2016 AMC 10B Problem 15

1. Color the $3 \times 3$ grid like a checkerboard, with the corners black. Then the $4$ corners and the center are black ($5$ squares), and the $4$ middle-of-edge squares are white.
2. Two squares that share an edge always have different colors.
3. The numbers $1, 2, 3, \ldots, 9$ walk from square to neighboring square, so the colors take turns: if $1$ is on one color, $2$ is on the other, $3$ is back on the first, and so on. So all the odd numbers share one color, and all the even numbers share the other.
4. There are $5$ odd numbers ($1, 3, 5, 7, 9$) and $4$ even numbers. Only the black color has $5$ squares, so the odd numbers fill the black squares: the four corners and the center.
5. All five odd numbers add to $1 + 3 + 5 + 7 + 9 = 25$. The corners hold four of them, adding to $18$. The center holds the one left over:
$$25 - 18 = 7.$$
6. This really can happen, for example:
$$\begin{array}{ccc} 1 & 2 & 3 \\ 8 & 7 & 4 \\ 9 & 6 & 5 \end{array}$$
Here the corners add to $1 + 3 + 9 + 5 = 18$ and the center is $7$.

The answer is $\boxed{\textbf{(C)}\ 7}$.

A path that moves one square at a time on a checkerboard switches color at every step, so odd steps land on one color and even steps on the other. When one color has more squares than the other, the larger group of numbers is forced onto it. Remember this coloring trick for any grid problem where consecutive numbers must be neighbors.

Color the $3 \times 3$ board like a checkerboard with the corners black. Then the four corners and the center are black ($5$ squares) and the four edge-middle squares are white ($4$ squares). Squares sharing an edge always have opposite colors.

Consecutive numbers sit on edge-adjacent squares, so as we go $1, 2, 3, \ldots, 9$ the colors alternate. Hence all the odd numbers are on one color and all the even numbers on the other. There are five odd numbers ($1, 3, 5, 7, 9$) and only four even ones, so the odd numbers must occupy the five black squares: the four corners and the center.

The odd numbers add to $1 + 3 + 5 + 7 + 9 = 25$. The corners hold four of them with sum $18$, so the center holds $25 - 18 = 7$.

Such a filling exists, for example
$$\begin{array}{ccc} 1 & 2 & 3 \\ 8 & 7 & 4 \\ 9 & 6 & 5 \end{array}$$
with corners $1 + 3 + 9 + 5 = 18$.

The answer is $\boxed{\textbf{(C)}\ 7}$.