2016 AMC 12B Problem 12

Attempt Problem 12 of the 2016 AMC 12B 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 2016 AMC 12B solutions, or check the answer key.

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

12.

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 number is in the center?
55
66
77
88
99
Answer: C
Concepts:paritylogical deduction
Small hint:
Color the grid like a checkerboard; consecutive numbers must lie on opposite colors, so they alternate
Big hint:
The five same-colored cells (four corners plus center) must hold the five odd numbers, which sum to 2525
Solution:
Color the grid like a checkerboard so the four corners and the center share one color. Since consecutive numbers occupy adjacent (opposite colored) squares, the numbers alternate parity along the chain, so the five same-colored cells contain the five odd numbers 1,3,5,7,9,1,3,5,7,9, which sum to 25.25. The four corners add to 18,18, so the center is 25−18=7.25-18=7.
Thus, the correct answer is C.
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