2016 AMC 10B Problem 15

Below is the professionally curated solution for Problem 15 of the 2016 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 10B solutions, or check the answer key.

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Concepts:paritylogical deduction

Difficulty rating: 1420

15.

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?

 5 \ 5

 6 \ 6

 7 \ 7

 8 \ 8

 9 \ 9

Solution:

We firstly claim that everything that is either in the center or a corner is odd. This is because every number is next to a consecutive number and therefore has the opposite parity as it.

Therefore, all of the points that are the center or a corner are the same parity. The are 55 such points, but only 44 even numbers, so their parity isn't even. Thus, they are all odd. This means the sum of the corners plus the 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 .

Problem 15 in Other Years