2016 AMC 10B Problem 15

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 $\boxed{\textbf{(C)}\ 7}$.

A path that moves one edge at a time alternates colors on a checkerboard, so parity of the step number matches the color. When one color class has more squares than the other, the larger class is forced to hold the larger parity class. This coloring invariant answers the question without constructing anything, and it is the standard tool for "consecutive numbers must be adjacent" grid problems.

• Trying to build an explicit arrangement and hunting through possibilities, when the coloring argument settles the center without any construction.
• Assuming the corners hold the even numbers, or forgetting that the center shares the corners' color and so must also hold an odd number.