AMC 10 Step by Step

2020 AMC 10A Problem 7

P1-10~2 minPrint

The 2525 integers from −10-10 to 14,14, inclusive, can be arranged to form a 55 -by- 55 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?

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Key insight

The five rows together use every number exactly once. So five times the row sum equals the total of all 25 numbers, which is 50. The row sum is 10.

Solution

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

1. The 55 rows share out all 2525 numbers, each number landing in exactly one row. So
(row sum)+(row sum)+⋯ (5 times)=total of all 25 numbers.\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: −10-10 with 1010, −9-9 with 99, down to −1-1 with 11. Each pair adds to 00, and 00 adds nothing.
3. What is left over is 11+12+13+14=5011 + 12 + 13 + 14 = 50.
4. So 5×(row sum)=505 \times \text{(row sum)} = 50, and the row sum is 50÷5=1050 \div 5 = 10.

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

Why this works

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-k with kk is a quick way to add a list that runs across zero.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The five rows use each of the 2525 integers exactly once, so five times the common row sum equals the total of all the integers.

Pair −10-10 with 1010, −9-9 with 99, and so on down to −1-1 with 11: these cancel along with 00. What remains is
11+12+13+14=50.11 + 12 + 13 + 14 = 50.
Hence the common sum is 505=10\frac{50}{5} = 10.

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

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.