# 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 | + | 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>. |

+ | |||

+ | ==See also== | ||

+ | *[[2013 Mock AIME I Problems/Problem 13|Preceded by Problem 13]] | ||

+ | *[[2013 Mock AIME I Problems/Problem 15|Followed by Problem 15]] | ||

+ | |||

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

+ | {{MAA Notice}} |

## Latest revision as of 18:04, 15 December 2020

## Problem

Let If are its roots, then compute the remainder when is divided by 997.

## Solution

Since is prime, we have equals to mod , which by Vieta's equals . Thus our answer is .

## See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.