2024 AMC 10B Problem 24

Put everything over $8$:
$$P(m) = \frac{4m + 2m^2 + m^4 + m^8}{8},$$
so $P(m)$ is an integer exactly when $8 \mid 4m + 2m^2 + m^4 + m^8$. Only the parity of $m$ matters.

$m$ even. Write $m = 2k$: $4m = 8k$, $2m^2 = 8k^2$, $m^4 = 16k^4$, $m^8 = 256k^8$. Every term is a multiple of $8$, so $P(m)$ is an integer. This covers $2022$ and $2024$.

$m$ odd. The square of any odd number is $1 \bmod 8$ (since $(2j+1)^2 = 4j(j+1) + 1$ and $j(j+1)$ is even). Hence $m^2 \equiv m^4 \equiv m^8 \equiv 1 \pmod 8$, and
$$4m + 2m^2 + m^4 + m^8 \equiv 4m + 2 + 1 + 1 = 4(m + 1) \pmod 8.$$
Since $m + 1$ is even, $4(m+1)$ is a multiple of $8$, so $P(m)$ is an integer. This covers $2023$ and $2025$.

All four values are integers.

The answer is $\boxed{\textbf{(E)}\ 4}$.

A sum of fractions with denominators $2, 4, 8$ should be tested as a single fraction over $8$; the question becomes a congruence mod $8$, where the fact "odd squares are $1 \bmod 8$" makes high powers trivial. The specific years are a distraction: $P(m)$ is an integer for every integer $m$.

Track fractional parts for odd $m$: $\frac{m}{2}$ contributes $\frac12$, $\frac{m^2}{4}$ contributes $\frac14$ (odd squares are $1 \bmod 4$), and $\frac{m^4}{8}$, $\frac{m^8}{8}$ each contribute $\frac18$. The fractional parts total $\frac12 + \frac14 + \frac18 + \frac18 = 1$, an integer.

• Testing only whether each fraction is an integer separately; for odd m none of them is, yet the fractional parts 1/2 + 1/4 + 1/8 + 1/8 add up to exactly 1.
• Reducing $2023 \equiv 7$ and computing $7^8 \bmod 8$ by hand instead of using $7^2 \equiv 1$, which invites arithmetic errors.