# 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 $\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$ $[asy] unitsize(12); //Force a white background in middle even when transparent fill((3,1)--(12,1)--(12,4)--(3,4)--cycle,white); //Black Squares, Gray Border (blends better than white) for(int a=0; a<7; ++a) { filldraw((2a,0)--(2a+1,0)--(2a+1,1)--(2a,1)--cycle,black,gray); } for(int b=7; b<15; ++b) { filldraw((b,14-b)--(b+1,14-b)--(b+1,15-b)--(b,15-b)--cycle,black,gray); } for(int c=1; c<7; ++c) { filldraw((c,c)--(c+1,c)--(c+1,c+1)--(c,c+1)--cycle,black,gray); } filldraw((6,4)--(7,4)--(7,5)--(6,5)--cycle,black,gray); filldraw((7,5)--(8,5)--(8,6)--(7,6)--cycle,black,gray); filldraw((8,4)--(9,4)--(9,5)--(8,5)--cycle,black,gray); //White Squares, Black Border filldraw((7,4)--(8,4)--(8,5)--(7,5)--cycle,white,black); for(int a=0; a<7; ++a) { filldraw((2a+1,0)--(2a+2,0)--(2a+2,1)--(2a+1,1)--cycle,white,black); } for(int b=9; b<15; ++b) { filldraw((b-1,14-b)--(b,14-b)--(b,15-b)--(b-1,15-b)--cycle,white,black); } for(int c=1; c<7; ++c) { filldraw((c+1,c)--(c+2,c)--(c+2,c+1)--(c+1,c+1)--cycle,white,black); } label("same",(6.3,2.45),N); label("pattern here",(7.5,1.4),N); [/asy]$

## 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)

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