AIME 2008 II · 第 4 题

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 $1+3+3^2 + \cdots + 3^{n} =\frac{3^{n+1}-1}{2}.$

Solution 2

Notice $2008$ has $7$ digits in base 3. Therefore, we can add $1093=1111111_3$ to get $3101=11020212_3$. Subtracting $1093$ back from this, we find $2008 = 1 0 (-1) 1 (-1) 1 0 1_3$. We need to sum the exponents of the non-zero digits of this number, so our answer is $7+5+4+3+2+0=\boxed{021}$