jmc

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 $7+5+4+3+2+0=\boxed{21}$. Note: Solution by bounding is also possible, namely using the fact that $1+3+3^2+\cdots +3^n=\frac{3^{n+1}-1}{2}.$