2020 AMC 10A Problem 7

You do not need to build the square. Just use the rows.

1. The $5$ rows share out all $25$ numbers, each number landing in exactly one row. So
$$\text{(row sum)} + \text{(row sum)} + \cdots \ (5 \text{ times}) = \text{total of all 25 numbers}.$$
2. Find that total. Pair each negative number with its positive match: $-10$ with $10$, $-9$ with $9$, down to $-1$ with $1$. Each pair adds to $0$, and $0$ adds nothing.
3. What is left over is $11 + 12 + 13 + 14 = 50$.
4. So $5 \times \text{(row sum)} = 50$, and the row sum is $50 \div 5 = 10$.

The answer is $\boxed{\textbf{(C)}\ 10}$.

If every row has the same sum and the rows use up all the numbers, then the row sum must be the grand total divided by the number of rows. This trick works for any magic square, of any size, without ever filling it in. Canceling $-k$ with $k$ is a quick way to add a list that runs across zero.

The five rows use each of the $25$ integers exactly once, so five times the common row sum equals the total of all the integers.

Pair $-10$ with $10$, $-9$ with $9$, and so on down to $-1$ with $1$: these cancel along with $0$. What remains is
$$11 + 12 + 13 + 14 = 50.$$
Hence the common sum is $\frac{50}{5} = 10$.

The answer is $\boxed{\textbf{(C)}\ 10}$.