2008 AIME II Problems/Problem 4
Problem
There exist unique nonnegative integers
and
unique integers
(
) with each
either
or
such that
Find
.
Solution
In base , we find that
. In other words,
![$2008 = 2 \cdot 3^{6} + 2 \cdot 3^{5} + 2 \cdot 3^3 + 1 \cdot 3^2 + 1 \cdot 3^0$](http://latex.artofproblemsolving.com/f/f/5/ff5fb8a140c5075181b078cc65d366822e75fdcc.png)
In order to rewrite as a sum of perfect powers of , we can use that
:
![$2008 = (3^7 - 3^6) + (3^6-3^5) + (3^4 - 3^3) + 3^2 + 3^0 = 3^7 - 3^5 + 3^4 - 3^3 + 3^2 + 3^0$](http://latex.artofproblemsolving.com/5/6/d/56d57e927fd20bdcb3784e34c65cfb009d164747.png)
The answer is .
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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All AIME Problems and Solutions |