Difference between revisions of "2009 AMC 8 Problems/Problem 18"
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==See Also== | ==See Also== | ||
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Latest revision as of 18:37, 15 April 2023
Problem
The diagram represents a -foot-by--foot floor that is tiled with -square-foot black tiles and white tiles. Notice that the corners have white tiles. If a -foot-by--foot floor is to be tiled in the same manner, how many white tiles will be needed?
Solutions
Solution 1
In a -foot-by--foot floor, there is white tile. In a -by-, there are . Continuing on, you can deduce the positive odd integer floor has white tiles. is the odd integer, so there are white tiles.
Solution 2
After testing a couple of cases, we find that the number of white squares is , where s is the side length of the square and is rounded up to the next whole number. Therefore, rounded up is 8, so , or .
Video Solution
https://www.youtube.com/watch?v=VThwtOomWYQ ~David
See Also
2009 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.