2020 AMC 10A Problem 7

Attempt Problem 7 of the 2020 AMC 10A below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 10A solutions, or check the answer key.

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

7.

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 five row sums together equal the sum of all 2525 integers
Big hint:
Compute the arithmetic-sequence sum from −10-10 through 1414
Video solution:
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Written solution:
The sum of the integers from −10-10 to 1414 is 25⋅−10+142=5025\cdot\dfrac{-10+14}{2}=50. If every row has common sum SS, then the five row sums add to 5050, so 5S=505S=50 and S=10S=10. Thus, C is the correct answer.
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