Difference between revisions of "Mock AIME 3 2006-2007 Problems/Problem 3"
(Created page with " == Problem == Let <math>P(x) = 2x^{59} - x^2 - x - 6</math>. If <math>Q(x)</math> is a polynomial whose roots are the 59th powers of the roots of <math>P(x)</math>, then find t...") |
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| − | Let <math>a_1,a_2,\ldots,a_{59}</math> denote the roots of P. Then, <math>\sum_{i} a_i = 0, and \sum_{i\neq j} = 0</math>. Therefore, <math>\sum_{i} a_i^2 = 0</math>. | + | Let <math>a_1,a_2,\ldots,a_{59}</math> denote the roots of P. Then, <math>\sum_{i} a_i = 0</math>, and <math>\sum_{i\neq j} = 0</math>. Therefore, <math>\sum_{i} a_i^2 = 0</math>. |
| − | Denote<math> S_n(a) = \sum_{i} a_i^n</math>. So <math>S_1(a) = 0, S_2(a) = 0</math>. To find <math>S_{59}(a)</math>, use Newton's recursive sums: | + | Denote <math> S_n(a) = \sum_{i} a_i^n</math>. So <math>S_1(a) = 0, S_2(a) = 0</math>. To find <math>S_{59}(a)</math>, use Newton's recursive sums: |
<math>S_{59}(a)\cdot 2 + S_{58}(a)\cdot 0 + S_{57}(a)\cdot 0 + \cdots + S_2(a)\cdot(-1) + S_1(a)\cdot(-1) + 59\cdot(-6) = 0</math>. | <math>S_{59}(a)\cdot 2 + S_{58}(a)\cdot 0 + S_{57}(a)\cdot 0 + \cdots + S_2(a)\cdot(-1) + S_1(a)\cdot(-1) + 59\cdot(-6) = 0</math>. | ||
Solving, we get <math>S_{59}(a) = 59\cdot 3 = \boxed{177}</math>. | Solving, we get <math>S_{59}(a) = 59\cdot 3 = \boxed{177}</math>. | ||
Latest revision as of 23:38, 15 February 2015
Problem
Let 
. If 
is a polynomial whose roots are the 59th powers of the roots of 
, then find the sum of the roots of .
Solution
Let 
denote the roots of P. Then, 
, and 
. Therefore, 
.
Denote 
. So 
. To find 
, use Newton's recursive sums:
.
Solving, we get 
.
