Difference between revisions of "2013 Mock AIME I Problems/Problem 14"
Sugar rush (talk | contribs) |
Sugar rush (talk | contribs) |
||
| Line 4: | Line 4: | ||
==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}} | ||
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
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
