AMC 10 · 2021 · #1
Grade 7 arithmeticPick an answer.
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.
Turn the absolute value into a range
The distance becomes a two-sided range.
Absolute value is distance from 0 — Grade 7 "rational number distance on the number line".
An absolute value is a distance from zero, so a bound on it is a range around zero.
▸ Why?
A number and its opposite are the same distance from zero, so the range reaches equally far both ways.
▸ Why?
Being within that distance is exactly two comparisons at once, one on each side.
Estimate three pi
Three pi sits between nine and ten.
Knowing π ≈ 3.14 is the Grade 7 circle-formula standard — used here just for size.
7.G.B.4Draw A DiagramList the integers
Both ends reach nine.
Ordering integers on the number line — Grade 6 standard.
6.NS.C.7Make A Systematic ListDo not forget zero
Do not drop the zero in the middle.
Add three small whole numbers — Grade 4 standard algorithm.
4.NBT.B.4Make A Systematic ListMatch the choice
The count is 19.
Even/odd parity argument — Grade 4 "generate and analyze patterns".
4.OA.C.5Eliminate PossibilitiesThis AMC 12 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!