AMC 10 · 2023 · #6
Grade 4 arithmeticPick an answer.
AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Counting parities for 2023 terms is hopeless by brute force, so reach for Tool #9 (Easier Problem): compute the first few parities by hand. As soon as the odd/even tags repeat the starting pair (odd, odd), Tool #5 (Pattern) takes over — the cycle must continue forever. Tool #2 (Systematic List) keeps the parity record tidy. Once the period is known, the original 2023 shrinks to a one-line division problem: how many cycle-ending slots fit in 2023. Algebra (Tool #13) is unnecessary; the parity rules already do all the work.
Tag parities from the two given odds using Odd+Odd=Even, Odd+Even=Odd: they run O, O, E, O, O, E, O, O.
Working with a much smaller sample of the sequence makes the structure visible — that is Grade 2 odd/even labeling.
2.OA.C.3Solve An Easier Related ProblemPair (L₄, L₅) = (O, O) repeats (L₁, L₂), and each parity depends only on the prior two — so (O, O, E) recurs with period 3.
Spotting a repeating block of three and arguing it must continue is exactly Grade 4 pattern-rule reasoning.
4.OA.C.5Look For A PatternThe even slot is third in each block, so L_n is even exactly when n is a multiple of 3 — now just count multiples of 3 up to 2023.
Recognizing the even slots are exactly the multiples of 3 is Grade 4 multiples thinking.
4.OA.B.4Make A Systematic ListSince 3 × 674 = 2022 fits but 3 × 675 = 2025 overshoots, there are 674 multiples of 3 up to 2023 — so 674 even terms, choice (E).
One division with remainder gives the count — Grade 4 multi-digit division.
4.NBT.B.6Look For A PatternThis AMC 10 problem only needs Grade 4 pattern and division skills you already know — write down odd or even for the first few terms, spot the (odd, odd, even) block of three, then ask how many of those blocks fit inside 2023.