Difference between revisions of "2010 AIME II Problems/Problem 10"

Revision as of 16:17, 24 March 2013

Problem

Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$ .

Solution

Solution 1

Let $f(x) = a(x-r)(x-s)$ . Then $ars=2010=2\cdot3\cdot5\cdot67$ . First consider the case where $r$ and $s$ (and thus $a$ ) are positive. There are $3^4 = 81$ ways to split up the prime factors between $a$ , $r$ , and $s$ . However, $r$ and $s$ are indistinguishable. In one case, $(a,r,s) = (2010,1,1)$ , we have $r=s$ . The other $80$ cases are double counting, so there are $40$ .

We must now consider the various cases of signs. For the $40$ cases where $|r|\neq |s|$ , there are a total of four possibilities, For the case $|r|=|s|=1$ , there are only three possibilities, $(r,s) = (1,1); (1,-1); (-1,-1)$ as $(-1,1)$ is not distinguishable from the second of those three.

Thus the grand total is $4\cdot40 + 3 = \boxed{163}$ .

Solution 2

We use Burnside's Lemma. The set being acted upon is the set of integer triples $(a,r,s)$ such that $ars=2010$ . Because $r$ and $s$ are indistinguishable, the permutation group consists of the identity and the permutation that switches $r$ and $s$ . In cycle notation, the group consists of $(a)(r)(s)$ and $(a)(r \: s)$ . There are $4 \cdot 3^4$ fixed points of the first permutation and $2$ fixed points of the second permutation. By Burnside's Lemma there are $\frac{1}{2} (4 \cdot 3^4+2)= \boxed{163}$ distinguishable triples $(a,r,s)$ .

Note: The permutation group is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ .

@@ Line 12: / Line 12: @@
 ===Solution 2===
-Due to [[Burnside's Lemma]], the answer is 2 <math>\frac{3^4+1}{2}</math>+2 <math>\frac{3^4}{2}=\boxed{163}</math> &mdash; it is group action of <math>\mathbb{Z}^2</math>.
+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 ==

2010 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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