Integer Quadratic

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mgao

1599 posts

mgao

#1 h Apr 1, 2010, 5:07 PM • 1 Y

Y by Adventure10

Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$ .

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BarbieRocks

1102 posts

BarbieRocks

#2 h Apr 1, 2010, 5:09 PM • 1 Y

Y by Adventure10

EDIT: You know what, disregard this post.

This post has been edited 1 time. Last edited by BarbieRocks, Apr 1, 2010, 5:21 PM

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AIME15

7892 posts

AIME15

#3 h Apr 1, 2010, 5:16 PM • 3 Y

Y by abracadabra1234, Adventure10, and 1 other user

$f(x) = a(x-r_1)(x-r_2)$ , $r_1, r_2 \in \mathbb{Z}$ .

And $f(0) = a \cdot r_1 \cdot r_2 = 2010$ .

So we want to split $2010$ into three factors, which is easily done.

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SnowEverywhere

801 posts

SnowEverywhere

#4 h Apr 1, 2010, 5:19 PM • 2 Y

Y by Adventure10, Mango247

I got $123 = 3/2(82)$ . I guess that I was wrong...

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RminusQ

171 posts

RminusQ

#5 h Apr 1, 2010, 5:21 PM • 4 Y

Y by Binomial-theorem, Adventure10, Mango247, and 1 other user

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.
This, sadly, is where I stopped during the contest, and wrote 041. However negative numbers and numbers, too.

We must now consider the various cases of signs. For the 40 cases where $|r|\neq |s|$ , there are a total of four possibilities, e.g. $(r,s) = (2,3); (-2,3); (2,-3); (-2,-3)$ . 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 $4\cdot40 + 3 = \underline{163}$

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BarbieRocks

1102 posts

BarbieRocks

#6 h Apr 1, 2010, 5:25 PM • 2 Y

Y by Adventure10, Mango247

can you (or someone else) post some other problems?

Do I have to wait all the way 'till tomorrow's Math Jam?

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PI-Dimension

824 posts

PI-Dimension

#7 h Apr 1, 2010, 5:53 PM • 2 Y

Y by Adventure10, Mango247

Can anyone find my mistake?

Click to reveal hidden text

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RminusQ

171 posts

RminusQ

#8 h Apr 1, 2010, 6:04 PM • 2 Y

Y by Adventure10, Mango247

If I'm understanding correctly, your 2 * (2*2*2*2/2) * 2 comes from [2 options for a] * [2*2*2*2 factors of 2010/a] / [but they come in pairs] * [but b can be + or -], correct?

When a = 2010, there is one factor for 2010/a, namely one. It DOESN'T come in pairs, so you WOULDN'T have the /2 factor. You would still have the *2 at the end, BUT, one of your sum/differences is 0, which can't be + and -. In short, in the last case, you have three solutions, not two and not four.
2010x^2 + 4020x + 2010
2010x^2 - 4020x + 2010
-2010x^2 + 2010

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FightTheTide

140 posts

FightTheTide

#9 h Apr 1, 2010, 8:43 PM • 2 Y

Y by Adventure10, Mango247

Shouldn't there be an equal number with a negative lead coefficient as there are with a positive lead coefficient? I don't see how the answer can be odd.

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Yongyi781

2142 posts

Yongyi781

#10 h Apr 2, 2010, 1:42 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

$-2010(x-1)(x+1)$ and $-2010(x+1)(x-1)$ are the same polynomial but $2010(x-1)^2$ and $2010(x+1)^2$ are different polynomials.

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FightTheTide

140 posts

FightTheTide

#11 h Apr 2, 2010, 1:56 AM • 2 Y

Y by Adventure10, Mango247

I had the same solution as PI-Dimension and I will be extremely ticked if I end up not making USAMO because I missed this problem by 1. I also guessed 219 on #15.

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Vworldv

276 posts

Vworldv

#12 h Apr 2, 2010, 5:56 AM • 4 Y

Y by jh235, Adventure10, Mango247, and 1 other user

Just to comment, Burnside's Lemma makes it more obvious (almost no possibility of a casework mistake) that the answer is $\frac{3^4+1}{2}+\frac{3^4+1}{2}+3^4$ -- it's just the group action of $\mathbb{Z}_2$ . If you don't know what Burnside's Lemma is, you should go learn it now.

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yugrey

2326 posts

yugrey

#13 h Apr 4, 2010, 9:28 PM • 1 Y

Y by Adventure10

wait, what if a is negative????
Oh, a is just a function of r and s; you can identify a solution by its r and s, you don't need to know what a is!

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Solid Snake

5 posts

Solid Snake

#14 h Apr 4, 2010, 9:55 PM • 2 Y

Y by Adventure10, Mango247

Yeah, that's right.

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bigman

557 posts

bigman

#15 h Jan 16, 2011, 5:51 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

RminusQ wrote:

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.
This, sadly, is where I stopped during the contest, and wrote 041. However negative numbers and numbers, too.

We must now consider the various cases of signs. For the 40 cases where $|r|\neq |s|$ , there are a total of four possibilities, e.g. $(r,s) = (2,3); (-2,3); (2,-3); (-2,-3)$ . 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 $4\cdot40 + 3 = \underline{163}$

Ca you explain the $81$ and then the $40$ ?

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viperstrike

1198 posts

viperstrike

#16 h Feb 10, 2014, 4:14 AM • 2 Y

Y by dasdhajskhd, Adventure10

This is hopefully a more understandable solution:

Click to reveal hidden text

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Amad271

186 posts

Amad271

#17 h Jul 21, 2015, 9:48 PM • 2 Y

Y by Adventure10, Mango247

Why are $p$ and $q$ indistinguishable? They are different roots aren't they?

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macandcheese

70 posts

macandcheese

#18 h Jul 22, 2015, 1:29 AM • 2 Y

Y by Adventure10, Mango247

If you switch the $p$ and $q$ values, you end up with the same polynomial. For example, if $a=1, p= 201, q= 10$ we have the polynomial $(x-201)(x-10)$ . If we switch p and q we get $(x-10)(x-201)$ . Since these are the same polynomial, and the question asks for the number of polynomials, $p$ and $q$ are indistinguishable.

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Wave-Particle

3690 posts

Wave-Particle

#19 h Jul 22, 2015, 1:43 AM • 1 Y

Y by Adventure10

Amad271 wrote:

Why are $p$ and $q$ indistinguishable? They are different roots aren't they?

great answer by macandcheese but just incase it wasn't clear, here is my answer:

Yes $p$ and $q$ are indistinguishable (they are both roots, it doesn't matter which you label p and q) but that does not mean the polynomial is going to be different. As stated above by the communitive property we have that $(x-201)(x-10)=(x-10)(x-201)=(x-p)(x-q)$ but they asked how many different polynomials there are and those two both give the same polynomial though $p$ and $q$ are interchanged.

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Amad271

186 posts

Amad271

#20 h Jul 22, 2015, 5:30 PM • 2 Y

Y by Adventure10, Mango247

Got it, thanks mac&cheese and anadiyer12.

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jasperE3

11449 posts

jasperE3

#21 h Mar 30, 2022, 2:35 PM

Y by

Let $f(x)=a(x+r)(x+s)=ax^2+bx+2010$ , then $b=a(r+s)$ and $ars=2010$ . WLOG $r\ge s$ . Finding $a,r,s$ uniquely determines the polynomial, so we just need the number of such $a,r,s$ .

Claim: the number of solutions to $ijk=2010$ in positive integers is $81$ , of which $41$ satisfy $j\ge k$ .
Note that $2010=2\cdot3\cdot5\cdot67$ , so there are $3^4$ total solutions in positive integers.
If $j=k$ , then since $2010$ is squarefree $j=k=1$ , so $i=2010$ . This is one solution. By symmetry half of the $81$ remaining solutions satisfy $j>k$ .

Case 1: $r>0$ , $s>0$
Note $a>0$ . Let $(i,j,k)=(a,r,s)$ , then we have $41$ solutions.

Case 2: $r<0$ , $s<0$
Note $a>0$ . Let $(i,j,k)=(a,-s,-r)$ , then we have $41$ solutions.

Case 3: $r>0$ , $s<0$
Note $a<0$ . Let $(i,j,k)=(-a,r,-s)$ , then we have $81$ solutions.

In total, there are $81+40+40=\boxed{163}$ such polynomials.

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OlympusHero

17020 posts

OlympusHero

#22 h Jan 4, 2023, 10:58 PM • 3 Y

Y by Mango247, Mango247, Mango247

Denote $f(x) = a(x-b)(x-c)$ for integers $a, b, c$ and $abc = 2010$ . We have $2010 = 2 \cdot 3 \cdot 5 \cdot 67$ , so via stars and bars there are $\binom{3}{2}^4 = 81$ positive integer solutions. For the signs, they are either all negative (one way) or we select two to be negative (three ways) for a total of four ways. That gives $81 \cdot 4 = 324$ , but the order of $b$ and $c$ doesn't matter and if we have $b = c$ then we overcounted, and we do with $(2010,-1,-1)$ and $(2010,1,1)$ . Therefore, the answer is $\frac{322}{2}+2 = \boxed{163}$ (we add back the 2 pairs at the end).

This post has been edited 1 time. Last edited by OlympusHero, Jan 4, 2023, 10:58 PM

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AngeloChu

472 posts

AngeloChu

#23 h Jan 4, 2023, 11:53 PM • 3 Y

Y by Mango247, Mango247, Mango247

$f(x)=a_1(x-a_2)(x-a_3)$ $a_1a_2a_3=2010$ $2010=2\cdot3\cdot5\cdot67$ There are $3^4=81$ ways to distribute the $4$ prime factors to the three $a$ 's
We can choose all of $a_1$ , $a_2$ , and $a_3$ to be positive or exactly $2$ to be negative.
We then have $324$ triples, and we divide by $2$ because $a_2$ and $a_3$ are distinguishable.
We add back $\frac{2}{2}=1$ for $(a_1,a_2,a_3)=(2010,1,1)$ and $(a_1,a_2,a_3)=(2010,-1,-1)$ where $a_2=a_3$
we end up with $\frac{324+2}{2}=163$

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MathWizard10

1428 posts

MathWizard10

#24 h Jan 5, 2023, 1:30 AM

Y by

This is the most trivial AIME P10

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CoolJupiter

925 posts

CoolJupiter

#25 h Jan 5, 2023, 1:33 AM • 3 Y

Y by Mango247, Mango247, Mango247

MathWizard10 wrote:

This is the most trivial AIME P10

Agreed

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megahertz13

3201 posts

megahertz13

#26 h May 23, 2025, 9:28 PM

Y by

mgao wrote:

Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$ .

Let $f(x)=a(x-r)(x-s)$ . The condition is equivalent to $ars=2010=2\cdot 3\cdot 5\cdot 67$ .

Case 1: $r=s$ . Thus $(r,s)=(1,1)$ or $(r,s)=(-1,-1)$ . Both of these yield one solution, so there are $2$ cases here.

Case 2: $r\ne s$ . There are $3^4$ ways to pick $(a,r,s)$ so that all of them are positive (ignoring the condition specifying this case), and we subtract $1$ since $r\ne s$ . Therefore, there are $80$ ways if $r,s>0$ . However, it's also possible that some of the numbers are negative: if $r,s<0$ , then there's $80$ cases by symmetry. If $r<0$ and $s>0$ or vice versa, then there are $2\cdot 81=162$ cases here. However, $r$ and $s$ are symmetrical (flipping them does not change the quadratic), so we divide by $2$ to yield $161$ .

The final answer is $2+161=163$ .

This post has been edited 1 time. Last edited by megahertz13, May 23, 2025, 9:29 PM

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