AIME 2026 I · 第 8 题

Solution 1

First, observe that $17017^{17}=7^{17}\cdot 11^{17}\cdot 13^{17}\cdot 17^{17}$. Thus, every factor of this must be in the form $7^a\cdot 11^b\cdot 13^c\cdot 17^d$. For a factor to be $5\pmod{12}$, it must be both $2\pmod{3}$ and $1\pmod{4}$. Now, since

$$7^a\cdot 11^b\cdot 13^c\cdot 17^d\equiv (-1)^{b+d}\pmod{3}$$ and

$$7^a\cdot 11^b\cdot 13^c\cdot 17^d\equiv (-1)^{a+b}\pmod{4},$$ we must have that $b+d$ odd and $a+b$ is even.

Notice that since $c\in \{0, 1, ..., 17\}$, there are $18$ choices for $c$. Also, for every $b\in\{0, 1, ..., 17\}$ there are always $9$ choices for $a$ and $9$ choices for $d$. Hence, the answer is $18\cdot 18\cdot 9\cdot 9\equiv \boxed{\textbf{244}}\pmod{1000}$.

~Mathkiddie

Solution 2

Note $17017=17\cdot1001=7\cdot11\cdot13\cdot17$.

By taking each of the factors mod $12$ we get:

$7\equiv-5\pmod{12}$, $11\equiv-1\pmod{12}$, $13\equiv1\pmod{12}$, $17\equiv5\pmod{12}$.

Let a factor $n=7^a11^b13^c17^d$.

By noting $5^2\equiv1\pmod{12}$ and $(-1)^2\equiv1\pmod{12}$ and the fact that we want $n\equiv5\pmod{12}$, we need $a+d$ to be odd and $a+b$ to be even. From here, we compute that for a fixed $a$, there are $9$ options for $b$ and $9$ options for $d$. As $a$ and $c$ may be $18$ options, the answer is $9\cdot9\cdot18\cdot18=26244$. The requested remainder is $244$.

~SilverRush

Solution 3 (Easy to understand, only basic mod needed)

We have $17017 = 17 * 1001 = 7 * 11 * 13 * 17$ Take all powers $\pmod{ 12}$. We get some nice patterns. For example, taking all powers of 7 mod 12 we get:

(7, 1, 7, 1, 7, 1....) so we have a pattern!

Now for 11, we get that 11 is $-1$ mod 12, so:

(11, 1, 11, 1, 11, 1...) is a pattern

For 13, it is 1 mod 12 so:

(1, 1, 1, 1, 1, ...)

For 17 it is 5 mod 12 so:

(5, 1, 5, 1....)

We realize to get 5 mod 12 we either have 1 * 1 * 1 * 5 or 7 * 11 * 1 * 1.

For both cases we have 9 (9 powers of 7 from $7^0$ to $7^{17}$ are congruent to 1 mod 12) * 9 (9 powers of 11 from $11^0$ to $11^{17}$ are congruent to 1 mod 12)* 9 (9 powers of 17 are congruent to 5 mod 12) * 18 (18 powers of 13 are 1 mod 12) ways to do it.

So we have 122*2 (mod 1000)

So finding this mod 1000 we get $\boxed{244}$

~Aarav22 (Did not make AIME).

Solution 4

We factor $17017 = 7 \cdot 11 \cdot 13 \cdot 17$.

We convert the factors into modulo $12$, $7 \equiv -5 \pmod{12}$, $11 \equiv -1 \pmod{12}$, $13 \equiv 1 \pmod{12}$, and $17 \equiv 5 \pmod{12}$.

Since $5$ raised to an odd power is congruent to $5 \pmod{12}$, we split into cases.

Case 1: odd exponent of $7$ and even exponent of $17$.

There are $9$ choices for the odd exponent of $7$, $9$ choices for the odd exponent of $11$, $18$ choices for the exponent of $13$ as it is $1 \pmod{12}$, and $9$ choices for the even exponent of $17$.

This yields $9 \cdot 9 \cdot 18 \cdot 9 = 13{,}122$ values.

Case 2: even exponent of $7$ and odd exponent of $17$.

There are $9$ choices for the even exponent of $7$, $9$ choices for the even exponent of $11$, $18$ choices for the exponent of $13$ as it is $1 \pmod{12}$, and $9$ choices for the odd exponent of $17$. This also yields $9 \cdot 9 \cdot 18 \cdot 9 = 13{,}122$ values.

Thus, the total number is $13{,}122 + 13{,}122 = 26{,}244 \equiv \boxed{244} \pmod{1000}$.

~plin

Video Solution (Fast and Easy 🔥🚀)

2026 AIME I #8

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