2010 AIME II Problems/Problem 10

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

Due to Burnside's Lemma, the answer is $\frac{3^4+1}{2}+\frac{3^4+1}{2}+3^4=\boxed{163}$ — it is group action of $\mathbb{Z}^2$ .

2010 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2010 AIME II Problems/Problem 10

Problem

Contents

Solution

Solution 1

Solution 2

See also