Printable

2020 AMC 10A Problem 7

2020 AMC 10A Problem 7AMC 10 Step by Step

Name: Date: September 13, 2026

Sequences & Series · difficulty 2 of 5 · about 2 min · position P1-10

The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?

A)
$2$
B)
$5$
C)
$10$
D)
$25$
E)
$50$

Answer

Key insight

Whatever the arrangement, the five rows together contain every number once, so the common sum is (total of -10 through 14)/5 = 50/5 = 10.

Solution

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 $\boxed{\textbf{(C)}\ 10}$ .

Why this works

You never need the actual magic square. Rows partition the grid, so any "all rows sum to the same value" condition pins that value to (total)/(number of rows). For an arithmetic sequence, cancel symmetric pairs around zero or use (count)(average) $= 25 \cdot 2 = 50$ .

The trap

Trying to construct the magic square, or dividing the total 50 by 25 cells instead of by 5 rows.

Common mistakes

Trying to construct the magic square, or dividing the total 50 by 25 cells instead of by 5 rows.
Miscounting the integers (there are $25$ , not $24$ ) or dropping $0$ or $14$ from the total.

Techniques

Set up the equation/formula and compute; no special trick needed · Exploit symmetry to reduce work or pair up objects