Difference between revisions of "2013 Mock AIME I Problems/Problem 14"

Revision as of 12:34, 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 $\boxed{993}$ .

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 ==Solution==
-Since <math>997</math> is prime, we have <math>a_1^{997}+a_2^{997}+\cdots + a_{2013}^{997}</math> equals to <math>a_1+a_2+\cdots + a_{2013}</math> mod <math>997</math>, which by Vieta's equals <math>-4</math>. Thus our answer is <math>993\pmod{997}</math>.
+Since <math>997</math> is prime, by [[Fermat's Little Theorem]], we have <math>a_1^{997}+a_2^{997}+\cdots + a_{2013}^{997} \equiv a_1+a_2+\cdots + a_{2013} \pmod{997}</math>, which, by [[Vieta's Formulas]], equals <math>-4 \equiv 993 \pmod{997}</math>. Thus our answer is <math>\boxed{993}</math>.
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Difference between revisions of "2013 Mock AIME I Problems/Problem 14"

Revision as of 12:34, 1 August 2024

Problem

Solution

See also