Difference between revisions of "2010 AIME II Problems/Problem 10"
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| − | The answer is <math>\frac{3^4+1}{2}+\frac{3^4+1}{2}+3^4=\boxed{163}</math> -- it is group action of <math>\mathbb{Z} | + | The answer is <math>\frac{3^4+1}{2}+\frac{3^4+1}{2}+3^4=\boxed{163}</math> -- it is group action of <math>\mathbb{Z}^2</math>. |
== See also == | == See also == | ||
{{AIME box|year=2010|num-b=9|num-a=11|n=II}} | {{AIME box|year=2010|num-b=9|num-a=11|n=II}} | ||
Revision as of 19:16, 3 April 2010
Problem 10
Find the number of second-degree polynomials 
with integer coefficients and integer zeros for which 
.
Solution
Solution 1
Let 
. Then 
. First consider the case where 
and 
(and thus 
) are positive.
There are 
ways to split up the prime factors between a, r, and s. However, r and s are indistinguishable. In one case, 
, we have 
. The other 
cases are double counting, so there are 
.
We must now consider the various cases of signs. For the cases where 
, there are a total of four possibilities, For the case 
, there are only three possibilities, 
as 
is not distinguishable from the second of those three.
Thus the grand total is 
Solution 2
Burnside's Lemma:
The answer is 
-- it is group action of 
.
See also
| 2010 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
