2020 AMC 12A Problem 5

Attempt Problem 5 of the 2020 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

5.

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?
22
55
1010
2525
5050
Answer: C
Concepts:magic squarearithmetic sequence
Small hint:
The total of all 2525 numbers is split equally among the 55 rows
Big hint:
Sum the arithmetic series from −10-10 to 14,14, then divide by 55
Solution:
The sum of the 2525 integers is (−10+14)⋅252=50.\dfrac{(-10 + 14) \cdot 25}{2} = 50.
The five rows each have the same sum and together account for the total, so each row sums to 50÷5=10.50 \div 5 = 10.
Thus, C is the correct answer.
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