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1988 AJHSME Problems/Problem 4

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Problem

The figure consists of alternating light and dark squares. The number of dark squares exceeds the number of light squares by

\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11

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)); }...

Solution

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 1 + 2 + \cdots + 8.

Same goes for the white squares, except it starts a row later, making it 1 + 2 + \cdots + 7.

Subtracting the number of white squares from the number of black squares... 1 + 2 + \cdots + 7 + 8 - (1 + 2 + \cdots + 7) = 8

See Also

1988 AJHSME (ProblemsResources)
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
Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.
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