# 2001 AIME I Problems/Problem 8

## Problem

Call a positive integer $N$ a 7-10 double if the digits of the base- $7$ representation of $N$ form a base- $10$ number that is twice $N$. For example, $51$ is a 7-10 double because its base- $7$ representation is $102$. What is the largest 7-10 double?

## Solution

We let $N_7 = \overline{a_na_{n-1}\cdots a_0}_7$; we are given that $$2(a_na_{n-1}\cdots a_0)_7 = (a_na_{n-1}\cdots a_0)_{10}$$m (This is because the digits in $N$ ' s base 7 representation make a number with the same digits in base 10 when multiplied by 2)

Expanding, we find that $$2 \cdot 7^n a_n + 2 \cdot 7^{n-1} a_{n-1} + \cdots + 2a_0 = 10^na_n + 10^{n-1}a_{n-1} + \cdots + a_0$$

or re-arranging, $$a_0 + 4a_1 = 2a_2 + 314a_3 + \cdots + (10^n - 2 \cdot 7^n)a_n$$

Since the $a_i$s are base- $7$ digits, it follows that $a_i < 7$, and the LHS is less than or equal to $30$. Hence our number can have at most $3$ digits in base- $7$. Letting $a_2 = 6$, we find that $630_7 = \boxed{315}_{10}$ is our largest 7-10 double.

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