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

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==Solution 2==
 
==Solution 2==
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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.
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Let <math>x</math> be the number of <math>1</math> by <math>1</math> tiles. There are <math>21</math> squares and each <math>2</math> by <math>2</math> or <math>1</math> by <math>4</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>2</math> by <math>2</math> and <math>1</math> by <math>4</math> shape takes up an equal number of blue and red squares, so there must be <math>3</math> more <math>1</math> by <math>1</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
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-Benedict T (countmath1)
 
-Benedict T (countmath1)
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==Video Solution by Math-X (First fully understand the problem!!!)==
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https://youtu.be/BaE00H2SHQM?si=eg8T8wApi2j83Ia_&t=1497
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~Math-X
  
 
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 
==Video Solution (A Clever Explanation You’ll Get Instantly)==
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https://youtu.be/16YYti_pDUg?si=KjRhUdCOAx10kgiW&t=59
 
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!)==
 
==Video Solution by NiuniuMaths (Easy to understand!)==
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~Thesmartgreekmathdude
 
~Thesmartgreekmathdude
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==Video Solution by WhyMath==
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https://youtu.be/LCCDCb1EHg0
  
 
==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}}

Latest revision as of 06:29, 15 November 2024

Problem

A $3 \times 7$ rectangle is covered without overlap by 3 shapes of tiles: $2 \times 2$, $1\times4$, and $1\times1$, shown below. What is the minimum possible number of $1\times1$ tiles used?

2024-AMC8-q7.png

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

Solution 1

We can eliminate B, C, and D, because they are not $21$ subtracted by 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 $1$ by $1$ tiles. There are $21$ squares and each $2$ by $2$ or $1$ by $4$ 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 $2$ by $2$ and $1$ by $4$ shape takes up an equal number of blue and red squares, so there must be $3$ more $1$ by $1$ 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 by Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=eg8T8wApi2j83Ia_&t=1497

~Math-X

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=WogR3xd8zMwJgzRf&t=712

~hsnacademy

Video Solution 1 (easy to digest) by Power Solve

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


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

Video Solution by Daily Dose of Math (Certified, Simple, and Logical)

https://youtu.be/8GHuS5HEoWc

~Thesmartgreekmathdude

Video Solution by WhyMath

https://youtu.be/LCCDCb1EHg0

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. AMC logo.png

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