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

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Thus the grand total is <math>4\cdot40 + 3 = \boxed{163}</math>.
 
Thus the grand total is <math>4\cdot40 + 3 = \boxed{163}</math>.
  
−
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:
+
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:
  
 
<math>(a,r,s)\\
 
<math>(a,r,s)\\
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(-a,r,-s)</math>
 
(-a,r,-s)</math>
  
−
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.
+
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.
  
−
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, <math>-a(r+s)</math> and <math>a(r+s)</math> 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.
+
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.
  
−
However with the (-a,-r,s) and (-a,r,-s), we have the coefficients of the x 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 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 <math>41*4=164</math>.
+
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>.
  
−
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.
+
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===

Revision as of 16:28, 26 September 2019

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 $\mathbb{Z}/2\mathbb{Z}$.

See also

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