2024 AMC 10B Problem 24

Problem 24 of 25HarderNumber Theory

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?

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Solution

Put everything over

8:

P(m) = \dfrac{m^8 + m^4 + 2m^2 + 4m}{8}.

m

is even, every term up top is divisible by

8.

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.