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2023 AMC 10A Problems/Problem 21

Revision as of 19:43, 9 November 2023 by Walmartbrian (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 that $0$, $9$, and $4$ are roots. Because of this, we can factor $P(x)$ as $x(x - 9)(x - 4)(x - a)$, where $a$ is the unknown root. Plugging in $x = 1$ gives $1(-8)(-3)(1 - a) = 1$, so $24(1 - a) = 1/24 \implies 1 - a = 24 \implies a = 23/24$. Therefore, our answer is $23 + 24 =$ $47$, or $C$

~aiden22gao

~cosinesine

~walmartbrian


Video Solution 1 by OmegaLearn

https://youtu.be/aOL04sKGyfU

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