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2020 AMC 10A Problem 7

Problem 7 of 25EasierAlgebraLogic

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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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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Tagged: magic square · arithmetic sequence

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