2013 Mock AIME I Problems/Problem 14

Revision as of 02:24, 31 January 2016 by Eisirrational (talk | contribs) (I am new to latex, but briefed the solution.)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Let \begin{align*}P(x) = x^{2013}+4x^{2012}+9x^{2011}+16x^{2010}+\cdots + 4052169x + 4056196 = \sum_{j=1}^{2014}j^2x^{2014-j}.\end{align*} 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, we have $a_1^{997}+a_2^{997}+\cdots + a_{2013}^{997}$ equals to $a_1+a_2+\cdots + a_{2013}$ mod 997, which by Vieta's equals -4. Thus our answer is 993 mod 997.