aime26-07

Proposed solution

I need to find the number of positive divisors of $17017^{17}$ that leave a remainder of 5 when divided by 12.

First, let me find the prime factorization of 17017.

$17017 = 7 \times 2431$

Let me factor 2431:
$2431 = 11 \times 221$
$221 = 13 \times 17$

So $17017 = 7 \times 11 \times 13 \times 17$

Therefore: $17017^{17} = 7^{17} \times 11^{17} \times 13^{17} \times 17^{17}$

Any positive divisor of $17017^{17}$ has the form:
$d = 7^a \times 11^b \times 13^c \times 17^d$ where $0 \leq a, b, c, d \leq 17$

I need to find how many such divisors satisfy $d \equiv 5 \pmod{12}$.

Since $12 = 4 \times 3$, I'll use the Chinese Remainder Theorem. A number is congruent to 5 modulo 12 if and only if:
- $d \equiv 1 \pmod{4}$ (since $5 \equiv 1 \pmod{4}$)
- $d \equiv 2 \pmod{3}$ (since $5 \equiv 2 \pmod{3}$)

Now I need to determine the congruence classes of powers of each prime modulo 4 and modulo 3.

For powers of 7:
- $7 \equiv 3 \pmod{4}$, so $7^2 \equiv 9 \equiv 1 \pmod{4}$
- Powers of 7 modulo 4 cycle: $7^1 \equiv 3$, $7^2 \equiv 1$, $7^3 \equiv 3$, $7^4 \equiv 1$, ...
- So $7^a \equiv 1 \pmod{4}$ when $a$ is even, and $7^a \equiv 3 \pmod{4}$ when $a$ is odd
- For $7^a \equiv 1 \pmod{4}$: $a$ must be even (0, 2, 4, 6, 8, 10, 12, 14, 16) → 9 values
- For $7^a \equiv 3 \pmod{4}$: $a$ must be odd (1, 3, 5, 7, 9, 11, 13, 15, 17) → 9 values

For powers of 7 modulo 3:
- $7 \equiv 1 \pmod{3}$, so $7^a \equiv 1 \pmod{3}$ for all $a \geq 1$
- So $7^a \equiv 1 \pmod{3}$ always

For powers of 11:
- $11 \equiv 3 \pmod{4}$, so $11^2 \equiv 9 \equiv 1 \pmod{4}$
- Powers of 11 modulo 4 cycle: $11^1 \equiv 3$, $11^2 \equiv 1$, $11^3 \equiv 3$, $11^4 \equiv 1$, ...
- So $11^b \equiv 1 \pmod{4}$ when $b$ is even, and $11^b \equiv 3 \pmod{4}$ when $b$ is odd
- For $11^b \equiv 1 \pmod{4}$: $b$ must be even (0, 2, 4, 6, 8, 10, 12, 14, 16) → 9 values
- For $11^b \equiv 3 \pmod{4}$: $b$ must be odd (1, 3, 5, 7, 9, 11, 13, 15, 17) → 9 values

For powers of 11 modulo 3:
- $11 \equiv 2 \pmod{3}$, so $11^2 \equiv 4 \equiv 1 \pmod{3}$
- Powers of 11 modulo 3 cycle: $11^1 \equiv 2$, $11^2 \equiv 1$, $11^3 \equiv 2$, $11^4 \equiv 1$, ...
- So $11^b \equiv 1 \pmod{3}$ when $b$ is even, and $11^b \equiv 2 \pmod{3}$ when $b$ is odd

For powers of 13:
- $13 \equiv 1 \pmod{4}$, so $13^c \equiv 1 \pmod{4}$ for all $c$
- For powers of 13 modulo 3:
- $13 \equiv 1 \pmod{3}$, so $13^c \equiv 1 \pmod{3}$ for all $c$

For powers of 17:
- $17 \equiv 1 \pmod{4}$, so $17^d \equiv 1 \pmod{4}$ for all $d$
- For powers of 17 modulo 3:
- $17 \equiv 2 \pmod{3}$, so $17^d \equiv 2^d \pmod{3}$
- Powers of 17 modulo 3 cycle: $17^1 \equiv 2$, $17^2 \equiv 1$, $17^3 \equiv 2$, $17^4 \equiv 1$, ...
- So $17^d \equiv 1 \pmod{3}$ when $d$ is even, and $17^d \equiv 2 \pmod{3}$ when $d$ is odd

Now I need to count divisors where:
- $d \equiv 1 \pmod{4}$
- $d \equiv 2 \pmod{3}$

For the product $d = 7^a \times 11^b \times 13^c \times 17^d$:

Since $13^c \equiv 1 \pmod{4}$ and $17^d \equiv 1 \pmod{4}$, we need:
$7^a \times 11^b \equiv 1 \pmod{4}$

This happens when:
- Both $7^a \equiv 1 \pmod{4}$ and $11^b \equiv 1 \pmod{4}$, OR
- Both $7^a \equiv 3 \pmod{4}$ and $11^b \equiv 3 \pmod{4}$

Case 1: $a$ even and $b$ even
Case 2: $a$ odd and $b$ odd

For the modulo 3 condition:
Since $7^a \equiv 1 \pmod{3}$, $11^b \equiv 1 \pmod{3}$ when $b$ is even, and $11^b \equiv 2 \pmod{3}$ when $b$ is odd.
Since $13^c \equiv 1 \pmod{3}$, $17^d \equiv 1 \pmod{3}$ when $d$ is even, and $17^d \equiv 2 \pmod{3}$ when $d$ is odd.

So $d \equiv 1 \times 1 \times 1 \times 1 = 1 \pmod{3}$ when $b$ is even and $d$ is even.
And $d \equiv 1 \times 2 \times 1 \times 2 = 4 \equiv 1 \pmod{3}$ when $b$ is odd and $d$ is odd.

We want $d \equiv 2 \pmod{3}$, so we need to check if any combinations work.

Actually, let me reconsider the modulo 3 calculation more carefully.

$d = 7^a \times 11^b \times 13^c \times 17^d$

Modulo 3:
- $7^a \equiv 1^a \equiv 1 \pmod{3}$
- $11^b \equiv 2^b \pmod{3}$  
- $13^c \equiv 1^c \equiv