1988 AJHSME Problems/Problem 4
Contents
Problem
The figure consists of alternating light and dark squares. The number of dark squares exceeds the number of light squares by
![[asy] unitsize(12); for(int a=0; a<7; ++a) { fill((2a,0)--(2a+1,0)--(2a+1,1)--(2a,1)--cycle,black); draw((2a+1,0)--(2a+2,0)); } for(int b=7; b<15; ++b) { fill((b,14-b)--(b+1,14-b)--(b+1,15-b)--(b,15-b)--cycle,black); } for(int c=1; c<7; ++c) { fill((c,c)--(c+1,c)--(c+1,c+1)--(c,c+1)--cycle,black); } for(int d=1; d<6; ++d) { draw((2d+1,1)--(2d+2,1)); } fill((6,4)--(7,4)--(7,5)--(6,5)--cycle,black); draw((5,4)--(6,4)); fill((7,5)--(8,5)--(8,6)--(7,6)--cycle,black); draw((7,4)--(8,4)); fill((8,4)--(9,4)--(9,5)--(8,5)--cycle,black); draw((9,4)--(10,4)); label("same",(6.3,2.45),N); label("pattern here",(7.5,1.4),N); [/asy]](http://latex.artofproblemsolving.com/e/d/f/edf5203610870ed881725c3f3dbbf2f2a89fd367.png)
Solution 1
If, for a moment, we disregard the white squares, we notice that the number of black squares in each row increases by 1 continuously as we go down the pyramid.
Thus, the number of black squares is 
.
Same goes for the white squares, except it starts a row later, making it 
.
Subtracting the number of white squares from the number of black squares...
![\[1 + 2 + \cdots + 7 + 8 - (1 + 2 + \cdots + 7) = 8 \Rightarrow (B)\]](http://latex.artofproblemsolving.com/0/9/5/09540fd2e683dcee0c3b7b3459cbbd29e0166c22.png)
Solution 2
It is trivial to notice that in each and every row, there is always one more black square than the white squares. Since there are 
rows, there are more black squares than the white squares. 
~sakshamsethi
See Also
| 1988 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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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