# Difference between revisions of "2008 AIME II Problems/Problem 4"

## Problem

There exist $r$ unique nonnegative integers $n_1 > n_2 > \cdots > n_r$ and $r$ unique integers $a_k$ ( $1\le k\le r$) with each $a_k$ either $1$ or $- 1$ such that $$a_13^{n_1} + a_23^{n_2} + \cdots + a_r3^{n_r} = 2008.$$ Find $n_1 + n_2 + \cdots + n_r$.

## Solution

In base $3$, we find that $\overline{2008}_{10} = \overline{2202101}_{3}$. In other words, $2008 = 2 \cdot 3^{6} + 2 \cdot 3^{5} + 2 \cdot 3^3 + 1 \cdot 3^2 + 1 \cdot 3^0$

In order to rewrite as a sum of perfect powers of $3$, we can use the fact that $2 \cdot 3^k = 3^{k+1} - 3^k$: $2008 = (3^7 - 3^6) + (3^6-3^5) + (3^4 - 3^3) + 3^2 + 3^0 = 3^7 - 3^5 + 3^4 - 3^3 + 3^2 + 3^0$

The answer is $7+5+4+3+2+0 = \boxed{021}$.

Note : Solution by bounding is also possible, namely using the fact that $1+3+3^2 + \cdots + 3^{n} = \displaystyle\frac{3^{n+1}-1}{2}.$

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