AMC 8 · 2014 · #6
Grade 4 arithmeticPick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Each rectangle's area is 2 × length, so the total area is 2 × 1 + 2 × 4 + … + 2 × 36. The shared factor 2 is a pattern (Tool #4) — every term has it — so we can factor it out and add the lengths just once instead of six times: total = 2 × (1 + 4 + 9 + 16 + 25 + 36). Tool #7 (Identify Subproblems) then splits the work into two clean pieces — first add the six numbers, then multiply by 2.
Write the total as a sum of six rectangle areas, each equal to width × length = 2 × its length.
Area = width × length for a rectangle is the Grade 3 area standard.
3.MD.C.7Identify SubproblemsEvery term shares a factor 2 — pull it out by the distributive property, so the lengths get added just once.
Factoring out a common factor is exactly the distributive property of multiplication over addition.
3.OA.B.5Use Matrix LogicAdd the six squares by easy pairs: (1 + 9) + (4 + 16) + (25 + 36) = 10 + 20 + 61 = 91.
Pairing numbers that add to round values (like 10 and 20) is a standard mental-math regrouping move.
4.NBT.B.4Identify SubproblemsMultiply the sum by the common width: 2 × 91 = 182 → (D).
A one-digit by two-digit multiplication is Grade 4 multi-digit arithmetic.
4.NBT.B.5Identify SubproblemsThis AMC 8 problem only needs Grade 4 arithmetic — once you factor out the shared width, it's just one addition and one multiplication!