Difference between revisions of "2024 AMC 8 Problems/Problem 7"
Rithvikvijay (talk | contribs) (→Solution 2) |
Rithvikvijay (talk | contribs) (→Solution 2) |
||
Line 11: | Line 11: | ||
==Solution 2== | ==Solution 2== | ||
− | Let <math>x</math> be the number of <math>1x1</math> tiles. There are <math>21</math> squares and each <math>2x2</math> or <math>1x4</math> tile takes up 4 squares, so <math>x \equiv 1 \pmod{4}</math>, so it is either <math>1</math> or <math>5</math>. Color the columns, starting with red, then blue, and alternating colors, ending with a red column. There are <math>12</math> red squares and <math>9</math> blue squares, but each <math>2x2</math> and <math>1x4</math> shape takes up an equal number of blue and red squares, so there must be <math>3</math> more <math>1x1</math> tiles on red squares | + | Let <math>x</math> be the number of <math>1x1</math> tiles. There are <math>21</math> squares and each <math>2x2</math> or <math>1x4</math> tile takes up 4 squares, so <math>x \equiv 1 \pmod{4}</math>, so it is either <math>1</math> or <math>5</math>. Color the columns, starting with red, then blue, and alternating colors, ending with a red column. There are <math>12</math> red squares and <math>9</math> blue squares, but each <math>2x2</math> and <math>1x4</math> shape takes up an equal number of blue and red squares, so there must be <math>3</math> more <math>1x1</math> tiles on red squares than on blue squares, which is impossible if there is just one, so the answer is <math>\boxed{\textbf{(E)\ 5}}</math>, which can easily be confirmed to work. |
~arfekete | ~arfekete |
Latest revision as of 14:22, 14 August 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Video Solution 1 (easy to digest) by Power Solve
- 6 Video Solution by Math-X (First fully understand the problem!!!)
- 7 Video Solution by NiuniuMaths (Easy to understand!)
- 8 Video Solution 2 by SpreadTheMathLove
- 9 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 10 Video Solution by Interstigation
- 11 See Also
Problem
A rectangle is covered without overlap by 3 shapes of tiles: , , and , shown below. What is the minimum possible number of tiles used?
Solution 1
We can eliminate B, C, and D, because they are not subtracted by any multiple of . Finally, we see that there is no way to have A, so the solution is .
Solution 2
Let be the number of tiles. There are squares and each or tile takes up 4 squares, so , so it is either or . Color the columns, starting with red, then blue, and alternating colors, ending with a red column. There are red squares and blue squares, but each and shape takes up an equal number of blue and red squares, so there must be more tiles on red squares than on blue squares, which is impossible if there is just one, so the answer is , which can easily be confirmed to work.
~arfekete
Solution 3
Suppose there are different tiles, different tiles and different tiles. Since the areas of these tiles must total up to (area of the whole grid), we have Reducing modulo gives , or or .
If , then . After some testing, there is no valid pair that works, so the answer must be , which can be constructed in many ways.
-Benedict T (countmath1)
Video Solution 1 (easy to digest) by Power Solve
https://youtu.be/16YYti_pDUg?si=KjRhUdCOAx10kgiW&t=59
Video Solution by Math-X (First fully understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=eg8T8wApi2j83Ia_&t=1497 ~Math-X
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=V-xN8Njd_Lc
~NiuniuMaths
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=L83DxusGkSY
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=L4ouVVVkFo4
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=578
See Also
2024 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.