1992 AHSME Problems/Problem 8


[asy] draw((-10,-10)--(-10,10)--(10,10)--(10,-10)--cycle,dashed+linewidth(.75)); draw((-7,-7)--(-7,7)--(7,7)--(7,-7)--cycle,dashed+linewidth(.75)); draw((-10,-10)--(10,10),dashed+linewidth(.75)); draw((-10,10)--(10,-10),dashed+linewidth(.75)); fill((10,10)--(10,9)--(9,9)--(9,10)--cycle,black); fill((9,9)--(9,8)--(8,8)--(8,9)--cycle,black); fill((8,8)--(8,7)--(7,7)--(7,8)--cycle,black); fill((-10,-10)--(-10,-9)--(-9,-9)--(-9,-10)--cycle,black); fill((-9,-9)--(-9,-8)--(-8,-8)--(-8,-9)--cycle,black); fill((-8,-8)--(-8,-7)--(-7,-7)--(-7,-8)--cycle,black); fill((10,-10)--(10,-9)--(9,-9)--(9,-10)--cycle,black); fill((9,-9)--(9,-8)--(8,-8)--(8,-9)--cycle,black); fill((8,-8)--(8,-7)--(7,-7)--(7,-8)--cycle,black); fill((-10,10)--(-10,9)--(-9,9)--(-9,10)--cycle,black); fill((-9,9)--(-9,8)--(-8,8)--(-8,9)--cycle,black); fill((-8,8)--(-8,7)--(-7,7)--(-7,8)--cycle,black); [/asy]

A square floor is tiled with congruent square tiles. The tiles on the two diagonals of the floor are black. The rest of the tiles are white. If there are 101 black tiles, then the total number of tiles is

$\text{(A) } 121\quad \text{(B) } 625\quad \text{(C) } 676\quad \text{(D) } 2500\quad \text{(E) } 2601$


$\fbox{E}$ Let the side length of the square be $s$. If $s$ is even, there are $s$ black tiles on each of the two diagonals, so the total number is $2s$. If $s$ is odd, we double-count the middle square which is on both diagonals, so the number is instead $2s-1$. In this case, since $101$ is odd, we clearly must have the second case, so $2s-1=101 \implies 2s=102 \implies s=51$, and indeed $s$ is odd as we expected. Thus the overall number of tiles is $51^2 = 2601$.

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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