Difference between revisions of "2024 AMC 8 Problems/Problem 7"

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==Solution 1==
 
==Solution 1==
  
We can eliminate B, C, and D, because they are not <math>21-</math> any multiple of <math>4</math>. Finally, we see that there is no way to have A, so the solution is <math>(E) \boxed{5}</math>.
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We can eliminate B, C, and D, because they are not <math>21-</math> any multiple of <math>4</math>. Finally, we see that there is no way to have A, so the solution is <math>\boxed{\textbf{(E)} 5}</math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 12:18, 26 January 2024

Problem

A $3$x$7$ rectangle is covered without overlap by 3 shapes of tiles: $2$x$2$, $1$x$4$, and $1$x$1$, shown below. What is the minimum possible number of $1$x$1$ tiles used?

$\textbf{(A) } 1\qquad\textbf{(B)} 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$

Solution 1

We can eliminate B, C, and D, because they are not $21-$ any multiple of $4$. Finally, we see that there is no way to have A, so the solution is $\boxed{\textbf{(E)} 5}$.

Solution 2

Let $x$ be the number of $1x1$ tiles. There are $21$ squares and each $2x2$ or $1x4$ tile takes up 4 squares, so $x \equiv 1 \pmod{4}$, so it is either $1$ or $5$. Color the columns, starting with red, then blue, and alternating colors, ending with a red column. There are $12$ red squares and $9$ blue squares, but each $2x2$ and $1x4$ shape takes up an equal number of blue and red squares, so there must be $3$ more $1x1$ tiles on red squares than on blue squares, which is impossible if there is just one, so the answer is $\boxed{\textbf{(E)\ 5}}$, which can easily be confirmed to work

~arfekete

Video Solution 1 (easy to digest) by Power Solve

https://youtu.be/16YYti_pDUg?si=KjRhUdCOAx10kgiW&t=59