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Difference between revisions of "2013 Mock AIME I Problems/Problem 14"
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==Solution==
 
==Solution==
  
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Since 997 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 997, which by Vieta's equals -4. Thus our answer is 993 mod 997.
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By [[Vieta's Formulas]], the product of the roots is <math>-2014^2</math>. Since <math>997</math> is prime with <math>997\nmid2014^2</math>, all the roots are relatively prime to <math>997</math>. Thus, 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, equals <math>-4 \equiv 993 \pmod{997}</math>. Thus our answer is <math>\boxed{993}</math>.
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==See also==
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*[[2013 Mock AIME I Problems]]
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*[[2013 Mock AIME I Problems/Problem 13|Preceded by Problem 13]]
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*[[2013 Mock AIME I Problems/Problem 15|Followed by Problem 15]]
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[[Category:Intermediate Number Theory Problems]]

Latest revision as of 10:51, 4 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

By Vieta's Formulas, the product of the roots is $-2014^2$. Since $997$ is prime with $997\nmid2014^2$, all the roots are relatively prime to $997$. Thus, 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, equals $-4 \equiv 993 \pmod{997}$. Thus our answer is $\boxed{993}$.

See also

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