Difference between revisions of "2010 AIME II Problems/Problem 10"
m |
JohnPeanuts (talk | contribs) (→Solution 2: Rewrote burnside's lemma solution with more explanation.) |
||
Line 12: | Line 12: | ||
===Solution 2=== | ===Solution 2=== | ||
− | + | We use [[Burnside's Lemma]]. The set being acted upon is the set of integer triples <math>(a,r,s)</math> such that <math>ars=2010</math>. Because <math>r</math> and <math>s</math> are indistinguishable, the permutation group consists of the identity and the permutation that switches <math>r</math> and <math>s</math>. In cycle notation, the group consists of <math>(a)(r)(s)</math> and <math>(a)(r \: s)</math>. There are <math>4 \cdot 3^4</math> fixed points of the first permutation and <math>2</math> fixed points of the second permutation. By Burnside's Lemma there are <math>\frac{1}{2} (4 \cdot 3^4+2)= \boxed{163}</math> distinguishable triples <math>(a,r,s)</math>. | |
+ | |||
+ | Note: The permutation group is isomorphic to <math>\mathbb{Z}/2\mathbb{Z}</math>. | ||
== See also == | == See also == |
Revision as of 16:17, 24 March 2013
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
We use Burnside's Lemma. The set being acted upon is the set of integer triples such that
. Because
and
are indistinguishable, the permutation group consists of the identity and the permutation that switches
and
. In cycle notation, the group consists of
and
. There are
fixed points of the first permutation and
fixed points of the second permutation. By Burnside's Lemma there are
distinguishable triples
.
Note: The permutation group is isomorphic to .
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 |