Difference between revisions of "2013 Mock AIME I Problems/Problem 14"
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− | Problem | + | ==Problem== |
+ | Let <math>P(x) = x^{2013}+4x^{2012}+9x^{2011}+16x^{2010}+\cdots + 4052169x + 4056196 = \sum_{j=1}^{2014}j^2x^{2014-j}.</math> If <math>a_1, a_2, \cdots a_{2013}</math> are its roots, then compute the remainder when <math>a_1^{997}+a_2^{997}+\cdots + a_{2013}^{997}</math> is divided by 997. | ||
− | + | ==Solution== | |
− | + | 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}} |
Revision as of 19:04, 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.