Difference between revisions of "1988 AJHSME Problems/Problem 4"
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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. | 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. | ||
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Subtracting the number of white squares from the number of black squares... | Subtracting the number of white squares from the number of black squares... | ||
<cmath>1 + 2 + \cdots + 7 + 8 - (1 + 2 + \cdots + 7) = 8 \Rightarrow (B)</cmath> | <cmath>1 + 2 + \cdots + 7 + 8 - (1 + 2 + \cdots + 7) = 8 \Rightarrow (B)</cmath> | ||
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+ | ==Solution 2== | ||
==See Also== | ==See Also== |
Revision as of 21:59, 8 July 2020
Contents
Problem
The figure consists of alternating light and dark squares. The number of dark squares exceeds the number of light squares by
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...
Solution 2
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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