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

Revision as of 12:43, 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

@@ Line 12: / Line 12: @@
 ~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

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