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

$2(a_na_{n-1}\cdots a_0)_7 = (a_na_{n-1}\cdots a_0)_{10}$ (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.

Solution 2 (Bash/Guess and Check)

Let $A$ be the base $10$ representation of our number, and let $B$ be its base $7$ representation.

Given this is an AIME problem, $A<1000$ . If we look at $B$ in base $10$ , it must be equal to $2A$ , so $B<2000$ when $B$ is looked at in base $10.$

If $B$ in base $10$ is less than $2000$ , then $B$ as a number in base $7$ must be less than $2*7^2=686$ .

$686$ is non-existent in base $7$ , so we're gonna have to bump that down to $666_7$ .

This suggests that $A$ is less than $\frac{666}{2}=333$ .

Guess and check shows that $310<A<320$ , and checking values in that range produces $\boxed{315}_{10}$ .

2001 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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2001 AIME I Problems/Problem 8

Contents

Problem

Solution

Solution 2 (Bash/Guess and Check)

See also