Question

There exist $r$ unique nonnegative integers $n_1 \gt n_2 \gt \cdots \gt n_r$ and $r$ unique integers $a_k$ $(1 \le k \le r)$ with each $a_k$ either $1$ or $-1$ such that $a_1 3^{n_1} + a_2 3^{n_2} + \cdots + a_r 3^{n_r} = 2008.$ Find $n_1 + n_2 + \cdots + n_r.$

Answer

Solution:

In base $3,$ $2008 = 2202101_3,$ that is, $2008 = 2 \cdot 3^6 + 2 \cdot 3^5 + 2 \cdot 3^3 + 3^2 + 3^0.$ To convert the digits $2$ into coefficients $\pm 1,$ use $2 \cdot 3^k = 3^{k+1} - 3^k.$ The two adjacent digits $2$ collapse neatly: $2 \cdot 3^6 + 2 \cdot 3^5 = (3^7 - 3^6) + (3^6 - 3^5) = 3^7 - 3^5,$ and $2 \cdot 3^3 = 3^4 - 3^3.$

Therefore $2008 = 3^7 - 3^5 + 3^4 - 3^3 + 3^2 + 3^0,$ which has distinct exponents and coefficients $\pm 1,$ as required. The sum of the exponents is $7 + 5 + 4 + 3 + 2 + 0 = 21.$

2008 AIME II Problem 4

Solution:

Problem 4 in Other Years