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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?
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<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>
  
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==Solution==
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==Solutions==
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===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.
  
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==Solution 2 (Pattern Spotting)==
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===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 <math>8^2=64</math>, or <math>\framebox{(C) 64}</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>.
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==Video Solution==
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https://www.youtube.com/watch?v=VThwtOomWYQ  ~David
  
 
==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 18:37, 15 April 2023

Problem

The diagram represents a $7$-foot-by-$7$-foot floor that is tiled with $1$-square-foot black tiles and white tiles. Notice that the corners have white tiles. If a $15$-foot-by-$15$-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]

$\textbf{(A)}\ 49 \qquad \textbf{(B)}\ 57 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 96 \qquad \textbf{(E)}\ 126$

Solutions

Solution 1

In a $1$-foot-by-$1$-foot floor, there is $1$ white tile. In a $3$-by-$3$, there are $4$. Continuing on, you can deduce the $n^{th}$ positive odd integer floor has $n^2$ white tiles. $15$ is the $8^{th}$ odd integer, so there are $\boxed{\textbf{(C)}\ 64}$ white tiles.

Solution 2

After testing a couple of cases, we find that the number of white squares is $(s/2)^2$, where s is the side length of the square and $s/2$ is rounded up to the next whole number. Therefore, $15/2$ rounded up is 8, so $8^2=64$, or $\framebox{(C) 64}$.

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

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