2023 AMC 10A Problems/Problem 21

Revision as of 17:57, 9 November 2023 by Aiden22gao (talk | contribs)

Let $P(x)$ be the unique polynomial of minimal degree with the following properties:

  • $P(x)$ has a leading coefficient $1$,
  • $1$ is a root of $P(x)-1$,
  • $2$ is a root of $P(x-2)$,
  • $3$ is a root of $P(3x)$, and
  • $4$ is a root of $4P(x)$.

The roots of $P(x)$ are integers, with one exception. The root that is not an integer can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime integers. What is $m+n$?

Solution 1 From the problem statement, we know P(1)=1, P(2-2)=0, P(9)=0 and 4P(4)=0 therefore we know P(x) must at least have the factors x(x-9)(x-4) and we can assume the last factor to be (x-a) where a is the fractional factor. Then we can use the fact that P(1)=1 to obtain that a 1-a must be 1/24 and a is 23/24. The answer is then 47. ~aiden22gao


Video Solution 1 by OmegaLearn

https://youtu.be/aOL04sKGyfU