2013 Mock AIME I Problems/Problem 14: Difference between revisions
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==Solution== | ==Solution== | ||
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 | 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>. | ||
==See also== | ==See also== | ||
*[[2013 Mock AIME I Problems]] | |||
*[[2013 Mock AIME I Problems/Problem 13|Preceded by Problem 13]] | *[[2013 Mock AIME I Problems/Problem 13|Preceded by Problem 13]] | ||
*[[2013 Mock AIME I Problems/Problem 15|Followed by Problem 15]] | *[[2013 Mock AIME I Problems/Problem 15|Followed by Problem 15]] | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
Latest revision as of 10:51, 4 August 2024
Problem
Let 
If 
are its roots, then compute the remainder when 
is divided by 997.
Solution
By Vieta's Formulas, the product of the roots is 
. Since 
is prime with 
, all the roots are relatively prime to . Thus, by Fermat's Little Theorem, we have 
, which, by Vieta, equals 
. Thus our answer is 
.
