# 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 1

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 $29a+14c=13b$. Combining the two gives $a=4, b=10, c=1$. Then we see that $222 \times 4+37 \times1=\boxed{925}$.

## 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 $4\cdot225+1\cdot15+10=\boxed{925}$.

## Solution 3 (Official MAA)

The problem is equivalent to finding a solution to the system of Diophantine equations $196a+14b+c=225a+15c+b$ and $225a+15c+b=216a+36c+6a+c,$ where $1\le a\le 5,\,0\le b\le 13,$ and $0\le c\le 5.$ Simplifying the second equation gives $b=22c-3a.$ Substituting for $b$ in the first equation and simplifying then gives $a=4c,$ so $a = 4$ and $c = 1,$ and the base- $10$ representation of $n$ is $222 \cdot 4 + 37 \cdot 1 = 925.$ It may be verified that $b=10\le 13.$

## Video Solution

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