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 $23 + 24 =\boxed{\textbf{(D) }47}$

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~cosinesine

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Video Solution 1 by OmegaLearn

https://youtu.be/aOL04sKGyfU

Video Solution 2

https://youtu.be/Jan9feBsPEg

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

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2023 AMC 10A Problems/Problem 21

Contents

Problem

Solution 1

Video Solution 1 by OmegaLearn

Video Solution 2

See Also