AMC 10 · 2024 · #5
Grade 6 number-theoryPick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The condition 2024 ∣ n! looks intimidating until we split it. Tool #7 (Identify Subproblems) says: factor 2024 into primes, then check each prime separately — n! is a multiple of 2024 exactly when n! contains every prime power in that factorization. The largest prime in the factorization sets a hard floor on n. Tool #3 (Eliminate Possibilities) then walks the answer choices: any candidate smaller than that floor cannot work, so they drop out immediately and the smallest candidate that clears the floor is the answer.
Subproblem 1 — factor 2024 by peeling off the 2s, leaving 2³ × 11 × 23 with 11 and 23 both prime.
Recognizing 253 = 11 × 23 is the Grade 4 "find factor pairs" move; both 11 and 23 are prime.
4.OA.B.4Identify SubproblemsSubproblem 2 — n! holds every prime ≤ n, so needing both 11 and 23 forces n ≥ 23 (the bigger prime wins).
Asking when a factorial contains a given prime is the same idea as asking for the LCM-style "smallest container" — a Grade 6 GCF/LCM mindset.
6.NS.B.4Identify SubproblemsSubproblem 3 — just 2 · 4 · 6 · 8 already contributes 2⁷, far above the 2³ that 2024 needs, so the power of 2 never binds.
Adding the exponents of like bases is the Grade 6 exponent rule 2^a · 2^b = 2^a+b — and 7 ＞ 3, so the 2-part is comfortable.
6.EE.A.1Identify SubproblemsChoices under 23 miss the prime 23 and 253 is not least; only 23 clears every requirement, so the answer is 23 → (D).
Once the largest prime sets the floor, eliminating every smaller choice is a one-line check — the classic "largest prime is the bottleneck" move for least-multiple problems.
6.NS.B.4Eliminate PossibilitiesSplitting 2024 = 2³ × 11 × 23 turns this AMC 10 problem into a Grade 6 question — once you spot the biggest prime 23, the answer has nowhere left to hide.