1988 AJHSME Problems/Problem 4

Problem

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

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

[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]

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 $1 + 2 + \cdots + 8$.

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... \[1 + 2 + \cdots + 7 + 8 - (1 + 2 + \cdots + 7) = 8 \Rightarrow (B)\]


Solution 2

It is simple to notice that in each and every row, there is always one more black square than the white squares. Since there are $8$ rows, there are $8$ more black squares than the white squares. $8\rightarrow \boxed{\text{B}}$

~sakshamsethi (Edited by Zack2008)

See Also

1988 AJHSME (ProblemsAnswer KeyResources)
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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