2024 AMC 10B Problem 24

Below is the professionally curated solution for Problem 24 of the 2024 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 10B solutions, or check the answer key.

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Concepts:modular arithmeticparitycasework

Difficulty rating: 2380

24.

Let

P(m)=m2+m24+m48+m88.P(m) = \frac{m}{2} + \frac{m^2}{4} + \frac{m^4}{8} + \frac{m^8}{8}.

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

00

11

22

33

44

Solution:

Put everything over 8:8: P(m)=m8+m4+2m2+4m8.P(m) = \dfrac{m^8 + m^4 + 2m^2 + 4m}{8}. If mm is even, every term up top is divisible by 8.8. If mm is odd, then m2m4m81m^2 \equiv m^4 \equiv m^8 \equiv 1 and 4m4(mod8),4m \equiv 4 \pmod 8, so the numerator is 1+1+2+4=80(mod8).1 + 1 + 2 + 4 = 8 \equiv 0 \pmod 8. Either way P(m)P(m) is an integer, so all 44 values are integers. Therefore, the answer is E.

Problem 24 in Other Years