Difference between revisions of "2016 AMC 10B Problems/Problem 15"
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<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) | 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) | ||
| + | ==Solution 2== | ||
| + | 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. | ||
==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}} | ||
Revision as of 19:41, 21 February 2016
Problem
All the numbers 
are written in a 
array of squares, one number in each square, in such a way that if two numbers of 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 - Trial Error
Quick testing shows that
![\[3 2 1\]](http://latex.artofproblemsolving.com/c/8/2/c82dafba05fb067127ee84b31824e8c320efaa14.png)
![\[4 7 8\]](http://latex.artofproblemsolving.com/6/6/8/668c4cb7b02fa2765b39a5fcc9ba5e6dcdec1338.png)
![\[5 6 9\]](http://latex.artofproblemsolving.com/e/8/0/e80c798d48833cd1f9ed8cd08bfc900ebfe2394c.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)
Solution 2
First let the numbers be
![\[1 8 7\]](http://latex.artofproblemsolving.com/3/8/4/3848b23500cd0cab26dce5163785055c302f38c0.png)
![\[2 9 6\]](http://latex.artofproblemsolving.com/2/6/8/268e8fb6726e99502056260adfb93a749b243180.png)
![\[3 4 5\]](http://latex.artofproblemsolving.com/3/8/9/38916c1dd7599404177cf3b915cd91fa5905b784.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.
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. 
