AMC 10 Step by Step

2021 AMC 10B Problem 1

P1-10~1 minPrint

How many integer values of xx satisfy ∣x∣<3π|x| < 3\pi ?

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Key insight

3π is about 9.42, so x can be any whole number from −9 up to 9. That is 9 negatives, 9 positives, and zero.

Solution

1. First find out how big 3π3\pi is. Since π≈3.14\pi \approx 3.14, we get 3π≈9.423\pi \approx 9.42.
2. ∣x∣|x| is the distance from xx to 00 on the number line. So ∣x∣<9.42|x| < 9.42 means xx is less than 9.429.42 steps away from 00, on either side.
3. Check the edge: ∣9∣=9|9| = 9 is less than 9.429.42, so 99 works. But ∣10∣=10|10| = 10 is too big. The same holds for −9-9 and −10-10.
4. So the whole numbers that work are
−9,−8,…,−1,0,1,2,…,9.-9, -8, \ldots, -1, \quad 0, \quad 1, 2, \ldots, 9.
5. Count them: 99 negative numbers, 99 positive numbers, and 00. That is 9+9+1=199 + 9 + 1 = 19.

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

Why this works

"∣x∣|x| is less than some number" always gives a stretch of the number line that is the same on both sides of 00. So count the positive numbers, double that, and add one for 00. The only real work is deciding which two whole numbers 3π3\pi sits between, because that tells you where the list stops.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Since π≈3.14\pi \approx 3.14, we have 3π≈9.423\pi \approx 9.42. The inequality ∣x∣<9.42|x| < 9.42 means −9.42<x<9.42-9.42 < x < 9.42.

The integers in this range are −9,−8,…,−1,0,1,…,8,9-9, -8, \dots, -1, 0, 1, \dots, 8, 9: nine negative integers, nine positive integers, and zero. That is 9+9+1=199 + 9 + 1 = 19 integers.

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

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.