Difference between revisions of "2007 AMC 8 Problems/Problem 11"
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=10|num-a=12}} | {{AMC8 box|year=2007|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:53, 2 July 2024
Contents
[hide]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

Video Solution
https://www.youtube.com/watch?v=66MVYKyetGc
Video Solution by Why Math
~savannahsolver
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=omFpSGMWhFc
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.