=4, which is 2^2. But that doesn't affect the count. The question is not about n being a perfect power, but about 2^24 being a perfect n-th power.

Therefore, regardless of n's properties, as long as n divides 24, it's okay.

Therefore, the answer is 7.

But let me check again. Let me think of the problem in another way.

Suppose we have 2^24. Let me think of exponents. For 2^24 to be a perfect n-th power, then the exponent 24 must be a multiple of n. Therefore, n divides 24. Therefore, the number of n>1 is equal to the number of divisors of 24 greater than 1.

Therefore, since 24 has 8 divisors, subtract 1, gives 7.

Therefore, the answer is 7.

But wait, hold on. Wait, the problem says "positive integers n>1". So, n can be any positive integer greater than 1. Therefore, the answer is 7.

But I have a doubt here because sometimes problems like this can have a trick. For example, maybe considering exponents that are not divisors but still... Wait, but as we saw, if n does not divide 24, then 24/n is not integer, so k would have to be 2^{24/n}, which is not integer. Therefore, no solution.

Therefore, the answer is 7.

But let me check with n= something like  something else. For example, n=  something like  24.

Yes, n=24 is allowed.

Alternatively, n=  something like  1. But n>1.

Therefore, I think the answer is 7.

But wait, hold on. Wait, let me check the original problem again.

"For how many positive integers n>1 is it true that 2^{24} is a perfect n^th power?"

So, the problem is in English, and sometimes "perfect n-th power" can be ambiguous. But in mathematics, a perfect n-th power is a number that can be expressed as some integer raised to the n-th power. Therefore, the reasoning is correct.

Therefore, the answer is 7.

But let me check once more. Let me list all the divisors of 24 greater than 1:

2, 3, 4, 6, 8, 12, 24.

That's 7 numbers.

Yes. Therefore, the answer is 7.

But wait, hold on. Wait, I just thought of something. What if n is a divisor of 24, but also n is a multiple of some other number? For example, n=  something like  2. But that's already considered.

Alternatively, maybe there's a case where n is not a divisor of 24, but 2^24 is still a perfect n-th power. For example, if n is  something like  24/k, but no.

Wait, for example, suppose n=  something like  24/ something. Wait, but if n divides 24, then 24/n is integer. If n does not divide 24, then 24/n is not integer. Therefore, k would have to be 2^{24/n}, which is not integer. Therefore, no.

Therefore, the answer is 7.

Therefore, I think the answer is 7.

But wait, hold on. Wait, I just checked the problem again. The problem says "perfect n-th power". Maybe the question is referring to the number being a perfect power in the sense of being a square, cube, etc., but not necessarily that it's a perfect n-th power for that specific n. Wait, but the wording is "is it true that 2^{24} is a perfect n-th power". So, for each n>1, check if 2^24 is a perfect n-th power.

Therefore, the answer is 7.

But maybe the answer is different? Wait, let me check with n= something like  something else.

Wait, for example, n=  something like  2. 2^24 is a perfect square. Yes.

n=3: 2^24 is a perfect cube. Yes.

n=4: 2^24 is a perfect fourth power. Yes.

n=6: 2^24 is a perfect sixth power. Yes.

n=8: 2^24 is a perfect eighth power. Yes.

n=12: 2^24 is a perfect twelfth power. Yes.

n=24: 2^24 is a perfect 24th power. Yes.

So, all of these work. Therefore, 7 values.

Therefore, the answer is 7.

But wait, hold on. Wait, I just thought of something. What if n is  something like  something else. For example, n=  something like  2^something.

But n is already considered in the list.

Alternatively, maybe the problem is in the way the question is phrased. For example, maybe "perfect n-th power" is meant to be a perfect power with exponent n, but not necessarily that n divides 24. But as we saw, that's not possible.

Alternatively, maybe the problem is in the definition of perfect power. For example, sometimes perfect power is defined as a number that can be expressed as m^k where k ≥ 2. But in this case, the question is specifically asking for which n>1 is 2^24 a perfect n-th power. So, for each n>1, check if there exists an integer m such that m^n = 2^24.

Which, as we saw, requires that n divides 24.

Therefore, the answer is 7.

Therefore, I think the answer is 7.

But let me check once more.

Divisors of 24: 1, 2, 3, 4, 6, 8, 12, 24.

Number of divisors greater than 1: 7.

Therefore, answer is 7.

But wait, hold on. Wait, I just thought of something. What if n is  something like  something else. For example, n=  something like  24.

But n=24 is allowed.

Alternatively, maybe the problem is in the fact that 2^24 is also a perfect power for exponents that are not divisors of 24? For example, if n= something like  something else.

Wait, for example, n=  something like  24/ something.

Wait, suppose n=  something like  24/ something. For example, n=  something like  24/ something.

But if n is not a divisor of 24, then 24/n is not integer. Therefore, m would have to be 2^{24/n}, which is not integer. Therefore, no.

Therefore, the answer is 7.

Therefore, I think the answer is 7.

But I just want to check if there's any other possible way. For example, if n is  something like  2^something.

But n is already considered in the list.

Alternatively, maybe the problem is in the fact that exponents can be negative? But n is a positive integer greater than 1.

Therefore, no.

Therefore, I think the answer is 7.

But wait, hold on. Wait, I just thought of something. What if n is  something like  24.

But n=24 is allowed.

Alternatively, maybe the problem is in the fact that 2^24 is also a perfect power for exponents that are not divisors of 24. For example, if n= something like  something else.

Wait, for example, n=  something like  24/ something.

But if n is not a divisor of 24, then 24/n is not integer. Therefore, m would have to be 2^{24/n}, which is not integer. Therefore, no.

Therefore, the answer is 7.

Therefore, I think the answer is 7.

But let me check once more.

List of n>1 where 2^24 is a perfect n-th power:

n=2: yes

n=3: yes

n=4: yes

n=6: yes

n=8: yes

n=12: yes

n=24: yes

That's 7 numbers.

Yes.

Therefore, the answer is 7.

But wait, hold on. Wait, the problem says "positive integers n>1". So, n=2,3,4,6,8,12,24.

Yes, 7 numbers.

Therefore, the answer is 7.

But I just want to check if there's a possibility that n= something like  something else. For example, n= something like  2^something.

But n=2,4,8 are already in the list.

Therefore, no.

Therefore, I think the answer is 7.

But I just want to check if there's a different interpretation. For example, maybe the question is asking for how many n>1 such that 2^24 is a perfect n-th power, but not necessarily that n divides 24. But as we saw, that's not possible.

Alternatively, maybe the question is asking for how many n>1 such that 2^24 is a perfect n-th power, but n is not necessarily a divisor of 24. But in that case, as we saw, there are no such n.

Therefore, the answer is 7.

Therefore, I think the answer is 7.

But wait, hold on. Wait, I just thought of something. What if n is  something like  something else. For example, n=  something like  24.

But n=24 is allowed.

Alternatively, maybe the problem is in the fact that exponents can be fractions? But n is an integer.

Therefore, no.

Therefore, I think the answer is 7.

Therefore, I think the answer is 7.

**Final Answer**
\boxed{7}

To determine how many positive integers $ n > 1 $ make $ 2^{24} $ a perfect $ n $-th power, we begin by understanding what it means for a number to be a perfect $ n $-th power.

---

### Understanding the Problem

A number is a **perfect $ n $-th power** if it can be written in the form $ k^n $, where $ k $ is an integer and $ n > 1 $. In this case, we are given:

$$
2^{24}
$$

We want to find all integers $ n > 1 $ such that:

$$
2^{24} = k^n
$$

for some integer $ k $.

---

### Prime Factorization Insight

Since $ 2^{24} $ is a power of a prime number (2), any integer $ k $ that satisfies $ k^n = 2^{24} $ must also be a power of 2. Let’s suppose:

$$
k = 2^m
$$

Then:

$$
k^n = (2^m)^n = 2^{mn}
$$

So we have:

$$
2^{mn} = 2^{24} \Rightarrow mn = 24
$$

This implies that $ n $ must be a **positive divisor of 24**. Therefore, the values of $ n $ that satisfy the condition are exactly the **positive divisors of 24** that are **greater than 1**.

---

### Finding the Divisors of 24

We factor 24:

$$
24 = 2^3 \cdot 3^1
$$

The number of positive divisors is:

$$
(3 + 1)(1 + 1) = 4 \cdot 2 = 8
$$

The divisors of 24 are:

$$
1, 2, 3, 4, 6, 8, 12, 24
$$

Excluding 1 (since $ n > 1 $), the valid values of $ n $ are:

$$
2, 3, 4, 6, 8, 12, 24
$$

---

### Final Count

There are **7** such values of $ n $.

---

### Final Answer

$$
\boxed{7}
$$

Solution: Note that $2^{24}$ is a perfect $n$th power if and only if $n$ is a divisor of 24. The factors of 24 which are greater than 1 are 2, 3, 4, 6, 8, 12, and 24, so we have $\boxed{7}$ possible values of $n$.


Explanation:
Okay, so I need to find p(0) + p(4) where p(x) is a monic polynomial of degree 4, and three of its roots are 1, 2, and 3. Let me start by recalling what a monic polynomial is. A monic polynomial is one where the leading coefficient (the coefficient of the highest power of x) is 1. Since it's a degree 4 polynomial, it has four roots (counting multiplicities). They told me three of the roots are 1, 2, and 3, so there must be a fourth root. Let me call that fourth root 'a'.

So, the polynomial can be written in factored form as p(x) = (x - 1)(x - 2)(x - 3)(x - a). Since it's monic, the leading coefficient is 1, which matches this form. Now, I need to figure out what 'a' is. Wait, but the problem doesn't give me any other information about the polynomial. Hmm. Maybe I don't need to find 'a' explicitly? Let me think.

The question asks for p(0) + p(4). Let me compute each term separately. Let's start with p(0). If I plug in x = 0 into the polynomial, I get p(0) = (0 - 1)(0 - 2)(0 - 3)(0 - a) = (-1)(-2)(-3)(-a). Let me compute that. Multiplying the constants first: (-1)(-2) is 2, then 2*(-3) is -6, then -6*(-a) is 6a. So p(0) = 6a.

Now, let's compute p(4). Plugging in x = 4, we get p(4) = (4 - 1)(4 - 2)(4 - 3)(4 - a) = (3)(2)(1)(4 - a). Multiplying those constants: 3*2 is 6, 6*1 is 6, so p(4) = 6*(4 - a) = 24 - 6a.

Therefore, p(0) + p(4) = 6a + 24 - 6a = 24. Wait, that's interesting. The 6a and -6a cancel each other out, leaving just 24. So regardless of the value of 'a', p(0) + p(4) is always 24? That seems surprising, but maybe that's correct. Let me check my steps again to make sure I didn't make a mistake.

First, p(x) is monic quartic with roots 1, 2, 3, and a. So factored form is correct. Then p(0) is product of (-1)(-2)(-3)(-a). Let me compute that again. (-1)*(-2) is 2, then 2*(-3) is -6, then -6*(-a) is 6a. Yes, that's correct. Then p(4) is (3)(2)(1)(4 - a) which is 6*(4 - a) = 24 - 6a. Then adding them together: 6a + 24 - 6a = 24. So yes, the answer is 24.

But wait, is there a possibility that the polynomial has a repeated root? The problem says "three of the roots are 1, 2, and 3". It doesn't specify if they are distinct or not. But since it's a degree 4 polynomial, and three roots are given, the fourth root could be any number, possibly one of the existing roots. However, even if there is a repeated root, the calculation for p(0) + p(4) would still hold because the fourth root is still 'a', whether it's equal to 1, 2, 3, or something else. For example, if a = 1, then p(x) would be (x - 1)^2(x - 2)(x - 3). But in that case, p(0) would be (-1)^2*(-2)*(-3) = 1*6 = 6, and p(4) would be (3)^2*(2)*(1) = 9*2 = 18. Then p(0) + p(4) = 6 + 18 = 24. Same result. Similarly, if a = 2, p(0) would be (-1)(-2)^2*(-3)(-2) = (-1)*(4)*(-3)*(-2). Wait, let me compute that again. Wait, if a = 2, then p(x) = (x - 1)(x - 2)^2(x - 3). Then p(0) = (-1)(-2)^2(-3) = (-1)*(4)*(-3) = 12. Then p(4) = (3)(2)^2(1) = 3*4*1 = 12. Then p(0) + p(4) = 12 + 12 = 24. Same result. Similarly, if a = 3, p(0) would be (-1)(-2)(-3)^2(-a) = (-1)(-2)(9)(-3) = (-1)(-2) is 2, 2*9 is 18, 18*(-3) is -54? Wait, hold on. Wait, if a = 3, then p(x) = (x - 1)(x - 2)(x - 3)^2. Then p(0) = (-1)(-2)(-3)^2 = (-1)(-2)(9) = 2*9 = 18. Then p(4) = (3)(2)(1)^2 = 3*2*1 = 6. Then p(0) + p(4) = 18 + 6 = 24. Still 24. So even if there are repeated roots, the answer is still 24. Therefore, regardless of the value of 'a', p(0) + p(4) is 24. Therefore, the answer is 24.

But let me think again. Is there a way that the problem could have more constraints? For example, maybe the polynomial is supposed to have distinct roots? But the problem didn't specify that. It just said three of the roots are 1, 2, 3. So the fourth root could be anything. Therefore, regardless of that, the answer is 24.

Alternatively, maybe there's another approach. Let me try expanding the polynomial. Let me write p(x) = (x - 1)(x - 2)(x - 3)(x - a). Let me first compute (x - 1)(x - 2)(x - 3). Let me compute that first. Let me compute (x - 1)(x - 2) first. That is x^2 - 3x + 2. Then multiply by (x - 3): (x^2 - 3x + 2)(x - 3) = x^3 - 3x^2 - 3x^2 + 9x + 2x - 6. Wait, let me do that step by step. Multiply x^2 by (x - 3): x^3 - 3x^2. Then -3x times (x - 3): -3x^2 + 9x. Then 2 times (x - 3): 2x - 6. Now add them all together: x^3 - 3x^2 - 3x^2 + 9x + 2x - 6. Combine like terms: x^3 - 6x^2 + 11x - 6. So (x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6. Therefore, p(x) = (x^3 - 6x^2 + 11x - 6)(x - a). Let me expand this. Multiply each term by (x - a):

First, x^3*(x - a) = x^4 - a x^3.

Then -6x^2*(x - a) = -6x^3 + 6a x^2.

Then 11x*(x - a) = 11x^2 - 11a x.

Then -6*(x - a) = -6x + 6a.

Now combine all these terms:

x^4 - a x^3 - 6x^3 + 6a x^2 + 11x^2 - 11a x - 6x + 6a.

Combine like terms:

x^4 + (-a - 6)x^3 + (6a + 11)x^2 + (-11a - 6)x + 6a.

So p(x) = x^4 + (-a - 6)x^3 + (6a + 11)x^2 + (-11a - 6)x + 6a.

Now, if I need to compute p(0) + p(4), let me compute p(0) first. p(0) is the constant term, which is 6a. Then p(4) is the value of the polynomial at x = 4. Let me compute p(4):

p(4) = (4)^4 + (-a - 6)(4)^3 + (6a + 11)(4)^2 + (-11a - 6)(4) + 6a.

Compute each term:

4^4 = 256.

(-a - 6)(4)^3 = (-a - 6)(64) = -64a - 384.

(6a + 11)(4)^2 = (6a + 11)(16) = 96a + 176.

(-11a - 6)(4) = -44a - 24.

And the last term is 6a.

So adding all these together:

256 + (-64a - 384) + (96a + 176) + (-44a - 24) + 6a.

Let me combine like terms step by step.

First, constants: 256 - 384 + 176 - 24.

Compute 256 - 384: that's -128.

Then -128 + 176: that's 48.

Then 48 - 24: that's 24.

Now, the terms with 'a':

-64a + 96a - 44a + 6a.

Compute -64a + 96a = 32a.

32a - 44a = -12a.

-12a + 6a = -6a.

So overall, p(4) = 24 - 6a.

Therefore, p(0) + p(4) = 6a + 24 - 6a = 24. Same result as before. So regardless of the value of 'a', the answer is 24. Therefore, the answer is 24.

But let me check once more with a specific example. Let me choose a specific value for 'a' and compute p(0) + p(4). Let's say a = 0. Then the polynomial is (x - 1)(x - 2)(x - 3)(x - 0) = x(x - 1)(x - 2)(x - 3). Then p(0) = 0, since there's an x term. Then p(4) = 4*(4 - 1)(4 - 2)(4 - 3) = 4*3*2*1 = 24. Therefore, p(0) + p(4) = 0 + 24 = 24. Correct.

Another example: let a = 4. Then the polynomial is (x - 1)(x - 2)(x - 3)(x - 4). Then p(0) = (-1)(-2)(-3)(-4) = 24. p(4) = (4 - 1)(4 - 2)(4 - 3)(4 - 4) = 3*2*1*0 = 0. Therefore, p(0) + p(4) = 24 + 0 = 24. Correct again.

Another example: let a = 5. Then p(0) = (-1)(-2)(-3)(-5) = 30. p(4) = (3)(2)(1)(-1) = -6. Then p(0) + p(4) = 30 - 6 = 24. Correct.

So regardless of the value of 'a', the answer is always 24. Therefore, the answer is 24.

But wait, the problem says "three of the roots of p(x) are 1, 2, and 3". So maybe there's a possibility that the fourth root is something else? But even so, as shown, the answer is always 24. Therefore, the answer is 24.

Alternatively, maybe there's a different approach. Let me think. Since p(x) is a monic quartic with roots 1, 2, 3, and a. Then p(x) = (x - 1)(x - 2)(x - 3)(x - a). Then p(0) is the product of the roots multiplied by (-1)^4, since for a monic polynomial, the constant term is (-1)^n times the product of the roots, where n is the degree. Wait, for a monic polynomial of degree 4, p(x) = x^4 + ... + (-1)^4 * product of roots. Wait, actually, the constant term is (-1)^4 times the product of the roots? Wait, no. Wait, the constant term is the product of the roots multiplied by (-1)^degree. Wait, let me recall Vieta's formula. For a monic polynomial p(x) = x^4 + c_3 x^3 + c_2 x^2 + c_1 x + c_0, the product of the roots is c_0 * (-1)^4 = c_0. Wait, Vieta's formula says that the product of the roots is (-1)^n * c_0, where n is the degree. Wait, for quartic polynomial, product of roots is (-1)^4 * c_0 = c_0. Therefore, the constant term is equal to the product of the roots. Therefore, p(0) = product of roots. Wait, but in our case, the roots are 1, 2, 3, and a. Therefore, p(0) = 1 * 2 * 3 * a = 6a. Which matches our previous calculation.

Similarly, if we think about p(4), it's not directly related to Vieta's formula, but we can think of it as evaluating the polynomial at x = 4. However, since we already did that, and found that p(0) + p(4) = 24 regardless of a, that seems to be the case.

Alternatively, maybe there's a trick here. Let me think. Suppose we consider p(x) = (x - 1)(x - 2)(x - 3)(x - a). Then p(0) + p(4) = [(-1)(-2)(-3)(-a)] + [(3)(2)(1)(4 - a)] = [6a] + [6(4 - a)] = 6a + 24 - 6a = 24. So that's another way to see it.

Therefore, regardless of the value of a, the answer is 24. Therefore, the answer is 24.

But let me check once more if there's any possible mistake. For example, if the polynomial had more roots or something else? But the problem says it's a degree 4 polynomial with three roots given, so the fourth is arbitrary. Therefore, the answer is always 24.

Alternatively, maybe the problem is designed so that even without knowing the fourth root, you can find p(0) + p(4). Which is exactly what we did.

Therefore, I think the answer is 24.

**Final Answer**
\boxed{24}

To solve the problem, we are given that $ p(x) $ is a **monic polynomial of degree 4**, and that **three of its roots are 1, 2, and 3**. We are asked to compute $ p(0) + p(4) $.

---

### Step 1: General Form of the Polynomial

Since $ p(x) $ is a monic polynomial of degree 4 and has roots 1, 2, 3, and an unknown fourth root $ a $, we can write:

$$
p(x) = (x - 1)(x - 2)(x - 3)(x - a)
$$

This is the factored form of the polynomial, and since it's monic, the leading coefficient is 1, which matches the given condition.

---

### Step 2: Compute $ p(0) $

Substitute $ x = 0 $ into the polynomial:

$$
p(0) = (0 - 1)(0 - 2)(0 - 3)(0 - a) = (-1)(-2)(-3)(-a)
$$

Now compute the constants:

$$
(-1)(-2) = 2,\quad 2 \cdot (-3) = -6,\quad -6 \cdot (-a) = 6a
$$

So:

$$
p(0) = 6a
$$

---

### Step 3: Compute $ p(4) $

Substitute $ x = 4 $ into the polynomial:

$$
p(4) = (4 - 1)(4 - 2)(4 - 3)(4 - a) = (3)(2)(1)(4 - a)
$$

Compute the constants:

$$
3 \cdot 2 = 6,\quad 6 \cdot 1 = 6,\quad 6 \cdot (4 - a) = 24 - 6a
$$

So:

$$
p(4) = 24 - 6a
$$

---

### Step 4: Add $ p(0) + p(4) $

$$
p(0) + p(4) = 6a + (24 - 6a) = 24
$$

The terms involving $ a $ cancel out, and we are left with:

$$
\boxed{24}
$$

---

### Final Answer

$$
\boxed{24}
$$

[Assistant]
Explanation:
Okay, so I need to find p(0) + p(4) where p(x) is a monic polynomial of degree 4, and three of its roots are 1, 2, and 3. Let me start by recalling what a monic polynomial is. A monic polynomial is one where the leading coefficient (the coefficient of the highest power of x) is 1. Since it's a degree 4 polynomial, it has four roots (counting multiplicities). They told me three of the roots are 1, 2, and 3, so there must be a fourth root. Let me call that fourth root 'a'.

So, the polynomial can be written in factored form as p(x) = (x - 1)(x - 2)(x - 3)(x - a). Since it's monic, the leading coefficient is 1, which matches this form. Now, I need to figure out what 'a' is. Wait, but the problem doesn't give me any other information about the polynomial. Hmm. Maybe I don't need to find 'a' explicitly? Let me think.

The question asks for p(0) + p(4). Let me compute each term separately. Let's start with p(0). If I plug in x = 0 into the polynomial, I get p(0) = (0 - 1)(0 - 2)(0 - 3)(0 - a) = (-1)(-2)(-3)(-a). Let me compute that. Multiplying the constants first: (-1)(-2) is 2, then 2*(-3) is -6, then -6*(-a) is 6a. So p(0) = 6a.

Now, let's compute p(4). Plugging in x = 4, we get p(4) = (4 - 1)(4 - 2)(4 - 3)(4 - a) = (3)(2)(1)(4 - a). Multiplying those constants: 3*2 is 6, 6*1 is 6, so p(4) = 6*(4 - a) = 24 - 6a.

Therefore, p(0) + p(4) = 6a + 24 - 6a = 24. Wait, that's interesting. The 6a and -6a cancel each other out, leaving just 24. So regardless of the value of 'a', p(0) + p(4) is always 24? That seems surprising, but maybe that's correct. Let me check my steps again to make sure I didn't make a mistake.

First, p(x) is monic quartic with roots 1, 2, 3, and a. So factored form is correct. Then p(0) is product of (-1)(-2)(-3)(-a). Let me compute that again. (-1)*(-2) is 2, then 2*(-3) is -6, then -6*(-a) is 6a. Yes, that's correct. Then p(4) is (3)(2)(1)(4 - a) which is 6*(4 - a) = 24 - 6a. Then adding them together: 6a + 24 - 6a = 24. So yes, the answer is 24.

But wait, is there a possibility that the polynomial has a repeated root? The problem says "three of the roots are 1, 2, and 3". It doesn't specify if they are distinct or not. But since it's a degree 4 polynomial, and three roots are given, the fourth root could be any number, possibly one of the existing roots. However, even if there is a repeated root, the calculation for p(0) + p(4) would still hold because the fourth root is still 'a', whether it's equal to 1, 2, 3, or something else. For example, if a = 1, then p(x) would be (x - 1)^2(x - 2)(x - 3). But in that case, p(0) would be (-1)^2*(-2)*(-3) = 1*6 = 6, and p(4) would be (3)^2*(2)*(1) = 9*2 = 18. Then p(0) + p(4) = 6 + 18 = 24. Same result. Similarly, if a = 2, p(0) would be (-1)(-2)^2*(-3)(-2) = (-1)*(4)*(-3)*(-2). Wait, let me compute that again. Wait, if a = 2, then p(x) = (x - 1)(x - 2)^2(x - 3). Then p(0) = (-1)(-2)^2(-3) = (-1)*(4)*(-3) = 12. Then p(4) = (3)(2)^2(1) = 3*4*1 = 12. Then p(0) + p(4) = 12 + 12 = 24. Same result. Similarly, if a = 3, p(0) would be (-1)(-2)(-3)^2(-a) = (-1)(-2)(9)(-3) = (-1)(-2) is 2, 2*9 is 18, 18*(-3) is -54? Wait, hold on. Wait, if a = 3, then p(x) = (x - 1)(x - 2)(x - 3)^2. Then p(0) = (-1)(-2)(-3)^2 = (-1)(-2)(9) = 2*9 = 18. Then p(4) = (3)(2)(1)^2 = 3*2*1 = 6. Then p(0) + p(4) = 18 + 6 = 24. Still 24. So even if there are repeated roots, the answer is still 24. Therefore, regardless of the value of 'a', p(0) + p(4) is 24. Therefore, the answer is 24.

But let me think again. Is there a way that the problem could have more constraints? For example, maybe the polynomial is supposed to have distinct roots? But the problem didn't specify that. It just said three of the roots are 1, 2, 3. So the fourth root could be anything. Therefore, regardless of that, the answer is 24.

Alternatively, maybe there's another approach. Let me try expanding the polynomial. Let me write p(x) = (x - 1)(x - 2)(x - 3)(x - a). Let me first compute (x - 1)(x - 2)(x - 3). Let me compute that first. Let me compute (x - 1)(x - 2) first. That is x^2 - 3x + 2. Then multiply by (x - 3): (x^2 - 3x + 2)(x - 3) = x^3 - 3x^2 - 3x^2 + 9x + 2x - 6. Wait, let me do that step by step. Multiply x^2 by (x - 3): x^3 - 3x^2. Then -3x times (x - 3): -3x^2 + 9x. Then 2 times (x - 3): 2x - 6. Now add them all together: x^3 - 3x^2 - 3x^2 + 9x + 2x - 6. Combine like terms: x^3 - 6x^2 + 11x - 6. So (x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6. Therefore, p(x) = (x^3 - 6x^2 + 11x - 6)(x - a). Let me expand this. Multiply each term by (x - a):

First, x^3*(x - a) = x^4 - a x^3.

Then -6x^2*(x - a) = -6x^3 + 6a x^2.

Then 11x*(x - a) = 11x^2 - 11a x.

Then -6*(x - a) = -6x + 6a.

Now combine all these terms:

x^4 - a x^3 - 6x^3 + 6a x^2 + 11x^2 - 11a x - 6x + 6a.

Combine like terms:

x^4 + (-a - 6)x^3 + (6a + 11)x^2 + (-11a - 6)x + 6a.

So p(x) = x^4 + (-a - 6)x^3 + (6a + 11)x^2 + (-11a - 6)x + 6a.

Now, if I need to compute p(0) + p(4), let me compute p(0) first. p(0) is the constant term, which is 6a. Then p(4) is the value of the polynomial at x = 4. Let me compute p(4):

p(4) = (4)^4 + (-a - 6)(4)^3 + (6a + 11)(4)^2 + (-11a - 6)(4) + 6a.

Compute each term:

4^4 = 256.

(-a - 6)(4)^3 = (-a - 6)(64) = -64a - 384.

(6a + 11)(4)^2 = (6a + 11)(16) = 96a + 176.

(-11a - 6)(4) = -44a - 24.

And the last term is 6a.

So adding all these together:

256 + (-64a - 384) + (96a + 176) + (-44a - 24) + 6a.

Let me combine like terms step by step.

First, constants: 256 - 384 + 176 - 24.

Compute 256 - 384: that's -128.

Then -128 + 176: that's 48.

Then 48 - 24: that's 24.

Now, the terms with 'a':

-64a + 96a - 44a + 6a.

Compute -64a + 96a = 32a.

32a - 44a = -12a.

-12a + 6a = -6a.

So overall, p(4) = 24 - 6a.

Therefore, p(0) + p(4) = 6a + 24 - 6a = 24. Same result as before. So regardless of the value of 'a', the answer is 24. Therefore, the answer is 24.

But let me check once more with a specific example. Let me choose a specific value for 'a' and compute p(0) + p(4). Let's say a = 0. Then the polynomial is (x - 1)(x - 2)(x - 3)(x - 0) = x(x - 1)(x - 2)(x - 3). Then p(0) = 0, since there's an x term. Then p(4) = 4*(4 - 1)(4 - 2)(4 - 3) = 4*3*2*1 = 24. Therefore, p(0) + p(4) = 0 + 24 = 24. Correct.

Another example: let a = 4. Then the polynomial is (x - 1)(x - 2)(x - 3)(x - 4). Then p(0) = (-1)(-2)(-3)(-4) = 24. p(4) = (4 - 1)(4 - 2)(4 - 3)(4 - 4) = 3*2*1*0 = 0. Therefore, p(0) + p(4) = 24 + 0 = 24. Correct again.

Another example: let a = 5. Then p(0) = (-1)(-2)(-3)(-5) = 30. p(4) = (3)(2)(1)(-1) = -6. Then p(0) + p(4) = 30 - 6 = 24. Correct.

So regardless of the value of 'a', the answer is always 24. Therefore, the answer is 24.

But wait, the problem says "three of the roots of p(x) are 1, 2, and 3". So maybe there's a possibility that the fourth root is something else? But even so, as shown, the answer is always 24. Therefore, the answer is 24.

Alternatively, maybe there's a different approach. Let me think. Since p(x) is a monic quartic with roots 1, 2, 3, and a. Then p(x) = (x - 1)(x - 2)(x - 3)(x - a). Then p(0) is the product of the roots multiplied by (-1)^4, since for a monic polynomial, the constant term is (-1)^n times the product of the roots, where n is the degree. Wait, for a monic polynomial of degree 4, p(x) = x^4 + ... + (-1)^4 * product of roots. Wait, actually, the constant term is (-1)^4 times the product of the roots? Wait, no. Wait, the constant term is the product of the roots multiplied by (-1)^degree. Wait, let me recall Vieta's formula. For a monic polynomial p(x) = x^4 + c_3 x^3 + c_2 x^2 + c_1 x + c_0, the product of the roots is c_0 * (-1)^4 = c_0. Wait, Vieta's formula says that the product of the roots is (-1)^n * c_0, where n is the degree. Wait, for quartic polynomial, product of roots is (-1)^4 * c_0 = c_0. Therefore, the constant term is equal to the product of the roots. Therefore, p(0) = product of roots. Wait, but in our case, the roots are 1, 2, 3, and a. Therefore, p(0) = 1 * 2 * 3 * a = 6a. Which matches our previous calculation.

Similarly, if we think about p(4), it's not directly related to Vieta's formula, but we can think of it as evaluating the polynomial at x = 4. However, since we already did that, and found that p(0) + p(4) = 24 regardless of a, that seems to be the case.

Alternatively, maybe there's a trick here. Let me think. Suppose we consider p(x) = (x - 1)(x - 2)(x - 3)(x - a). Then p(0) + p(4) = [(-1)(-2)(-3)(-a)] + [(3)(2)(1)(4 - a)] = [6a] + [6(4 - a)] = 6a + 24 - 6a = 24. So that's another way to see it.

Therefore, regardless of the value of a, the answer is 24. Therefore, the answer is 24.

But let me check once more if there's any possible mistake. For example, if the polynomial had more roots or something else? But the problem says it's a degree 4 polynomial with three roots given, so the fourth is arbitrary. Therefore, the answer is always 24.

Alternatively, maybe the problem is designed so that even without knowing the fourth root, you can find p(0) + p(4). Which is exactly what we did.

Therefore, I think the answer is 24.

**Final Answer**
\boxed{24}

To solve the problem, we are given that $ p(x) $ is a **monic polynomial of degree 4**, and that **three of its roots are 1, 2, and 3**. We are asked to compute $ p(0) + p(4) $.

---

### Step 1: General Form of the Polynomial

Since $ p(x) $ is a monic polynomial of degree 4 and has roots 1, 2, 3, and an unknown fourth root $ a $, we can write:

$$
p(x) = (x - 1)(x - 2)(x - 3)(x - a)
$$

This is the factored form of the polynomial, and since it's monic, the leading coefficient is 1, which matches the given condition.

---

### Step 2: Compute $ p(0) $

Substitute $ x = 0 $ into the polynomial:

$$
p(0) = (0 - 1)(0 - 2)(0 - 3)(0 - a) = (-1)(-2)(-3)(-a)
$$

Now compute the constants:

$$
(-1)(-2) = 2,\quad 2 \cdot (-3) = -6,\quad -6 \cdot (-a) = 6a
$$

So:

$$
p(0) = 6a
$$

---

### Step 3: Compute $ p(4) $

Substitute $ x = 4 $ into the polynomial:

$$
p(4) = (4 - 1)(4 - 2)(4 - 3)(4 - a) = (3)(2)(1)(4 - a)
$$

Compute the constants:

$$
3 \cdot 2 = 6,\quad 6 \cdot 1 = 6,\quad 6 \cdot (4 - a) = 24 - 6a
$$

So:

$$
p(4) = 24 - 6a
$$

---

### Step 4: Add $ p(0) + p(4) $

$$
p(0) + p(4) = 6a + (24 - 6a) = 24
$$

The terms involving $ a $ cancel out, and we are left with:

$$
\boxed{24}
$$

---

### Final Answer

$$
\boxed{24}
$$

Solution:
Since three of the roots of $p(x)$ are 1, 2, and 3, we can write
\[p(x) = (x - 1)(x - 2)(x - 3)(x - r).\]Then
\begin{align*}
p(0) + p(4) &= (-1)(-2)(-3)(-r) + (3)(2)(1)(4 - r) \\
&= 6r + 24 - 6r = \boxed{24}.
\end{align*}