Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1992 AHSME Problems/Problem 8

Problem

[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

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

Solution

$\fbox{E}$ 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 (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_8&oldid=92117"