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Difference between revisions of "2016 AMC 10B Problems/Problem 15"
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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);
 
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>
 
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>
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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>.
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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==
 
==Solution 2 - Trial and Error==
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<cmath>4~7~8</cmath>
 
<cmath>4~7~8</cmath>
 
<cmath>5~6~9</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)
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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 3 (not rigorous)==
 
==Solution 3 (not rigorous)==
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<cmath>2  ~ 9  ~6</cmath>
 
<cmath>2  ~ 9  ~6</cmath>
 
<cmath>3  ~ 4~  5</cmath>
 
<cmath>3  ~ 4~  5</cmath>
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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>
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<cmath>3 ~ 4~ 5</cmath>
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 +
<math>\boxed {\textbf{(C) }7}</math>
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 +
~JH. L
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2016|ab=B|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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

See Also

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
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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