2022 AMC 8 Problem 20

Attempt Problem 20 of the 2022 AMC 8 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 2022 AMC 8 solutions, or check the answer key.

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

20.

The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number xx in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of x?x?
−1-1
55
66
88
99
Answer: D
Concepts:magic squareinequality
Small hint:
The completed row and column sums must match the top row
Big hint:
Express the other missing entries in terms of xx
Solution:
Adding the numbers in the top row shows that every row and column must have sum 1212.
In the first column, the missing number above xx is 14−x14-x. In the bottom row, the missing number to the right of xx is 4−x4-x. In the middle row, the remaining missing number is x−1x-1.
Since xx is larger than the other missing numbers, we need x>14−xx>14-x, x>4−xx>4-x, and x>x−1x> x-1. The strongest condition is x>7x>7, so the smallest possible integer value is 88.
Thus, the correct answer is D.
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