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>\frac{3^4+1}{2}+3^4=\boxed{163}< | + | Due to [[Burnside's Lemma]], the answer is 2<math>\frac{3^4+1}{2}+2</math>\frac{3^4}{2}=\boxed{163}<math> — it is group action of </math>\mathbb{Z}^2$. |
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== See also == | == See also == |
Revision as of 11:34, 12 March 2011
Problem
Find the number of second-degree polynomials with integer coefficients and integer zeros for which
.
Solution
Solution 1
Let . Then
. First consider the case where
and
(and thus
) are positive. There are
ways to split up the prime factors between
,
, and
. However,
and
are indistinguishable. In one case,
, we have
. The other
cases are double counting, so there are
.
We must now consider the various cases of signs. For the cases where
, there are a total of four possibilities, For the case
, there are only three possibilities,
as
is not distinguishable from the second of those three.
Thus the grand total is .
Solution 2
Due to Burnside's Lemma, the answer is 2\frac{3^4}{2}=\boxed{163}
\mathbb{Z}^2$.
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 |