2025 AIME I Problems/Problem 1

Problem

Find the sum of all integer bases $b>9$ for which $17_b$ is a divisor of $97_b.$

Solution 1

We have, $b + 7 \mid 9b + 7$ meaning $b + 7 \mid -56$ so taking divisors of $56$ under bounds to find $b = 49, 21$ meaning our answer is $49+21=\boxed{070}.$

~mathkiddus

Solution 2

This means that $a(b+7)=9b+7$ where $a$ is a natural number. Rearranging we get $(a-9)(b+7)=-56$ . Since $b>9$ , $b=49,21$ . Thus the answer is $49+21=\boxed{70}$

~zhenghua

Page

Toolbox

Search

2025 AIME I Problems/Problem 1

Problem

Solution 1

Solution 2