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Difference between revisions of "1988 AJHSME Problems/Problem 4"

m
m (Change to filldraw, so that all black squares get gray border, while all white squares get black border. Now Looks good with transparency on dark-mode.)
 
(15 intermediate revisions by 6 users not shown)
Line 7: Line 7:
 
<asy>
 
<asy>
 
unitsize(12);
 
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)
 
for(int a=0; a<7; ++a)
 
  {
 
  {
−
   fill((2a,0)--(2a+1,0)--(2a+1,1)--(2a,1)--cycle,black);
+
   filldraw((2a,0)--(2a+1,0)--(2a+1,1)--(2a,1)--cycle,black,gray);
−
  draw((2a+1,0)--(2a+2,0));
 
 
  }
 
  }
 
for(int b=7; b<15; ++b)
 
for(int b=7; b<15; ++b)
 
  {
 
  {
−
   fill((b,14-b)--(b+1,14-b)--(b+1,15-b)--(b,15-b)--cycle,black);
+
   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)
 
for(int c=1; c<7; ++c)
 
  {
 
  {
−
   fill((c,c)--(c+1,c)--(c+1,c+1)--(c,c+1)--cycle,black);
+
   filldraw((c,c)--(c+1,c)--(c+1,c+1)--(c,c+1)--cycle,black,gray);
 
  }
 
  }
−
for(int d=1; d<6; ++d)
+
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)
 
  {
 
  {
−
   draw((2d+1,1)--(2d+2,1));
+
   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);
 
  }
 
  }
−
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("same",(6.3,2.45),N);
 
label("pattern here",(7.5,1.4),N);
 
label("pattern here",(7.5,1.4),N);
 
</asy>
 
</asy>
  
−
==Solution==
+
==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.
 
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.
Line 39: Line 51:
  
 
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>
+
<cmath>1 + 2 + \cdots + 7 + 8 - (1 + 2 + \cdots + 7) = 8 \Rightarrow (B)</cmath>
 +
 
 +
 
 +
==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 <math>8</math> rows, there are <math>8</math> more black squares than the white squares. <math>8\rightarrow \boxed{\text{B}}</math>
 +
 
 +
~sakshamsethi
 +
(Edited by Zack2008)
  
 
==See Also==
 
==See Also==
Line 45: Line 64:
 
{{AJHSME box|year=1988|num-b=3|num-a=5}}
 
{{AJHSME box|year=1988|num-b=3|num-a=5}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 14:22, 17 January 2023

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)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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