2007 AMC 8 Problems/Problem 11: Difference between revisions
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==See Also== | |||
{{AMC8 box|year=2007|num-b=10|num-a=12}} | |||
Revision as of 22:37, 12 November 2012
Problem
Tiles 
and 
are translated so one tile coincides with each of the rectangles 
and 
. In the final arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle 
?
cannot be determined
Solution
We first notice that tile III has a 
on the bottom and a 
on the right side. Since no other tile has a or a , Tile III must be in rectangle . Tile III also has a 
on the left, so Tile IV must be in Rectangle .
The answer is 
See Also
| 2007 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
