Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 11"
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Let <math>\omega^3=1</math> with <math>\omega\neq 1</math>. We have | Let <math>\omega^3=1</math> with <math>\omega\neq 1</math>. We have | ||
− | \begin{align*} | + | <cmath> \begin{align*} \frac{f(1)+f(\omega)+f(\omega^2)}{3} &= \frac{(1-1)^{2004}+(\omega-1)^{2004}+(\omega^2-1)^{2004}}{3} \\ |
− | \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}) \\ | &= \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} | + | &= \sum_{n=0}^{668}(-1)^n \binom{2004}{3n}. |
− | \end{align*} | + | \end{align*} </cmath> |
where the last step follows because <math>1^k+\omega^k+\omega^{2k}</math> is 0 when <math>k</math> is not divisible by 3, and <math>3</math> when <math>k</math> is divisible by 3. | where the last step follows because <math>1^k+\omega^k+\omega^{2k}</math> is 0 when <math>k</math> is not divisible by 3, and <math>3</math> when <math>k</math> is divisible by 3. |
Latest revision as of 10:51, 2 July 2015
Problem
Let denote the value of the sum Determine the remainder obtained when is divided by .
Solution
Consider the polynomial
Let with . We have
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |