aime26-07

Proposed solution

To solve the problem, we start by analyzing the number \(17017^{17}\). First, we reduce \(17017\) modulo \(12\):

\[
17017 \div 12 = 1418 \text{ remainder } 1 \quad \Rightarrow \quad 17017 \equiv 1 \pmod{12}
\]

Thus,

\[
17017^{17} \equiv 1^{17} \equiv 1 \pmod{12}
\]

We need to find the number of positive integer divisors of \(17017^{17}\) that leave a remainder of \(5\) when divided by \(12\). Let \(d\) be such a divisor. Then,

\[
d \equiv 5 \pmod{12}
\]

Since \(d\) is a divisor of \(17017^{17}\), we express \(d\) in terms of its prime factorization. The prime factorization of \(17017\) is:

\[
17017 = 13 \times 1313
\]

Thus,

\[
17017^{17} = (13 \times 1313)^{17} = 13^{17} \times 1313^{17}
\]

A divisor \(d\) of \(17017^{17}\) can be written as \(d = 13^a \times 1313^b\) where \(0 \leq a \leq 17\) and \(0 \leq b \leq 17\). We need \(d \equiv 5 \pmod{12}\). We analyze the congruence of \(13\) and \(1313\) modulo \(12\):

\[
13 \equiv 1 \pmod{12}
\]
\[
1313 \equiv 1 \pmod{12}
\]

Therefore,

\[
13^a \equiv 1^a \equiv 1 \pmod{12}
\]
\[
1313^b \equiv 1^b \equiv 1 \pmod{12}
\]

So,

\[
d = 13^a \times 1313^b \equiv 1 \times 1 \equiv 1 \pmod{12}
\]

However, we need \(d \equiv 5 \pmod{12}\). This implies there is no combination of \(a\) and \(b\) such that \(13^a \times 1313^b \equiv 5 \pmod{12}\) because \(1 \equiv 1 \pmod{12}\) for any \(a\) and \(b\). Therefore, there are no divisors of \(17017^{17}\) that leave a remainder of \(5\) when divided by \(12\).

Hence, the number of such divisors \(N\) is \(0\). The remainder when \(0\) is divided by \(1000\) is:

\[
\boxed{0}
\]