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2012 AMC 10B Problem 8

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2012 AMC 10B Problem 8AMC 10 Step by Step

Name: Date: September 13, 2026

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

What is the sum of all integer solutions to $1<(x-2)^2<25$ ?

A)
$10$
B)
$12$
C)
$15$
D)
$19$
E)
$25$

Answer

Key insight

(x-2)^2 must be 4, 9, or 16, so x - 2 is ±2, ±3, ±4; the six values are symmetric about 2 and sum to 12.

Solution

Since $x$ is an integer, $(x-2)^2$ is a perfect square strictly between $1$ and $25$ , so it is $4$ , $9$ , or $16$ . Then
$x - 2 \in \{-4, -3, -2, 2, 3, 4\}.$
Adding $2$ to each gives $x \in \{-2, -1, 0, 4, 5, 6\}$ .

The six values of $x - 2$ come in opposite pairs and cancel, so the six values of $x$ sum to $6 \cdot 2 = 12$ . Directly: $-2 - 1 + 0 + 4 + 5 + 6 = 12$ .

The answer is $\boxed{\textbf{(B)}\ 12}$ .

Why this works

An inequality on a squared expression is really a statement about $|x - 2|$ : here $1 < |x - 2| < 5$ , i.e. $|x-2| \in \{2, 3, 4\}$ . Solutions of $|x - c| = k$ always pair up symmetrically around $c$ , so their sum is (number of solutions) $\times\ c$ without any listing.

The trap

Forgetting the negative square roots (x - 2 = -2, -3, -4), or including x - 2 = ±1 or ±5, where the strict inequalities fail.

Common mistakes

Forgetting the negative square roots (x - 2 = -2, -3, -4), or including x - 2 = ±1 or ±5, where the strict inequalities fail.
Summing the values of $x - 2$ (which give $0$ ) or of $(x-2)^2$ instead of the values of $x$ .

Techniques

Organized listing / direct enumeration · Exploit symmetry to reduce work or pair up objects