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== | |
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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. | 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. | ||
Revision as of 15:08, 15 December 2020
Problem
Let 
If 
are its roots, then compute the remainder when 
is divided by 997.
Solution
Since 997 is prime, we have equals to 
mod 997, which by Vieta's equals -4. Thus our answer is 993 mod 997.
