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

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== Problem ==
 
== 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>.
 
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>.
  
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__TOC__
 
 
==Solution==
 
==Solution==
 
===Solution 1===
 
===Solution 1===
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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.
 
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.
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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>.
 
Thus the grand total is <math>4\cdot40 + 3 = \boxed{163}</math>.
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Note: The only reason why we can be confident that <math>r = s</math> is the only case where the polynomials are being overcounted is because of this: We have the four configurations listed below:
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<math>(a,r,s)\\
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(a,-r,-s)\\
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(-a,-r,s)\\
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(-a,r,-s)</math>
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And notice, we start by counting all the positive solutions. So <math>r</math> and <math>s</math> must be strictly positive, no <math>0</math> or negatives allowed. The negative transformations will count those numbers.
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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 <math>x</math> term would have coefficients that are always different, <math>-a(r + s)</math> and <math>a(r + s)</math> because of the negative <math>r</math> and <math>s</math>. Since the <math>a</math> is never equal, these can never create equal <math>x</math> coefficients. We don't need to worry about this as <math>r</math> and <math>s</math> are positive and so that won't have any chance.
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However with the <math>(-a,-r,s)</math> and <math>(-a,r,-s)</math>, we have the coefficients of the <math>x</math> term as <math>a(s-r)</math> and <math>a(r-s)</math>. In other words, they are equal if <math>s-r=r-s</math> or <math>r=s</math>. Well if <math>r = 1</math>, then we have <math>s = 1</math> and in the <math>(r,-s)</math> case we have <math>(1,-1)</math> and if we transform using <math>(s,-r)</math>, then we have <math>(-1, 1)</math>. 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 <math>(-1,1)</math> and <math>(1,-1)</math> twice as they will create equal sums. This is why we subtract <math>1</math> from <math>41*4=164</math>.
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Each different transformation will give us different coordinates <math>(a,r,s)...</math> it is just that some of them create equal coefficients for the <math>x</math>-term, and we see that they are equal only in this case by our exploration, so we subtract <math>1</math> to account and get <math>163</math>.
  
 
===Solution 2===
 
===Solution 2===
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We use [[Burnside's Lemma]]. The set being acted upon is the set of integer triples <math>(a,r,s)</math> such that <math>ars=2010</math>. Because <math>r</math> and <math>s</math> are indistinguishable, the permutation group consists of the identity and the permutation that switches <math>r</math> and <math>s</math>. In cycle notation, the group consists of <math>(a)(r)(s)</math> and <math>(a)(r \: s)</math>. There are <math>4 \cdot 3^4</math> fixed points of the first permutation and <math>2</math> fixed points of the second permutation. By Burnside's Lemma, there are <math>\frac{1}{2} (4 \cdot 3^4+2)= \boxed{163}</math> distinguishable triples <math>(a,r,s)</math>.
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We use [[Burnside's Lemma]]. The set being acted upon is the set of integer triples <math>(a,r,s)</math> such that <math>ars=2010</math>. Because <math>r</math> and <math>s</math> are indistinguishable, the permutation group consists of the identity and the permutation that switches <math>r</math> and <math>s</math>. In cycle notation, the group consists of <math>(a)(r)(s)</math> and <math>(a)(r \: s)</math>. There are <math>4 \cdot 3^4</math> fixed points of the first permutation (after distributing the primes among <math>a</math>, <math>r</math>, <math>s</math> 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 <math>2</math> fixed points of the second permutation (<math>r=s=\pm 1</math>). By Burnside's Lemma, there are <math>\frac{1}{2} (4 \cdot 3^4+2)= \boxed{163}</math> distinguishable triples <math>(a,r,s)</math>.
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Note: The permutation group is isomorphic to <math>\mathbb{Z}/2\mathbb{Z}</math>.
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==Solution 3==
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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.
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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
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If <math>|k|\in \{2,3,5,67\}</math>. Take <math>|k|=2</math> as an example, follow the procedure above, there are <math>2^3+2^3=16</math> ways, and there are <math>\binom {4}{1}=4</math> ways to choose <math>|k|</math>, so there are <math>16\cdot 4=64</math> ways
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If <math>|k|</math> is the product of two factors of <math>2010</math>, there are <math>8</math> ways for each. Thus there are <math>\binom {4}{2}\cdot 8=48</math> ways
  
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Note: The permutation group is isomorphic to <math>\mathbb{Z}/2\mathbb{Z}</math>.
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If <math>|k|</math> is the product of three factors of <math>2010</math>, there are <math>\binom {4}{3}\cdot 4=16</math> ways
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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
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Thus, <math>32+64+48+16+3=\boxed{163}</math>
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~bluesoul
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

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

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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