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

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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>.
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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 (after distributing the primes among <math>a</math>, <math>r</math>, <math>s</math> and then considering their signs) 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>.
  
  

Revision as of 15:21, 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 (after distributing the primes among $a$, $r$, $s$ and then considering their signs) 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 $\mathbb{Z}/2\mathbb{Z}$.

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
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