Difference between revisions of "2009 AMC 8 Problems/Problem 18"
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<math> \textbf{(A)}\ 49 \qquad \textbf{(B)}\ 57 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 96 \qquad \textbf{(E)}\ 126</math> | <math> \textbf{(A)}\ 49 \qquad \textbf{(B)}\ 57 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 96 \qquad \textbf{(E)}\ 126</math> | ||
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+ | ==Solution== | ||
+ | In a <math>1</math>-foot-by-<math>1</math>-foot floor, there is <math>1</math> white tile. In a <math>3</math>-by-<math>3</math>, there are <math>4</math>. Continuing on, you can deduce the <math>n^{th}</math> positive odd integer floor has <math>n^2</math> white tiles. <math>15</math> is the <math>8^{th}</math> odd integer, so there are <math>\boxed{\textbf{(C)}\ 64}</math> white tiles. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=17|num-a=19}} | {{AMC8 box|year=2009|num-b=17|num-a=19}} |
Revision as of 16:27, 25 December 2012
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?
Solution
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.
See Also
2009 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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All AJHSME/AMC 8 Problems and Solutions |