Difference between revisions of "2016 AMC 10B Problems/Problem 15"
(→Problem) |
(→Solution added) |
||
| Line 6: | Line 6: | ||
| − | ==Solution== | + | ==Solution 1 - Trial Error== |
| − | + | Quick testing shows that | |
| + | <cmath>3 2 1</cmath> | ||
| + | <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) | ||
==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 13:27, 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)
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. 
