Question

There is a unique polynomial $P(x)$ of least degree with leading coefficient $1$ satisfying all of the following:

$1$ 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 $4P(x).$

All the roots of $P(x)$ except one are integers. If the one non-integer root can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, what is $m + n?$

$41$

$43$

$45$

$47$

$49$

Answer: D

Concepts:polynomial substitution

Difficulty rating: 2120

Solution:

Translate each condition into a value: $P(1) = 1,$ $P(0) = 0,$ $P(9) = 0,$ and $P(4) = 0.$ So $0, 4, 9$ are roots. Could a cubic do it? A monic cubic with those roots has $P(1) = (1)(-3)(-8) = 24 \ne 1,$ so no. The least-degree monic polynomial is degree $4:$ $P(x) = x(x - 4)(x - 9)(x - c).$ Now $P(1) = (1)(-3)(-8)(1 - c)$ $= 24(1 - c)$ $= 1,$ so $1 - c = \frac{1}{24}$ and $c = \frac{23}{24}.$ That's the lone non-integer root, so $m + n = 23 + 24 = 47.$ Thus, D is the correct answer.

2023 AMC 10A Problem 21

Answer: D

Difficulty rating: 2120

Solution:

Problem 21 in Other Years