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2009 AMC 8 Problems/Problem 18

Revision as of 20:55, 3 September 2019 by Jerry122805 (talk | contribs) (→‎Solution 2 (Pattern Spotting))

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]

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

Solution

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 (Pattern Spotting)

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 $\framebox{(C) 64}$.

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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