Question

Let

$\begin{aligned} P(m) &= \frac{m}{2} + \frac{m^2}{4} \\ &\quad {}+ \frac{m^4}{8} + \frac{m^8}{8}. \end{aligned}$

How many of the values of $P(2022),$ $P(2023),$ $P(2024),$ and $P(2025)$ are integers?

$0$

$1$

$2$

$3$

$4$

Answer: E

Concepts:modular arithmetic parity casework

Difficulty rating: 2380

Solution:

Put everything over $8:$ $P(m) = \dfrac{m^8 + m^4 + 2m^2 + 4m}{8}.$ If $m$ is even, every term up top is divisible by $8.$ If $m$ is odd, then $m^2 \equiv m^4 \equiv m^8 \equiv 1$ and $4m \equiv 4 \pmod 8,$ so the numerator is $1 + 1 + 2 + 4 = 8 \equiv 0 \pmod 8.$ Either way $P(m)$ is an integer, so all $4$ values are integers. Therefore, the answer is E.

2024 AMC 10B Problem 24

Answer: E

Difficulty rating: 2380

Solution:

Problem 24 in Other Years