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

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==Video Solution 1 (easy to digest) by Power Solve==
 
==Video Solution 1 (easy to digest) by Power Solve==
 
https://youtu.be/16YYti_pDUg?si=KjRhUdCOAx10kgiW&t=59
 
https://youtu.be/16YYti_pDUg?si=KjRhUdCOAx10kgiW&t=59
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==Video Solution 2 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=L83DxusGkSY
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2024|num-b=6|num-a=8}}
 
{{AMC8 box|year=2024|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:54, 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

Solution 3

Suppose there are $a$ different $2\times 2$ tiles, $b$ different $4\times 1$ tiles and $c$ different $1\times 1$ tiles. Since the areas of these tiles must total up to $21$ (area of the whole grid), we have \[4a + 4b + c = 21.\] Reducing modulo $4$ gives $c\equiv 1\pmod{4}$, or $c = 1$ or $c = 5$.

If $c = 1$, then $a + b = 5$. After some testing, there is no valid pair $(a, b)$ that works, so the answer must be $\boxed{\textbf{(E)\ 5}}$, 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 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=L83DxusGkSY

See Also

2024 AMC 8 (ProblemsAnswer KeyResources)
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

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