2008 AIME II 第 4 题

Start by writing $2008$ in base $3$

Replace each digit $2$ using $2 \cdot 3^k = 3^{k+1} - 3^k$ so that every coefficient becomes $1$ or $-1$

In base $3,$ $2008 = 2202101_3,$ that is, $$\begin{aligned} 2008 &= 2 \cdot 3^6 + 2 \cdot 3^5 + 2 \cdot 3^3 \\ &\quad {}+ 3^2 + 3^0. \end{aligned}$$ 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: $$\begin{aligned} 2 \cdot 3^6 + 2 \cdot 3^5 &= (3^7 - 3^6) \\ &\quad {}+ (3^6 - 3^5) \\ &= 3^7 - 3^5, \end{aligned}$$ and $2 \cdot 3^3 = 3^4 - 3^3.$

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