2010 AIME II Problems/Problem 10

Revision as of 11:34, 12 March 2011 by Elysion5 (talk | contribs) (Fixed a minor mistake in solution 2.)

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 2$\frac{3^4+1}{2}+2$\frac{3^4}{2}=\boxed{163}$— it is group action of$\mathbb{Z}^2$.


See also

2010 AIME II (ProblemsAnswer KeyResources)
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