2012 AMC 10B Problem 8

1. Since $x$ is a whole number, $x - 2$ is a whole number too, and $(x-2)^2$ is a perfect square.
2. 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.")
3. Each square comes from two numbers, one positive and one negative:

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

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

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.

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