2025 AIME I Problems/Problem 1

Revision as of 19:28, 13 February 2025 by Zhenghua (talk | contribs)

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

See also

2025 AIME I (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png