Difference between revisions of "2016 AMC 12B Problems/Problem 12"
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| + | Solution by Mlux: | ||
| + | Draw a <math>3\times3</math> matrix. Notice that no adjacent numbers could be in the corners since two consecutive numbers must share an edge. Now find 4 nonconsecutive numbers that add up to <math>18</math>. Trying <math>1+3+5+9 = 18</math> works. Place each odd number in the corner in a clockwise order. Then fill in the spaces. There has to be a <math>2</math> in between the <math>1</math> and <math>3</math>. There is a <math>4</math> between <math>3</math> and <math>5</math>. The final grid should similar to this. | ||
| + | <math>\newline | ||
| + | 1, 2, 3\newline | ||
| + | 8, 7, 4\newline | ||
| + | 9, 6, 5</math> | ||
| + | Solution by Mlux | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=11|num-a=13}} | {{AMC12 box|year=2016|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:58, 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
This problem needs a solution. If you have a solution for it, please help us out by adding it.
Solution by Mlux:
Draw a matrix. Notice that no adjacent numbers could be in the corners since two consecutive numbers must share an edge. Now find 4 nonconsecutive numbers that add up to . Trying 
works. Place each odd number in the corner in a clockwise order. Then fill in the spaces. There has to be a 
in between the 
and 
. There is a 
between and 
. The final grid should similar to this.
Solution by Mlux
See Also
| 2016 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 11 |
Followed by Problem 13 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
