2022 AMC 8

Complete problem set with solutions and individual problem pages

Problem 20 Hard

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 x in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of x?

  • A.

    -1

  • B.

    5

  • C.

    6

  • D.

    8

  • E.

    9

Answer:D

Solution 1

The sum of the numbers in each row is 12. Consider the second row. In order for the sum of the numbers in this row to equal 12, the two shaded numbers must add up to 13:

If two numbers add up to 13, one of them must be at least 7: If both shaded numbers are no more than 6, their sum can be at most 12. Therefore, for x to be larger than the three missing numbers, x must be at least 8. We can construct a working scenario where x=8:

So, our answer is \boxed{\textbf{(D) } 8}.

 

Solution 2

The sum of the numbers in each row is -2+9+5=12, and the sum of the numbers in each column is 5+(-1)+8=12.

Let y be the number in the lower middle. It follows that x+y+8=12, or x+y=4.

We express the other two missing numbers in terms of x and y, as shown below:

We have x>x-1, x>y+10, and x>y. Note that the first inequality is true for all values of x. We only need to solve the second inequality so that the third inequality is true for all values of x. By substitution, we get x>(4-x)+10, from which x>7.

Therefore, the smallest possible value of x is \boxed{\textbf{(D) } 8}.

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