Difference between revisions of "1988 AJHSME Problems/Problem 4"
(New page: ==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{(...) |
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The figure consists of alternating light and dark squares. The number of dark squares exceeds the number of light squares by | 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 | + | <math>\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11</math> |
<asy> | <asy> | ||
| Line 32: | Line 32: | ||
==Solution== | ==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. | 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 <math> 1 + 2 + | + | Thus, the number of black squares is <math> 1 + 2 + \cdots + 8 </math>. |
| − | Same goes for the white squares, except it starts a row later, making | + | Same goes for the white squares, except it starts a row later, making it <math> 1 + 2 + \cdots + 7</math>. |
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</cmath> |
| − | |||
==See Also== | ==See Also== | ||
[[1988 AJHSME Problems]] | [[1988 AJHSME Problems]] | ||
Revision as of 16:06, 14 April 2009
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
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...
![\[1 + 2 + \cdots + 7 + 8 - (1 + 2 + \cdots + 7) = 8\]](http://latex.artofproblemsolving.com/2/8/5/2853288952e8919e0f21822fb70a227692db3216.png)
