Difference between revisions of "2016 AMC 10B Problems/Problem 15"

Revision as of 10:49, 28 September 2021

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 $\boxed{7}$ .

@@ Line 8: / Line 8: @@
 Consecutive numbers share an edge. That means that it is possible to walk from <math>1</math> to <math>9</math> 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 <math>1+3+5+7+9=25.</math> Therefore if the sum of the numbers in the corners is <math>18</math>, the number in the center must be <math>7</math>, which is answer <math>\textbf{(C)}</math>.
+But since there are only four even numbers in the set, the five darker squares must contain the odd numbers, which sum to <math>1+3+5+7+9=25.</math> Therefore if the sum of the numbers in the corners is <math>18</math>, the number in the center must be <math>\boxed{7}</math>, which is answer <math>\textbf{(C)}</math>.
 ==Solution 2 - Trial and Error==

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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