AMC 10 · 2024 · #24

Grade 6 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
paritymodular-arithmeticexponentsdivisibility-rules caseworkeasier-related-problemidentify-subproblems ↑ Prerequisites: paritymodular-arithmeticexponentsfraction-arithmetic
📏 Medium solution 💡 3 insights
Problem
Define P(m) = m2\frac{m}{2} + m24\frac{m²}{4} + m48\frac{m⁴}{8} + m88\frac{m⁸}{8}. Of the four values P(2022), P(2023), P(2024), P(2025), how many are integers?

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
3
(E)
4

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The four inputs 2022, 2023, 2024, 2025 look special but the denominators 2, 4, 8, 8 only care about m mod 8. Tool #7 (Identify Subproblems) splits the question by parity: two even cases (2022, 2024) and two odd cases (2023, 2025). Tool #9 (Easier Related Problem) replaces the giant inputs with small representatives — m = 2 for the even case, m = 1 for the odd case — and computes P exactly there. Tool #5 (Look for a Pattern) shows the integer/non-integer status is the same for every even (resp. odd) input, so the small representatives generalize. Tool #6 (Guess and Check) provides a direct check by computing P(1) and P(2) in seconds.

1STEP 1

Split the four inputs by parity: 2022, 2024 are even and 2023, 2025 are odd, so whether P(m) is an integer depends only on parity.

even cases: 2022, 2024 ; odd cases: 2023, 2025
2STEP 2

Even m = 2k: every term becomes a whole number, and the smallest case gives P(2) = 36. So P(2022), P(2024) are integers.

P(2k) = k + k² + 2k⁴ + 32 k⁸ ∈ Z
3STEP 3

Odd m = 1: P(1) = 12\frac{1}{2} + 14\frac{1}{4} + 18\frac{1}{8} + 18\frac{1}{8} = 88\frac{8}{8} = 1 — the four fractional pieces sum to exactly one whole.

P(1) = 12\frac{1}{2} + 14\frac{1}{4} + 18\frac{1}{8} + 18\frac{1}{8} = 88\frac{8}{8} = 1
4STEP 4

Any odd m also gives m² ≡ m⁴ ≡ m⁸ ≡ 1 (mod 8), so its fractional pieces match m = 1 and again sum to 1 → P(2023), P(2025) are integers.

m odd → m ≡ 1 ( mod 2), m² ≡ m⁴ ≡ m⁸ ≡ 1 ( mod 8)
5STEP 5

Every input is even or odd, and both parities yield integers — so all four P-values are integers → (E).

#{m : P(m) ∈ Z} ∩ {2022, 2023, 2024, 2025} = 4 → (E)
Answer
4
Two free spot-checks. P(2) = 36 (Step 2) and P(1) = 1 (Step 3) are both integers, confirming the parity argument on the simplest representatives. For one more check, P(3) = 32\frac{3}{2} + 94\frac{9}{4} + 818\frac{81}{8} + 65618\frac{6561}{8} = 12+18+81+65618\frac{12 + 18 + 81 + 6561}{8} = 66728\frac{6672}{8} = 834 — integer, agreeing with the odd-case claim. The answer 4 is the upper end of the choice list, which feels right because the structure of P was clearly built so that the fractional pieces 12\frac{1}{2} + 14\frac{1}{4} + 18\frac{1}{8} + 18\frac{1}{8} = 1 tile to one whole — the problem-setter chose denominators 2, 4, 8, 8 deliberately for this.
💡Key takeaway

Big inputs like 2024 scare you for nothing: the denominators only care about parity. Test the smallest even m = 2 and the smallest odd m = 1 — both give whole-number P(m), so every integer m does, and the answer is all four.

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