AMC 10 · 2023 · #3
Grade 6 number-theoryPick an answer.
AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
After noticing that 5 ∣ k² forces 5 ∣ k (because 5 is prime), the qualifying squares are exactly 5², 10², 15², … — a perfectly regular arithmetic-style pattern in k. Tool #5 (Look for a Pattern) frames it: write k = 5m and the constraint becomes 25 m² ＜ 2023. Tool #2 (Systematic List) then finishes by listing m = 1, 2, 3, … in order and stopping at the first m that breaks the bound. Algebra (#13) would be heavier and a brute list of all 44 squares under 2023 is wasteful; the pattern + list combo is the natural fit.
A square's prime exponents are all even, so if 5 divides k² it must divide k too — k has to be a multiple of 5.
Reading a square through its prime factors shows that any prime dividing the square already had to divide the base — the Grade 6 "whole-number exponent" idea.
6.EE.A.1Look For A PatternSet k = 5m, so N = 25m²; the bound N ＜ 2023 collapses to m² ＜ = 80.92.
Renaming k as 5m turns the pattern "k is a multiple of 5" into a plain Grade 6 expression in m, with no divisibility left to track.
6.EE.B.6Look For A PatternList the squares 1², 2², 3², … in order and stop the moment one clears the bound 80.92.
Knowing 1² through 9² by heart is the Grade 3 multiplication-fluency standard — the list assembles itself.
3.OA.C.7Make A Systematic List8² = 64 stays under the bound but 9² = 81 clears it, so m runs 1 to 8 — eight squares, choice (A).
Each m gives one allowed k = 5m and hence one perfect square 25 m², so counting valid m is the same as counting valid N — a Grade 4 factor-pair / multiples-style count.
4.OA.B.4Make A Systematic ListThis AMC 10 problem only needs Grade 6 "5 inside a square means 5 inside its root" reasoning — the squares of 5, 10, …, 40 are exactly the eight that fit.