Difference between revisions of "2004 AMC 10A Problems/Problem 16"
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==Problem== | ==Problem== | ||
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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>. | ||
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+ | == Video Solutions == | ||
+ | *https://youtu.be/0W3VmFp55cM?t=4697 | ||
+ | *https://youtu.be/aMmF6jz6xA4 | ||
==See also== | ==See also== |
Latest 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
Thus, 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.