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.

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 $4\cdot40 + 3 = \boxed{163}$ .

Note: The only reason why we can be confident that r=s is the only case where the polynomials are being overcounted is because of this: We have the four configurations listed below:

$(a,r,s) (a,-r,-s) (-a,-r,s) (-a,r,-s)$

And notice, we start by counting all the positive solutions. So r and s must be strictly positive, no 0 or negatives allowed. The negative transformations will count those numbers.

So with these we can conclude that only the first and second together have a chance of being equal, and the third and fourth together. If we consider the first and second, the x term would have coefficients that are always different, $-a(r+s)$ and $a(r+s)$ because of the negative r and s. Since the a is never equal, these can never create equal x coefficients. We don't need to worry about this as r and s are positive and so that won't have any chance.

However with the (-a,-r,s) and (-a,r,-s), we have the coefficients of the x term as $a(s-r)$ and $a(r-s)$ . In other words, they are equal if $s-r=r-s$ or $r=s$ . Well if r=1, then we have s=1 and in the (r,-s) case we (1,-1) and if we transform using (s,-r), then we have (-1, 1). So this is the only way that we could possibly overcount the equal cases, and so we need to make sure we don't count (-1,1) and (1,-1) twice as they will create equal sums. This is why we subtract 1 from $41*4=164$ .

Each different transformation will give us different coordinates (a,r,s)... it is just that some of them create equal coefficients for the x-term, and we see that they are equal only in this case by our exploration, so we subtract 1 to account and get 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. We have 4 ways since we can keep them all positive, first 2 negative, first and third negative, or last two negative) 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 $\mathbb{Z}/2\mathbb{Z}$ .

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