2004 AMC 10A Problems/Problem 16: Difference between revisions
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===Solution 2=== | ===Solution 2=== | ||
We use complementary counting. There are only <math>2\times2</math> and <math>1\times1</math> squares that do not contain the black square. Counting, there are <math>12</math>-<math>2\times2</math> squares, and <math>25-1 = 24</math> <math>1\times1</math> squares that do not contain the black square. That gives <math>12+24=36</math> squares that don't contain it. There are a total of <math>25+16+9+4+1 = 55</math> squares possible <math>(25</math> - <math>1\times1</math> squares <math>16</math> - <math>2\times2</math> squares <math>9</math> - <math>3\times3</math> squares <math>4</math> - <math>4\times4</math> squares and <math>1</math> - <math>5\times5</math> square), therefore there are <math>55-36 = 19</math> squares that contain the black square, which is <math>\boxed{\mathrm{(D)}\ 19}</math>. | We use complementary counting. There are only <math>2\times2</math> and <math>1\times1</math> squares that do not contain the black square. Counting, there are <math>12</math>-<math>2\times2</math> squares, and <math>25-1 = 24</math> <math>1\times1</math> squares that do not contain the black square. That gives <math>12+24=36</math> squares that don't contain it. There are a total of <math>25+16+9+4+1 = 55</math> squares possible <math>(25</math> - <math>1\times1</math> squares <math>16</math> - <math>2\times2</math> squares <math>9</math> - <math>3\times3</math> squares <math>4</math> - <math>4\times4</math> squares and <math>1</math> - <math>5\times5</math> square), therefore there are <math>55-36 = 19</math> squares that contain the black square, which is <math>\boxed{\mathrm{(D)}\ 19}</math>. | ||
== Video Solutions == | |||
*https://youtu.be/0W3VmFp55cM?t=4697 | |||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
*https://youtu.be/aMmF6jz6xA4 | |||
https://youtu.be/aMmF6jz6xA4 | |||
Education, the Study of Everything | Education, the Study of Everything | ||
Revision as of 22:44, 16 January 2021
Problem
The 
grid shown contains a collection of squares with sizes from 
to . How many of these squares contain the black center square?
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 Solutions
~ pi_is_3.14
Education, the Study of Everything
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
