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
We have 8 rows, 7 of which the black squares exceed the white squares by one. We then add a extra square because in the first row there is no white square so now we have 
so therefore the answer is 
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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