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

Latest revision as of 15:01, 16 September 2023

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 have $(1,-1)$ 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 $\mathbb{Z}/2\mathbb{Z}$ .

Solution 3

The equation can be written in the form of $k(x-a)(x-b)$ , where $|k|$ is a divisor of $2010$ . Factor $2010=2\cdot 3\cdot 5\cdot 67$ . We divide into few cases to study.

Firstly, if $|k|=1$ , the equation can be $-(x-a)(x+b)$ or $(x-a)(x-b)$ or $(x+a)(x+b)$ , there are $2^4+2^4=32$ ways

If $|k|\in \{2,3,5,67\}$ . Take $|k|=2$ as an example, follow the procedure above, there are $2^3+2^3=16$ ways, and there are $\binom {4}{1}=4$ ways to choose $|k|$ , so there are $16\cdot 4=64$ ways

If $|k|$ is the product of two factors of $2010$ , there are $8$ ways for each. Thus there are $\binom {4}{2}\cdot 8=48$ ways

If $|k|$ is the product of three factors of $2010$ , there are $\binom {4}{3}\cdot 4=16$ ways

In the end, $|k|=2010$ , only $2010(x-1)(x-1); 2010(x+1)(x+1); -2010(x-1)(x+1)$ work, there are $3$ ways

Thus, $32+64+48+16+3=\boxed{163}$

~bluesoul

@@ Line 1: / Line 1: @@
+__TOC__
 == Problem ==
 Find the number of second-degree [[polynomial]]s <math>f(x)</math> with integer [[coefficient]]s and integer zeros for which <math>f(0)=2010</math>.
-__TOC__
 ==Solution==
 ===Solution 1===
@@ Line 34: / Line 35: @@
 ==Solution 3==
-The equation can be written in the form of <math>|k|(x-a)(x-b)</math>, where <math>|k|</math> is a divisor of <math>2010</math>. Factor <math>2010=2\cdot 3\cdot 5\cdot 67</math>. We divide into few cases to study.
+The equation can be written in the form of <math>k(x-a)(x-b)</math>, where <math>|k|</math> is a divisor of <math>2010</math>. Factor <math>2010=2\cdot 3\cdot 5\cdot 67</math>. We divide into few cases to study.
 Firstly, if <math>|k|=1</math>, the equation can be <math>-(x-a)(x+b)</math> or <math>(x-a)(x-b)</math> or <math>(x+a)(x+b)</math>, there are <math>2^4+2^4=32</math> ways
@@ Line 47: / Line 48: @@
 In the end, <math>|k|=2010</math>, only <math>2010(x-1)(x-1); 2010(x+1)(x+1); -2010(x-1)(x+1)</math> work, there are <math>3</math> ways
 Thus, <math>32+64+48+16+3=\boxed{163}</math>
+~bluesoul
 == 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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