AMC8 2024

AMC8 2024 · Q16

AMC8 2024 · Q16. It mainly tests Basic counting (rules of product/sum), Area & perimeter.

Minh enters the numbers $1$ through $81$ into the cells of a $9 \times 9$ grid in some order. She calculates the product of the numbers in each row and column. What is the least number of rows and columns that could have a product divisible by $3$?

Minh 将数字 $1$ 到 $81$ 按某种顺序填入 $9 \times 9$ 网格的单元格中。她计算每行和每列数字的乘积。可能有产品能被 $3$ 整除的最少行数和列数是多少？

(A) 8 8

(B) 9 9

(D) 11 11

(E) 12 12

Answer

Correct choice: (D)

正确答案：（D）

Solution

Note you can swap/rotate any configuration of rows, such that all the rows and columns that have a product of 3 are in the top left. Hence the points are bounded by a $a \times b$ rectangle. This has $ab$ area and $a+b$ rows and columns divisible by $3$. We want $ab\ge 27$ and $a+b$ minimized. If $ab=27$, we achieve minimum with $a+b=9+3=12$. If $ab=28$,our best is $a+b=7+4=11$. Note if $a+b=10$, $ab=25$. Because $25<27$, there is no smaller answer, and we get $\boxed{\textbf{(D)} 11}$.

注意你可以交换/旋转任意行配置，使得所有乘积能被 $3$ 整除的行和列都在左上角。因此这些点被一个 $a \times b$ 矩形包围。这个矩形有 $ab$ 个面积和 $a+b$ 个能被 $3$ 整除的行和列。我们希望 $ab\ge 27$ 且 $a+b$ 最小化。如果 $ab=27$，我们用 $a+b=9+3=12$ 达到最小。如果 $ab=28$，我们最好的是 $a+b=7+4=11$。注意如果 $a+b=10$，$ab=25$。因为 $25<27$，没有更小的答案，我们得到 $\boxed{\textbf{(D)} 11}$。

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