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2021 AMC 10B Problem 1

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2021 AMC 10B Problem 1AMC 10 Step by Step

Name: Date: September 13, 2026

Absolute Value & Inequalities · difficulty 1 of 5 · about 1 min · position P1-10

How many integer values of $x$ satisfy $|x| < 3\pi$ ?

A)
$9$
B)
$10$
C)
$18$
D)
$19$
E)
$20$

Answer

Key insight

Since 3π ≈ 9.42, the condition is −9 ≤ x ≤ 9, which contains 19 integers including zero.

Solution

Since $\pi \approx 3.14$ , we have $3\pi \approx 9.42$ . The inequality $|x| < 9.42$ means $-9.42 < x < 9.42$ .

The integers in this range are $-9, -8, \dots, -1, 0, 1, \dots, 8, 9$ : nine negative integers, nine positive integers, and zero. That is $9 + 9 + 1 = 19$ integers.

The answer is $\boxed{\textbf{(D)}\ 19}$ .

Why this works

$|x| < c$ is the same as $-c < x < c$ , a symmetric interval around $0$ . Counting integers in a symmetric interval is "twice the positive count plus one for zero." The only care needed is locating $3\pi$ between consecutive integers ( $9 < 3\pi < 10$ ), which decides whether $9$ itself is included.

The trap

Forgetting x = 0 and answering 18, or counting only the positive side and answering 9.

Common mistakes

Forgetting x = 0 and answering 18, or counting only the positive side and answering 9.
Rounding $3\pi$ up to $10$ and including $\pm 10$ , which gives $21$ , or including $x = 10$ on one side only.

Techniques

Set up the equation/formula and compute; no special trick needed