2016 AMC 10B Problems/Problem 15

Problem

All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ 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 $18$ . What is the number in the center?

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

Solution 1

Consecutive numbers share an edge. That means that it is possible to walk from $1$ to $9$ by single steps north, south, east, or west. Consequently, the squares in the diagram with different shades have different parity: $[asy]size(4cm); for(int i=0;i<3;++i)for(int j=0;j<3;++j)filldraw(box((i,j),(i+1,j+1)),gray((i+j)%2*.2+.7));[/asy]$ But since there are only four even numbers in the set, the five darker squares must contain the odd numbers, which sum to $1+3+5+7+9=25.$ Therefore if the sum of the numbers in the corners is $18$ , the number in the center must be $\boxed{7}$ , which is answer $\textbf{(C)}$ .

Solution 2 - Trial and Error

Quick testing shows that $3~2~1$ $4~7~8$ $5~6~9$ is a valid solution. $3+1+5+9 = 18$ , and the numbers follow the given condition. The center number is found to be $\boxed{7}$ . — @adihaya (talk) 12:27, 21 February 2016 (EST)

Solution 3 (not rigorous)

First let the numbers be $1 ~8~ 7$ $2 ~ 9 ~6$ $3 ~ 4~ 5$ with the numbers $1-8$ around the outsides and $9$ in the middle. We see that the sum of the four corner numbers is $16$ . If we switch $7$ and $9$ , then the corner numbers will add up to $18$ and the consecutive numbers will still be touching each other. The answer is $\boxed{7}$ .

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2016 AMC 10B Problems/Problem 15

Contents

Problem

Solution 1

Solution 2 - Trial and Error

Solution 3 (not rigorous)

See Also