Difference between revisions of "2009 AMC 8 Problems/Problem 18"
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+ | ==Problem== | ||
The diagram represents a <math> 7</math>-foot-by-<math> 7</math>-foot floor that is tiled with <math> 1</math>-square-foot black tiles and white tiles. Notice that the corners have white tiles. If a <math> 15</math>-foot-by-<math> 15</math>-foot floor is to be tiled in the same manner, how many white tiles will be needed? | The diagram represents a <math> 7</math>-foot-by-<math> 7</math>-foot floor that is tiled with <math> 1</math>-square-foot black tiles and white tiles. Notice that the corners have white tiles. If a <math> 15</math>-foot-by-<math> 15</math>-foot floor is to be tiled in the same manner, how many white tiles will be needed? | ||
+ | |||
<asy>unitsize(10); | <asy>unitsize(10); | ||
draw((0,0)--(7,0)--(7,7)--(0,7)--cycle); | draw((0,0)--(7,0)--(7,7)--(0,7)--cycle); | ||
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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> | ||
− | ==Solution== | + | ==Solutions== |
+ | ===Solution 1=== | ||
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. | 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. | ||
− | ==Solution 2== | + | ===Solution 2=== |
− | After testing a couple of cases, we find that the number of white squares is <math>(s/2)^2</math>, where s is the side length of the square and <math>s/2</math> is rounded up to the next whole number. Therefore, <math>15/2</math> rounded up is 8, so 8^2=64, or <math>\framebox{C}</math>. | + | After testing a couple of cases, we find that the number of white squares is <math>(s/2)^2</math>, where s is the side length of the square and <math>s/2</math> is rounded up to the next whole number. Therefore, <math>15/2</math> rounded up is 8, so <math>8^2=64</math>, or <math>\framebox{(C) 64}</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=17|num-a=19}} | {{AMC8 box|year=2009|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:46, 21 October 2020
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 .
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 |
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