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2024 AMC 10B Problem 3

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2024 AMC 10B Problem 3AMC 10 Step by Step

Name: Date: September 13, 2026

Absolute Value & Inequalities · difficulty 1 of 5 · about 2 min · position P1-10

For how many integer values of $x$ is $|2x| \leq 7 \pi$

A)
$16$
B)
$17$
C)
$19$
D)
$20$
E)
$21$

Answer

Key insight

7 pi is just under 22, so |x| <= 10.99, giving the 21 integers from -10 to 10.

Solution

The inequality is $|x| \le \frac{7\pi}{2}$ . Since $\pi \approx 3.1416$ , we have $7\pi \approx 21.99$ , so $\frac{7\pi}{2} \approx 10.995$ .

Therefore $-10.995 \le x \le 10.995$ , and the integers in that range are $-10, -9, \dots, 9, 10$ . That is $10$ negative values, $10$ positive values and $0$ , for a total of $21$ .

The answer is $\boxed{\textbf{(E)}\ 21}$ .

Why this works

An absolute value inequality is a symmetric interval around zero; the only real work is locating the endpoint. Knowing $\pi < \frac{22}{7}$ (equivalently $7\pi < 22$ ) settles whether $11$ is included. If you were unsure, note that $7\pi > 22$ would give $23$ integers, which is not a choice.

The trap

Forgetting the negative integers (answering 11) or forgetting zero (answering 20).

Common mistakes

Forgetting the negative integers (answering 11) or forgetting zero (answering 20).
Rounding $7\pi$ up to $22$ and including $x = \pm 11$ , giving $23$ , which is not even offered.

Techniques

Bound the quantity above/below or estimate to pin it down · Set up the equation/formula and compute; no special trick needed