# Difference between revisions of "2018 AIME I Problems/Problem 2"

## Problem

The number $n$ can be written in base $14$ as $\underline{a}\text{ }\underline{b}\text{ }\underline{c}$, can be written in base $15$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{b}$, and can be written in base $6$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }$, where $a > 0$. Find the base-$10$ representation of $n$.

## Solution

We have these equations: $196a+14b+c=225a+15c+b=222a+37c$. Taking the last two we get $3a+b=22c$. Because $c \neq 0$ otherwise $a \ngtr 0$, and $a \leq 5$, $c=1$.

Then we know $3a+b=22$. Taking the first two equations we see that $29a+14=13b$. Combining the two gives $a=4, b=10$. Then we see that $222 \times 4+37 \times1=\boxed{925}$.

-gorefeebuddie

## Solution 2

We know that $196a+14b+c=225a+15c+b=222a+37c$. Combining the first and third equations give that $196a+14b+c=222a+37c$, or $$7b=13a+18c$$ The second and third gives $222a+37c=225a+15c+b$, or $$22c-3a=b$$$$154c-21a=7b=13a+18c$$$$4c=a$$ We can have $a=4,8,12$, but only $a=4$ falls within the possible digits of base $6$. Thus $a=4$, $c=1$, and thus you can find $b$ which equals $10$. Thus, our answer is $4\cdot225+1\cdot15+0=\boxed{925}$.