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

Latest revision as of 20:13, 18 June 2022

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 $\boxed {\textbf{(C) }7}$ .

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 $\boxed {\textbf{(C) }7}$ .. — @adihaya (talk) 12:27, 21 February 2016 (EST) ~edited

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 $\boxed {\textbf{(C) }7}$ . ~edited

Solution 4 (answer choices)

Testing out the box with the center square taking on the value of 5 and 6, we find that they either do not satisfy the first or the second condition. Testing 7, we find that a valid configuration is $1 ~8~ 9$ $2 ~ 7 ~6$ $3 ~ 4~ 5$

$\boxed {\textbf{(C) }7}$

~JH. L

@@ 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 {\textbf{(C) }7}</math>.
 ==Solution 2 - Trial and Error==
@@ Line 15: / Line 15: @@
 <cmath>4~7~8</cmath>
 <cmath>5~6~9</cmath>
-is a valid solution. <math>3+1+5+9 = 18</math>, and the numbers follow the given condition. The center number is found to be <math>\boxed{7}</math>. [[User:Adihaya|— @adihaya]] ([[User talk:Adihaya|talk]]) 12:27, 21 February 2016 (EST)
+is a valid solution. <math>3+1+5+9 = 18</math>, and the numbers follow the given condition. The center number is found to be <math>\boxed {\textbf{(C) }7}</math>.. [[User:Adihaya|— @adihaya]] ([[User talk:Adihaya|talk]]) 12:27, 21 February 2016 (EST) ~edited
-==Solution 2 (not rigorous)==
+==Solution 3 (not rigorous)==
 First let the numbers be
 <cmath>1   ~8~   7 </cmath>
 <cmath>2  ~ 9   ~6</cmath>
 <cmath>3  ~ 4~   5</cmath>
-with the numbers <math>1-8</math> around the outsides and <math>9</math> in the middle. We see that the sum of the four corner numbers is <math>16</math>. If we switch <math>7</math> and <math>9</math>, then the corner numbers will add up to <math>18</math> and the consecutive numbers will still be touching each other. The answer is <math>\boxed{7}</math>.
+with the numbers <math>1-8</math> around the outsides and <math>9</math> in the middle. We see that the sum of the four corner numbers is <math>16</math>. If we switch <math>7</math> and <math>9</math>, then the corner numbers will add up to <math>18</math> and the consecutive numbers will still be touching each other. The answer is <math>\boxed {\textbf{(C) }7}</math>. ~edited
+==Solution 4 (answer choices)==
+Testing out the box with the center square taking on the value of 5 and 6, we find that they either do not satisfy the first or the second condition. Testing 7, we find that a valid configuration is
+<cmath>1 ~8~ 9 </cmath>
+<cmath>2 ~ 7 ~6</cmath>
+<cmath>3 ~ 4~ 5</cmath>
+<math>\boxed {\textbf{(C) }7}</math>
+~JH. L
 ==See Also==
 {{AMC10 box|year=2016|ab=B|num-b=14|num-a=16}}
 {{MAA Notice}}

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