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2016 AMC 10B Problem 15

Problem 15 of 25IntermediateNumber Theory

All the numbers 1,1, 2,2, 3,3, 4,4, 5,5, 6,6, 7,7, 8,8, 99 are written in a 3×33\times3 array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to 18.18. What is the number in the center?

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Solution

Color the array like a checkerboard, with the center and four corners one color and the four edge squares the other color. Consecutive numbers occupy adjacent squares, so the path 1,2,,91,2,\ldots,9 alternates colors. Therefore all five odd numbers occupy one color class and all four even numbers occupy the other. The color class consisting of the center and four corners has five squares, so it contains the five odd numbers. Hence the sum of the corners and center is 1+3+5+7+9=25.1+3+5+7+9=25. Since the corners have a sum of 18,18, the center has a value of 2518=7.25-18=7. Thus, the correct answer is C .

More practice

Concepts: parity · logical deduction

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.