Difference between revisions of "2009 AMC 8 Problems/Problem 18"
Jerry122805 (talk | contribs) (→Solution 2) |
Jerry122805 (talk | contribs) (→Solution 2) |
||
| Line 26: | Line 26: | ||
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 (Pattern Spotting)== |
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}</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}</math>. | ||
Revision as of 20:54, 3 September 2019
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?
![[asy]unitsize(10); draw((0,0)--(7,0)--(7,7)--(0,7)--cycle); draw((1,7)--(1,0)); draw((6,7)--(6,0)); draw((5,7)--(5,0)); draw((4,7)--(4,0)); draw((3,7)--(3,0)); draw((2,7)--(2,0)); draw((0,1)--(7,1)); draw((0,2)--(7,2)); draw((0,3)--(7,3)); draw((0,4)--(7,4)); draw((0,5)--(7,5)); draw((0,6)--(7,6)); fill((1,0)--(2,0)--(2,7)--(1,7)--cycle,black); fill((3,0)--(4,0)--(4,7)--(3,7)--cycle,black); fill((5,0)--(6,0)--(6,7)--(5,7)--cycle,black); fill((0,5)--(0,6)--(7,6)--(7,5)--cycle,black); fill((0,3)--(0,4)--(7,4)--(7,3)--cycle,black); fill((0,1)--(0,2)--(7,2)--(7,1)--cycle,black);[/asy]](http://latex.artofproblemsolving.com/b/9/3/b930ac4db49eb7fe094a36707bd316921d9d85db.png)
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.
Solution 2 (Pattern Spotting)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
