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

 
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==Solution==
 
==Solution==
  
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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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>.
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==See also==
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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]]
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{{MAA Notice}}

Latest revision as of 18:04, 15 December 2020

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, 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 $993\pmod{997}$.

See also

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