AMC 10 · 2021 · #1

Grade 7 arithmetic
absolute-valuesystematic-enumerationestimationinterval-arithmetic systematic-enumerationbound-inequality-then-enumerate ↑ Prerequisites: absolute-value
📏 Short solution 💡 2 insights
Problem
Count the integers x whose distance from 0 on the number line is less than 3π.

Pick an answer.

(A)
~9
(B)
~10
(C)
~18
(D)
~19
(E)
~20

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

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.

1STEP 1

Absolute value is distance from 0, so |x| < 3π becomes the range -3π < x < 3π.

|x| < 3π ⟺ -3π < x < 3π
2STEP 2

Since π ≈ 3.14, estimate 3π ≈ 9.42, so we need integers strictly between -9.42 and 9.42.

3π ≈ 3 × 3.14 = 9.42
3STEP 3

The integers strictly inside run from -9 up to 9, since 10 > 9.42 falls outside.

{-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
4STEP 4

Count them: 9 negatives + 9 positives + zero gives 9 + 9 + 1 = 19, choice (D).

9 + 9 + 1 = 19 → (D)
5STEP 5

A symmetric range around 0 forces an odd count, ruling out (C) 18 and (E) 20 — confirming (D).

odd count → (D)
Answer
~19
3π ≈ 9.42, so the range (-9.42, 9.42) definitely includes -9 through 9 and excludes -10, 10. The count of 19 is between (C) 18 and (E) 20, exactly the odd value that symmetry around 0 forces.
💡Key takeaway

This 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!