AMC 10 · 2024 · #3

Grade 8 arithmetic
interval-arithmeticestimationsystematic-enumeration convert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: interval-arithmeticorder-of-operationsestimation
📏 Short solution 💡 2 insights
Problem
Count how many integers x make the inequality |2x| ≤ 7π true.

Pick an answer.

(A)
16
(B)
17
(C)
19
(D)
20
(E)
21
How to solve
Strategy Convert to Algebra

The givens are an algebraic absolute-value inequality plus an irrational bound — Tool #13 (Convert to Algebra) is the natural lead. Translate |2x| ≤ 7π into -7π ≤ 2x ≤ 7π, then divide by 2 to get -7π/2 ≤ x ≤ 7π/2. Once we have a clean numerical interval, Tool #2 (Make a Systematic List) finishes the job: 7π/2 ≈ 10.996, so the integers -10, -9, …, 10 are exactly the ones inside the band — list them in order and count.

1STEP 1

Unfold the absolute value

It unfolds into a two-sided bound.

|2x| ≤ 7π ⟺ -7π ≤ 2x ≤ 7π
2STEP 2

Divide by two

The bound becomes seven pi over two.

-7π/2 ≤ x ≤ 7π/2
3STEP 3

Estimate as a decimal

About 10.996, just short of eleven.

7π/2 ≈ 21.991/2 ≈ 10.996
4STEP 4

List the integers

Everything from negative ten up to ten qualifies.

x ∈ {-10, -9, -8, …, -1, 0, 1, …, 9, 10}
5STEP 5

Count them

That is 21 integers.

10 - (-10) + 1 = 21 → (E)
Answer
21
Boundary check: does x = 11 really fail? |2 · 11| = 22 and 7π ≈ 21.99 < 22, so yes 11 fails. Does x = 10 really pass? |2 · 10| = 20 and 20 < 21.99, so yes 10 passes. The interval [-10, 10] has 21 integers, matching (E).
💡Key takeaway

This AMC 12 problem only needs Grade 8 rational approximations of π (plus Grade 6 absolute value) — once you know 7π/2 ≈ 10.996, you just count -10 through 10 on the number line!