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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{(...)
 
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.)
 
(17 intermediate revisions by 6 users not shown)
Line 3: Line 3:
 
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>
 
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)
 +
{
 +
  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)
 
  {
 
  {
−
   draw((2d+1,1)--(2d+2,1));
+
   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.
−
Thus, the number of black squares is <math> 1 + 2 + ... + 8 </math>
+
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 is <math> 1 + 2 + ... + 7</math>
+
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...
−
<math> 1 + 2 + ... + 7 + 8 - (1 + 2 + ... + 7) </math>
+
<cmath>1 + 2 + \cdots + 7 + 8 - (1 + 2 + \cdots + 7) = 8 \Rightarrow (B)</cmath>
−
<math> = 8</math>
+
 
 +
 
 +
==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==
  
−
[[1988 AJHSME Problems]]
+
{{AJHSME box|year=1988|num-b=3|num-a=5}}
 +
[[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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