2023 AMC 10A Problem 21

Unpack each condition as a statement about values of $P$:

• $1$ is a root of $P(x) - 1$ means $P(1) - 1 = 0$, so $P(1) = 1$.
• $2$ is a root of $P(x - 2)$ means $P(2 - 2) = P(0) = 0$.
• $3$ is a root of $P(3x)$ means $P(9) = 0$.
• $4$ is a root of $4P(x)$ means $4P(4) = 0$, so $P(4) = 0$.

Thus $P$ is monic with roots $0, 4, 9$. The smallest candidate is the cubic $x(x-4)(x-9)$, but it gives $P(1) = 1 \cdot (-3)(-8) = 24 \ne 1$. So the degree must be at least $4$, and with one more root $r$,
$$P(x) = x(x-4)(x-9)(x-r), \qquad P(1) = 24(1 - r) = 1.$$
Hence $1 - r = \frac{1}{24}$ and $r = \frac{23}{24}$. This degree-$4$ polynomial is unique, and its only non-integer root is $\frac{23}{24}$, so $m + n = 23 + 24 = 47$.

The answer is $\boxed{\textbf{(D)}\ 47}$.

"$a$ is a root of $P(\text{expression in } x)$" always means "$P$ vanishes at the expression evaluated at $a$," so the clues are just function values in disguise. A monic polynomial with prescribed roots is a product of linear factors, and one extra value condition ($P(1) = 1$) fixes exactly one extra unknown. Reading each condition literally, in order, is the whole problem.

• Reading '2 is a root of P(x-2)' as P(2) = 0 or P(-2) = 0 instead of P(0) = 0, and similarly getting P(3) = 0 instead of P(9) = 0.
• Trying to satisfy $P(1) = 1$ with the cubic by changing the leading coefficient, which violates the monic requirement.