2023 AMC 10A Problem 21

Each clue looks strange, but each one just says what $P$ equals at some number. To decode a clue, plug the given number in for $x$.

1. "$1$ is a root of $P(x) - 1$." Plug in $x = 1$: $P(1) - 1 = 0$, so $P(1) = 1$.
2. "$2$ is a root of $P(x - 2)$." Plug in $x = 2$: $P(2 - 2) = 0$, so $P(0) = 0$.
3. "$3$ is a root of $P(3x)$." Plug in $x = 3$: $P(3 \times 3) = 0$, so $P(9) = 0$.
4. "$4$ is a root of $4P(x)$." Plug in $x = 4$: $4P(4) = 0$, so $P(4) = 0$.
5. Build $P$. So $P$ has roots $0$, $4$ and $9$, and its leading coefficient is $1$. The simplest try is
$$P(x) = x(x - 4)(x - 9).$$
But then $P(1) = 1 \times (-3) \times (-8) = 24$, not $1$. So three factors are not enough.
6. Add one more root. Try degree $4$ with an unknown fourth root $r$:
$$P(x) = x(x - 4)(x - 9)(x - r).$$
Now $P(1) = 24 \times (1 - r)$, and this must equal $1$:
$$24(1 - r) = 1 \quad\Longrightarrow\quad 1 - r = \frac{1}{24} \quad\Longrightarrow\quad r = \frac{23}{24}.$$
7. The roots are $0, 4, 9$ and $\frac{23}{24}$. The one that is not a whole number is $\frac{23}{24}$, so $m + n = 23 + 24 = 47$.

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

"$a$ is a root of $P(\text{something})$" means: put $a$ into the something, and $P$ of the result is $0$. Once every clue is read this way, you have three roots and one extra value. A polynomial with leading coefficient $1$ is a product of $(x - \text{root})$ factors, and the one extra value fixes the one unknown root. The only danger is misreading a clue, so decode them slowly, one at a time.