AMC8 2024

AMC8 2024 · Q7

AMC8 2024 · Q7. It mainly tests Arithmetic misc, Area & perimeter.

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?

一个 $3 \times 7$ 的矩形被 3 种形状的瓷砖无重叠覆盖：$2 \times 2$、$1\times4$ 和 $1\times1$，如下图所示。使用 $1\times1$ 瓷砖的最小可能数量是多少？

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (E)

正确答案：（E）

Solution

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 (E) $5$, which can easily be confirmed to work.

设 $x$ 为 $1\times 1$ 小方砖的数量。总共有 $21$ 个小格，而每块 $2\times 2$ 或 $1\times 4$ 的砖都占 $4$ 个小格，因此 $x \equiv 1 \pmod{4}$，所以 $x$ 只能是 $1$ 或 $5$。给各列涂色：从红色开始，接着蓝色，交替进行，并以红色列结束。这样红色小格有 $12$ 个，蓝色小格有 $9$ 个；但每个 $2\times 2$ 与 $1\times 4$ 形状所覆盖的红蓝小格数量相等，所以落在红色小格上的 $1\times 1$ 砖必须比落在蓝色小格上的多 $3$ 块。若只有 $1$ 块 $1\times 1$ 砖则不可能满足，因此答案为 (E) $5$，并且容易验证其可行。

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