Mock AIME 1 Pre 2005 Problems/Problem 11
Problem
Let denote the value of the sum Determine the remainder obtained when is divided by .
Solution
Consider the polynomial
Let with . We have $\begin{align*} \frac{f(1)+f(\omega)+f(\omega^2)}{3} &= \frac{(1-1)^{2004}+(\omega-1)^{2004}+(\omega^2-1)^{2004}}{3} \\ &= \frac{1}{3}\sum_{n=0}^{2004}\binom{2004}{n}\cdot(-1)^n\cdot(1^{2004-n}+\omega^{2004-n}+(\omega^2)^{2004-n}) \\ &= \sum_{n=0}^{668}(-1)^n \binom{2004}{3n}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)
where the last step follows because is 0 when is not divisible by 3, and when is divisible by 3.
We now compute . WLOG, let . Then , and . These numbers are both of the form , where is a 12th root of unity, so both of these, when raised to the 2004-th power, become . Thus, our desired sum becomes .
To find , we notice that so that . Then . Thus, our answer is .
See also
Mock AIME 1 Pre 2005 (Problems, Source) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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