Difference between revisions of "2013 Mock AIME I Problems/Problem 14"
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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== |
Latest revision as of 11:51, 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
.