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

Revision as of 19:19, 12 March 2017

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.

You may ask: How can one of ${r, s}$ be positive and the other negative? $a$ will be negative as a result. That way, it's still $+2010$ that gets multiplied.

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 ( $r=s=\pm 1$ ). 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 8: / Line 8: @@
 We must now consider the various cases of signs. For the <math>40</math> cases where <math>|r|\neq |s|</math>, there are a total of four possibilities, For the case <math>|r|=|s|=1</math>, there are only three possibilities, <math>(r,s) = (1,1); (1,-1); (-1,-1)</math> as <math>(-1,1)</math> is not distinguishable from the second of those three.
+You may ask: How can one of <math>{r, s}</math> be positive and the other negative? <math>a</math> will be negative as a result. That way, it's still <math>+2010</math> that gets multiplied.
 Thus the grand total is <math>4\cdot40 + 3 = \boxed{163}</math>.

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