2004 AMC 10A Problems/Problem 16
Contents
Problem
The 
grid shown contains a collection of squares with sizes from 
to . How many of these squares contain the black center square?
![[asy] for (int i=0; i<5; ++i) { for (int j=0; j<5; ++j) { draw((-2.5+i, -2.5+j)--(-1.5+i, -2.5+j) -- (-1.5+i, -1.5+j) -- (-2.5+i, -1.5+j)--cycle); } fill((-0.5,-0.5)--(-0.5, 0.5)--(0.5,0.5) -- (0.5,-0.5)--cycle, black); } [/asy]](http://latex.artofproblemsolving.com/f/1/2/f12ad733cea58225fdba60c1c7867a76f3b29e80.png)
Solution
Solution 1
Since there are five types of squares: 
and 
We must find how many of each square contain the black shaded square in the center.
If we list them, we get that
- There is
of all 
squares, containing the black square - There are
of all 
squares, containing the black square - There are
of all 
squares, containing the black square - There are of all
squares, containing the black square - There is of all
squares, containing the black square
squares, containing the black square
squares, containing the black square
squares, containing the black squareThus, the answer is 
.
Solution 2
We use complementary counting. There are only 
and 
squares that do not contain the black square. Counting, there are 
- squares, and 
squares that do not contain the black square. That gives 
squares that don't contain it. There are a total of 
squares possible 
- squares 
- squares 
- 
squares 
- 
squares and 
- 
square), therefore there are 
squares that contain the black square, which is 
.
Video Solution by OmegaLearn
https://youtu.be/HhdpuJt78Hg?t=168
~ pi_is_3.14 ~VictorZhang
Video Solutions
See also
| 2004 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 AMC 10 Problems and Solutions | ||
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