Let N be the sum of all prime powers that can be written as 4^n+n^4 for some positive integer n.

What are the last 3 digits of N?

Details and assumptions:
A prime power is a number of the form pk, where p is a prime and k is a positive integer. Examples: 3,9,16.

To find the last three digits of N, we need to compute the sum of all prime powers that can be written as 4^n + n^4 for some positive integer n. Let's break down the process into steps:

Step 1: List out the prime powers that satisfy the condition.
We can start by calculating 4^n + n^4 for various positive integers n and identify which ones are prime powers. Let's do this for n = 1, 2, 3, 4, 5, and 6:

For n = 1: 4^1 + 1^4 = 4 + 1 = 5 (not a prime power)
For n = 2: 4^2 + 2^4 = 16 + 16 = 32 = 2^5 (a prime power)
For n = 3: 4^3 + 3^4 = 64 + 81 = 145 (not a prime power)
For n = 4: 4^4 + 4^4 = 256 + 256 = 512 = 2^9 (a prime power)
For n = 5: 4^5 + 5^4 = 1024 + 625 = 1649 (not a prime power)
For n = 6: 4^6 + 6^4 = 4096 + 1296 = 5392 = 2^4 * 337 (a prime power)

From this list, we can see that the prime powers satisfying the condition are 2^5 and 2^9.

Step 2: Compute the sum of these prime powers.
Now that we know the prime powers that satisfy the condition, let's add them up:

2^5 + 2^9 = 32 + 512 = 544

So, the sum of these prime powers is 544.

Step 3: Find the last three digits.
To determine the last three digits of the sum, we can take the sum modulo 1000. This means keeping only the last three digits of the sum.

544 % 1000 = 544

Therefore, the last three digits of N are 544.