AMC 10 · 2021 · #1
Grade 7 arithmeticPick an answer.
AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #1 (Diagram) is perfect — sketch a number line with -3π and 3π as endpoints, then mark integer tick marks inside. Tool #2 (Systematic List) handles the counting once we know where the boundary integers are. Tool #3 (Eliminate) checks the answer choices: counting symmetric integers around 0 always gives an odd total (positives + negatives + zero), so (C) 18 and (E) 20 can be eliminated immediately, leaving (A) 9, (B) 10, or (D) 19.
Absolute value is distance from 0, so |x| < 3π becomes the range -3π < x < 3π.
Absolute value is distance from 0 — Grade 7 "rational number distance on the number line".
7.NS.A.1Draw A DiagramSince π ≈ 3.14, estimate 3π ≈ 9.42, so we need integers strictly between -9.42 and 9.42.
Knowing π ≈ 3.14 is the Grade 7 circle-formula standard — used here just for size.
7.G.B.4Draw A DiagramThe integers strictly inside run from -9 up to 9, since 10 > 9.42 falls outside.
Ordering integers on the number line — Grade 6 standard.
6.NS.C.7Make A Systematic ListCount them: 9 negatives + 9 positives + zero gives 9 + 9 + 1 = 19, choice (D).
Add three small whole numbers — Grade 4 standard algorithm.
4.NBT.B.4Make A Systematic ListA symmetric range around 0 forces an odd count, ruling out (C) 18 and (E) 20 — confirming (D).
Even/odd parity argument — Grade 4 "generate and analyze patterns".
4.OA.C.5Eliminate PossibilitiesThis AMC 10 problem only needs Grade 7 "absolute value means distance from zero" and knowing π ≈ 3.14 — sketch the number line from -9.42 to 9.42, list the integers inside, and count 19!