# Difference between revisions of "2020 AIME I Problems/Problem 3"

Note: Please do not post problems here until after the AIME.

## Problem

A positive integer $N$ has base-eleven representation $\underline{a}\underline{b}\underline{c}$ and base-eight representation $\underline1\underline{b}\underline{c}\underline{a},$ where $a,b,$ and $c$ represent (not necessarily distinct) digits. Find the least such $N$ expressed in base ten.

## Solution

Since $a$, $b$, and $c$ are digits in base eight, they are all 7 or less. Now, from the given equation, $121a+11b+c=512+64b+8c+a \implies 120a-53b-7c=512$. Since $a$, $b$, and $c$ have to be positive, $a \geq 5$. If $a = 5$, then by casework we see that $b = 1$ and $c = 5$ is the only solution. Since the question asks for the lowest working value, we can stop here. Finally, $515_{11} = 621_{10}$, so our answer is $\boxed{621}$.

~ JHawk0224

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