Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "Problem (Unofficial, please update when official one comes out): <math>P(x) = x^3 + ax^2 + bx + c</math> is a polynomial with integer coefficients between <math>-20</math> an...")
 
Line 28: Line 28:
  
 
~mathboy100
 
~mathboy100
 +
 +
I believe this solution is wrong. The answer is <math>738</math>. ~r00tsOfUnity

Revision as of 12:39, 8 February 2023

Problem (Unofficial, please update when official one comes out):

$P(x) = x^3 + ax^2 + bx + c$ is a polynomial with integer coefficients between $-20$ and $20$, inclusive. There is exactly one integer $m$ such that $P(m) = P(2)$. How many possible values are there for the ordered triple $(a, b, c)$?

Solution

If $m$ is the only integral value that satisfies $P(m) = P(2)$, we can show that $m$ is the only real value that satisfies $P(m) = P(2)$.

Next, we have $P(m) = P(2)$, so therefore

\[m^3 + am^2 + bm + c - 8 - 4a - 2b - c = 0.\]

We can now simplify:

\[m^3 - 8 + am^2 - 4a + bm - 2b = 0\] \[(m-2)(m^2 + 2m + 4) + a(m-2)(m+2) + b(m-2) = 0\] \[(m-2)(m^2 + 2m + 4 + am + 2a + b) = 0.\]

Since $m \neq 2$,

\[m^2 + (a+2)m + 2a + b + 4 = 0.\]

We can now apply the quadratic formula, yielding

\[m = \frac{-a-2 \pm \sqrt{a^2 - 4a - 4b - 12}}{2}.\]

For this to have exactly $1$ solution, we must have $a^2 - 4a - 4b - 12 = 0$, and thus $(a-2)^2 = 4(b+2)$. This means that $|a| \leq 9$, yielding $9$ solutions for $a$. For any solution of $a$, $b$ can only attain one value. And, the value of $c$ doesn't matter. Our answer is thus $9 \cdot 41 = \boxed{369}$.

~mathboy100

I believe this solution is wrong. The answer is $738$. ~r00tsOfUnity

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AIME_I_Problems/Problem_9&oldid=188877"