# Difference between revisions of "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 - Trial 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 2

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 $\boxed{7}$.

## Solution 3

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 there are only four even numbers in the set, so 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 centre must be $7$, which is answer $\textbf{(C)}$.