Difference between revisions of "2004 AIME I Problems/Problem 7"
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== Solution == | == Solution == | ||
| + | Let our [[polynomial]] be <math>P(x)</math>. It is clear that the [[coefficient]] of <math>x</math> in <math>P(x)</math> is <math>-1 + 2 - 3 + \ldots + 14 - 15 = -8</math>, so <math>P(x) = 1 -8x + Cx^2 + Q(x)</math> where <math>Q(x)</math> is some polynomial [[divisibility | divisible]] by <math>x^3</math>. Then <math>P(-x) = 1 + 8x + Cx^2 + Q(-x)</math> and so <math>P(x)\cdot P(-x) = 1 + (2C - 64)x^2 + R(x)</math>, where <math>R(x)</math> is some polynomial divisible by <math>x^3</math>. However, we also know <math>P(x)\cdot P(-x) = (1 - x)(1 + x)(1 +2x)(1 - 2x) \cdots (1 - 15x)(1 + 15x) = (1 - x^2)(1 - 4x^2)\cdots(1 - 225x^2) = 1 - (1 + 4 + \ldots + 225)x^2 + R(x)</math>. | ||
| + | Equating coefficients, we have <math>2C - 64 = -(1 + 4 + \ldots + 225) = -1240</math>, so <math>-2C = 1176</math> and <math>|C| = 588</math>. | ||
== See also == | == See also == | ||
* [[2004 AIME I Problems]] | * [[2004 AIME I Problems]] | ||
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Revision as of 15:33, 20 July 2006
Problem
Let 
be the coefficient of 
in the expansion of the product 
Find 
Solution
Let our polynomial be 
. It is clear that the coefficient of 
in is 
, so 
where 
is some polynomial divisible by 
. Then 
and so 
, where 
is some polynomial divisible by . However, we also know 
.
Equating coefficients, we have 
, so 
and 
.
