Difference between revisions of "2024 AMC 8 Problems/Problem 7"
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| − | 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 | + | 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 | ||
Revision as of 13:18, 26 January 2024
Problem
A 
x
rectangle is covered without overlap by 3 shapes of tiles: 
x, 
x
, and x, shown below. What is the minimum possible number of x tiles used?
Solution 1
We can eliminate B, C, and D, because they are not 
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
