Difference between revisions of "2010 AIME II Problems/Problem 10"
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===Solution 2=== | ===Solution 2=== | ||
| − | Due to [[Burnside's Lemma]], the answer is <math> | + | Due to [[Burnside's Lemma]], the answer is <math>\frac{3^4+1}{2}+3^4=\boxed{163}</math> — it is group action of <math>\mathbb{Z}^2</math>. |
== See also == | == See also == | ||
Revision as of 17:43, 9 June 2010
Problem
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 , , and . However, and 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
Due to 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 | ||
