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2013 Mock AIME I Problems/Problem 14: Difference between revisions

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==See also==
==See also==
*[[2013 Mock AIME I Problems]]
*[[2013 Mock AIME I Problems/Problem 13|Preceded by Problem 13]]
*[[2013 Mock AIME I Problems/Problem 13|Preceded by Problem 13]]
*[[2013 Mock AIME I Problems/Problem 15|Followed by Problem 15]]
*[[2013 Mock AIME I Problems/Problem 15|Followed by Problem 15]]


[[Category:Intermediate Number Theory Problems]]
[[Category:Intermediate Number Theory Problems]]

Revision as of 12:35, 1 August 2024

Problem

Let $P(x) = x^{2013}+4x^{2012}+9x^{2011}+16x^{2010}+\cdots + 4052169x + 4056196 = \sum_{j=1}^{2014}j^2x^{2014-j}.$ If $a_1, a_2, \cdots a_{2013}$ are its roots, then compute the remainder when $a_1^{997}+a_2^{997}+\cdots + a_{2013}^{997}$ is divided by 997.

Solution

Since $997$ is prime, by Fermat's Little Theorem, we have $a_1^{997}+a_2^{997}+\cdots + a_{2013}^{997} \equiv a_1+a_2+\cdots + a_{2013} \pmod{997}$, which, by Vieta's Formulas, equals $-4 \equiv 993 \pmod{997}$. Thus our answer is $\boxed{993}$.

See also

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