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> | ||
| − | 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 | + | 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== | ||
| Line 22: | Line 22: | ||
<cmath>2 ~ 9 ~6</cmath> | <cmath>2 ~ 9 ~6</cmath> | ||
<cmath>3 ~ 4~ 5</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== | ==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
Contents
Problem
All the numbers 
are written in a 
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 
. What is the number in the center?
Solution 1
Consecutive numbers share an edge. That means that it is possible to walk from 
to 
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]](http://latex.artofproblemsolving.com/f/3/9/f39f0e7b752d26dd1dc58cc402af636029b64af9.png)
But since there are only four even numbers in the set, the five darker squares must contain the odd numbers, which sum to 
Therefore if the sum of the numbers in the corners is , the number in the center must be 
.
Solution 2 - Trial and Error
Quick testing shows that
![\[3~2~1\]](http://latex.artofproblemsolving.com/3/a/5/3a51c280f2c1f293c56ef5ade79e8cf929032f16.png)
![\[4~7~8\]](http://latex.artofproblemsolving.com/b/4/5/b451d8e5438bb977e84409ebde060296f919f95e.png)
![\[5~6~9\]](http://latex.artofproblemsolving.com/c/7/a/c7a0e6b99a06bf9a2eda08c3fd65ec5e6073ddc0.png)
is a valid solution. 
, and the numbers follow the given condition. The center number is found to be .. — @adihaya (talk) 12:27, 21 February 2016 (EST) ~edited
Solution 3 (not rigorous)
First let the numbers be
![\[1 ~8~ 7\]](http://latex.artofproblemsolving.com/7/b/c/7bcdb07a2e0f1753b2fcff3951d784fbc30eb50b.png)
![\[2 ~ 9 ~6\]](http://latex.artofproblemsolving.com/2/e/9/2e97fdf3f74cf5c7dc9362dfe41d551990a03eb5.png)
![\[3 ~ 4~ 5\]](http://latex.artofproblemsolving.com/e/b/9/eb90e4cb8f0fe8760958e2b5678f6d84044007dc.png)
with the numbers 
around the outsides and in the middle. We see that the sum of the four corner numbers is 
. If we switch 
and , then the corner numbers will add up to and the consecutive numbers will still be touching each other. The answer is . ~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
![\[1 ~8~ 9\]](http://latex.artofproblemsolving.com/b/3/0/b3004c9d8193debf2b2a5c48bdb4ae0f6ea527b9.png)
![\[2 ~ 7 ~6\]](http://latex.artofproblemsolving.com/7/a/0/7a0b346fc28c358ea8959b215cb918b953fb1502.png)
~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. 
