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

Latest revision as of 15:55, 20 February 2024

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 $3 \times 7$ 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?

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

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 $1x1$ 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 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

@@ Line 1: / Line 1: @@
+==Problem==
+A <math>3 \times 7</math> rectangle is covered without overlap by 3 shapes of tiles: <math>2 \times 2</math>, <math>1\times4</math>, and <math>1\times1</math>, shown below. What is the minimum possible number of <math>1\times1</math> tiles used?
+[[File:2024-AMC8-q7.png|center]]
+<math>\textbf{(A) } 1\qquad\textbf{(B)} 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5</math>
+==Solution 1==
+We can eliminate B, C, and D, because they are not <math>21</math> subtracted by 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==
+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
+==Solution 3==
+Suppose there are <math>a</math> different <math>2\times 2</math> tiles, <math>b</math> different <math>4\times 1</math> tiles and <math>c</math> different <math>1\times 1</math> tiles. Since the areas of these tiles must total up to <math>21</math> (area of the whole grid), we have
+<cmath>4a + 4b + c = 21.</cmath>
+Reducing modulo <math>4</math> gives <math>c\equiv 1\pmod{4}</math>, or <math>c = 1</math> or <math>c = 5</math>.
+If <math>c = 1</math>, then <math>a + b = 5</math>. After some testing, there is no valid pair <math>(a, b)</math> that works, so the answer must be <math>\boxed{\textbf{(E)\ 5}}</math>, 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==
+{{AMC8 box|year=2024|num-b=6|num-a=8}}
+{{MAA Notice}}

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

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