# 2023 AMC 10A Problems/Problem 21

## Problem

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 $m+n$?

$\textbf{(A) }41\qquad\textbf{(B) }43\qquad\textbf{(C) }45\qquad\textbf{(D) }47\qquad\textbf{(E) }49$

## Solution 1

From the problem statement, we know $P(2-2)=0$, $P(9)=0$ and $4P(4)=0$. Therefore, we know that $0$, $9$, and $4$ are roots. So, we can factor $P(x)$ as $x(x - 9)(x - 4)(x - a)$, where $a$ is the unknown root. Since $P(x) - 1 = 0$, we plug in $x = 1$ which gives $1(-8)(-3)(1 - a) = 1$, therefore $24(1 - a) = 1 \implies 1 - a = 1/24 \implies a = 23/24$. Therefore, our answer is $23 + 24 =\boxed{\textbf{(D) }47}$

~aiden22gao

~cosinesine

~walmartbrian

~sravya_m18

~ESAOPS

## Solution 2

We proceed similarly to solution one. We get that $x(x-9)(x-4)(x-a)=1$. Expanding, we get that $x(x-9)(x-4)(x-a)=x^4-(a+13)x^3+(13a+36)x^2-36ax$. We know that $P(1)=1$, so the sum of the coefficients of the cubic expression is equal to one. Thus $1+(a+13)+(13a+36)-36a=1$. Solving for a, we get that a=23/24. Therefore, our answer is $23 + 24 =\boxed{\textbf{(D) }47}$

~Aopsthedude

## Video Solution 2

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)