2024 AMC 10B Problem 3

1. First find the size of $7\pi$. Using $\pi \approx 3.1416$:
$$7\pi \approx 21.99.$$
So $7\pi$ is just under $22$.
2. The condition $|2x| \le 7\pi$ says that $2x$ is at most $21.99$ away from $0$, on either side. So
$$-21.99 \le 2x \le 21.99.$$
3. Divide by $2$: $x$ must be between about $-10.995$ and $10.995$.
4. Which whole numbers fit? $x = 11$ does not, since $2 \cdot 11 = 22$ is more than $21.99$. But $x = 10$ does, since $20 \le 21.99$. The same holds on the negative side.
5. So $x$ can be $-10, -9, \ldots, -1, 0, 1, \ldots, 10$. That is $10$ negative numbers, $10$ positive numbers and $0$:
$$10 + 10 + 1 = 21.$$

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

An absolute value condition like $|u| \le k$ covers everything from $-k$ to $k$, the same distance on both sides of zero. The only real work is deciding exactly where the edge is. Here that comes down to whether $7\pi$ is below or above $22$, and since $\pi$ is a little less than $\frac{22}{7}$, it is below.

If you do not remember whether $\pi$ is bigger or smaller than $\frac{22}{7}$, try both. If $7\pi$ were bigger than $22$, then $x = \pm 11$ would also work, giving $23$ values. But $23$ is not a choice, so $7\pi$ must be under $22$, and the answer is $21$.