2012 AMC 10B Problem 8

P1-10~2 minPrint

Absolute Value & Inequalities Number Properties Organized listing / direct enumeration Exploit symmetry to reduce work or pair up objects

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

Show solution without answering

Key insight

The square must be 4, 9 or 16, so x - 2 is 2, 3 or 4 or their negatives. The six answers pair up around 2.

Solution

Since $x$ is a whole number, $x - 2$ is a whole number too, and $(x-2)^2$ is a perfect square.
List the perfect squares: $0, 1, 4, 9, 16, 25, \ldots$ The ones strictly between $1$ and $25$ are $4$ , $9$ and $16$ . ( $1$ and $25$ themselves are not allowed, because the signs are "less than," not "less than or equal.")
Each square comes from two numbers, one positive and one negative:

$(x-2)^2$	$4$	$9$	$16$
$x - 2$	$2$ or $-2$	$3$ or $-3$	$4$ or $-4$
$x$	$4$ or $0$	$5$ or $-1$	$6$ or $-2$

The six solutions are $-2, -1, 0, 4, 5, 6$ .
Add them: $-2 - 1 + 0 + 4 + 5 + 6 = 12$ .

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

Why this works

The values of $x - 2$ come in pairs like $2$ and $-2$ that cancel when added. So the six values of $x$ add up to six copies of $2$ , which is $12$ , without any listing. Whenever you solve "the square of something equals a number," remember both the positive and the negative root.

Concise solution

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

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 $\boxed{\textbf{(B)}\ 12}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Introduction to Number Theory, Chapter 1: Integers: The Basics
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)