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Difference between revisions of "2023 AIME I Problems/Problem 9"

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~s214425
 
~s214425
  
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==Solution 2==
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==Solution 2 (factor the difference)==
  
 
<math>p(x)-p(2)</math> is a cubic with at least two integral real roots, therefore it has three real roots, which are all integers.
 
<math>p(x)-p(2)</math> is a cubic with at least two integral real roots, therefore it has three real roots, which are all integers.
  
 
There are exactly two distinct roots, so either <math>p(x)=p(2)+(x-2)^2(x-m)</math> or <math>p(x)=p(2)+(x-2)(x-m)^2</math>, with <math>m\neq 2</math>.
 
There are exactly two distinct roots, so either <math>p(x)=p(2)+(x-2)^2(x-m)</math> or <math>p(x)=p(2)+(x-2)(x-m)^2</math>, with <math>m\neq 2</math>.
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In the first case <math>p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)</math>, with <math>|4+4m|\leq 20</math> (which entails <math>|4+m|\leq 20</math>), so <math>m</math> can be <math>-6,-5,-4,-3,-2,-1,0,1,3,4</math> and <math>-4m+p(2)</math> can be any integer from <math>-20</math> to <math>20</math>, giving <math>410</math> polynomials.
 
  
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In the second case <math>p(x)=x^3-(1+2m)x^2+(4m+m^2)x-2m^2+p(2)</math>, and <math>m</math> can be <math>-6,-5,-4,-3,-2,-1,0,1</math> and <math>-4m+p(2)</math> can be any integer from <math>-20</math> to <math>20</math>, giving <math>328</math> polynomials.
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In the first case <math>p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)</math>, with <math>|4+4m|\leq 20</math> (which entails <math>|4+m|\leq 20</math>), so <math>m</math> can be <math>-6,-5,-4,-3,-2,-1,0,1, (\textbf{not 2}!), 3,4</math> and <math>-4m+p(2)</math> can be any integer from <math>-20</math> to <math>20</math>, giving <math>410</math> polynomials (since the coefficients are given by linear functions of <math>m</math> and thus are distinct).
 +
 
 +
In the second case <math>p(x)=x^3-(2+2m)x^2+(4m+m^2)x-2m^2+p(2)</math>, and <math>m</math> can be <math>-6,-5,-4,-3,-2,-1,0,1</math> and <math>-4m+p(2)</math> can be any integer from <math>-20</math> to <math>20</math>, giving <math>328</math> polynomials.
  
 
The total is <math>\boxed{738}</math>.
 
The total is <math>\boxed{738}</math>.
  
 
~EVIN-
 
~EVIN-
 +
 +
==Video Solution==
 +
https://youtu.be/-Asb_5nTgSg
 +
 +
~MathProblemSolvingSkills.com
 +
 +
  
 
==See also==
 
==See also==

Latest revision as of 19:40, 27 December 2023

Problem

Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c,$ where $a, b,$ and $c$ are integers in $\{-20,-19,-18,\ldots,18,19,20\},$ such that there is a unique integer $m \not= 2$ with $p(m) = p(2).$

Solution 1 (bash)

Plugging $2$ and $m$ into $P(x)$ and equating them, we get $8+4a+2b+c = m^3+am^2+bm+c$. Rearranging, we have \[(m^3-8) + (m^2 - 4)a + (m-2)b = 0.\] Note that the value of $c$ won't matter as it can be anything in the provided range, giving a total of $41$ possible choices for $c.$ So what we just need to do is to just find the number of ordered pairs $(a, b)$ that work, and multiply it by $41.$ We can start by first dividing both sides by $m-2.$ (Note that this is valid since $m\neq2:$ \[m^2 + 2m + 4 + (m+2)a + b = 0.\] We can rearrange this so it is a quadratic in $m$: \[m^2 + (a+2)m + (4 + 2a + b) = 0.\] Remember that $m$ has to be unique and not equal to $2.$ We can split this into two cases: case $1$ being that $m$ has exactly one solution, and it isn't equal to $2$; case $2$ being that $m$ has two solutions, one being equal to $2,$ but the other is a unique solution not equal to $2.$


$\textbf{Case 1:}$

There is exactly one solution for $m,$ and that solution is not $2.$ This means that the discriminant of the quadratic equation is $0,$ using that, we have $(a+2)^2 = 4(4 + 2a + b),$ rearranging in a neat way, we have \[(a-2)^2 = 4(4 + b)\Longrightarrow a = 2\pm2\sqrt{4+b}.\] Using the fact that $4+b$ must be a perfect square, we can easily see that the values for $b$ can be $-4, -3, 0, 5,$ and $12.$ Also since it's a "$\pm$" there will usually be $2$ solutions for $a$ for each value of $b.$ The two exceptions for this would be if $b = -4$ and $b = 12.$ For $b=-4$ because it would be a $\pm0,$ which only gives one solution, instead of two. And for $b=12$ because then $a = -6$ and the solution for $m$ would equal to $2,$ and we don't want this. (We can know this by putting the solutions back into the quadratic formula).

So we have $5$ solutions for $b,$ each of which give $2$ values for $a,$ except for $2,$ which only give one. So in total, there are $5*2 - 2 = 8$ ordered pairs of $(a,b)$ in this case.


$\textbf{Case 2:}$

$m$ has two solutions, but exactly one of them isn't equal to $2.$ This ensures that $1$ of the solutions is equal to $2.$

Let $r$ be the other value of $m$ that isn't $2.$ By Vieta: \begin{align*} r+2 &= -a-2\\ 2r &= 4+2a+b. \end{align*} From the first equation, we subtract both sides by $2$ and double both sides to get $2r = -2a - 8$ which also equals to $4+2a+b$ from the second equation. Equating both, we have $4a + b + 12 = 0.$ We can easily count that there would be $11$ ordered pairs $(a,b)$ that satisfy that.

However, there's an outlier case in which $r$ happens to also equal to $2,$ and we don't want that. We can reverse engineer and find out that $r=2$ when $(a,b) = (-6, 12),$ which we overcounted. So we subtract by one and we conclude that there are $10$ ordered pairs of $(a,b)$ that satisfy this case.


This all shows that there are a total of $8+10 = 18$ amount of ordered pairs $(a,b).$ Multiplying this by $41$ (the amount of values for $c$) we get $18\cdot41=\boxed{738}$ as our final answer.

~s214425

Solution 2 (factor the difference)

$p(x)-p(2)$ is a cubic with at least two integral real roots, therefore it has three real roots, which are all integers.

There are exactly two distinct roots, so either $p(x)=p(2)+(x-2)^2(x-m)$ or $p(x)=p(2)+(x-2)(x-m)^2$, with $m\neq 2$.

In the first case $p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)$, with $|4+4m|\leq 20$ (which entails $|4+m|\leq 20$), so $m$ can be $-6,-5,-4,-3,-2,-1,0,1, (\textbf{not 2}!), 3,4$ and $-4m+p(2)$ can be any integer from $-20$ to $20$, giving $410$ polynomials (since the coefficients are given by linear functions of $m$ and thus are distinct).

In the second case $p(x)=x^3-(2+2m)x^2+(4m+m^2)x-2m^2+p(2)$, and $m$ can be $-6,-5,-4,-3,-2,-1,0,1$ and $-4m+p(2)$ can be any integer from $-20$ to $20$, giving $328$ polynomials.

The total is $\boxed{738}$.

~EVIN-

Video Solution

https://youtu.be/-Asb_5nTgSg

~MathProblemSolvingSkills.com


See also

2023 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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