Difference between revisions of "2007 AMC 8 Problems/Problem 11"
(→Solution) |
(→Solution) |
||
Line 40: | Line 40: | ||
We first notice that tile III has a <math>0</math> on the bottom and a <math>5</math> on the right side. Since no other tile has a <math>0</math> or a <math>5</math>, Tile III must be in rectangle <math>D</math>. Tile III also has a <math>1</math> on the left, so Tile IV must be in Rectangle <math>C</math>. | We first notice that tile III has a <math>0</math> on the bottom and a <math>5</math> on the right side. Since no other tile has a <math>0</math> or a <math>5</math>, Tile III must be in rectangle <math>D</math>. Tile III also has a <math>1</math> on the left, so Tile IV must be in Rectangle <math>C</math>. | ||
− | The answer is <math>\boxed{\textbf{(D)} | + | The answer is <math>\boxed{\textbf{(D)}}</math> |
<center>[[Image:AMC8_2007_11S.png]]</center> | <center>[[Image:AMC8_2007_11S.png]]</center> |
Revision as of 13:19, 9 December 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 |