Question

正八边形 $ABCDEFGH$ 的中心为 $J$ 。要把数字 $1$ 到 $9$ 分别填在八个顶点和中心上，每个数字用一次，使得直线 $AJE$ 、 $BJF$ 、 $CJG$ 、 $DJH$ 上三个数的和都相等。有多少种填法？

The regular octagon $ABCDEFGH$ has its center at $J.$ Each of the vertices and the center are to be associated with one of the digits $1$ through $9,$ with each digit used once, in such a way that the sums of the numbers on the lines $AJE,$ $BJF,$ $CJG,$ and $DJH$ are all equal. In how many ways can this be done?

$384$

$576$

$1152$

$1680$

$3456$

答案：C

知识点：幻方模运算配对与分组乘法原理

难度评级：2060

解答：

设每条线的公共和为 $S$ 。 $\begin{aligned} S &= A+J+E \\ &=B+J+F \\ &=C+J+G \\ &=D+J+H \end{aligned}$ $\begin{aligned} 4S &= A+B+C+D+E \\ &\quad+F+G+H+4J \end{aligned}$ $4S = 45+3J$ $45+3J \equiv 0 \mod 4$ $3J \equiv 3 \mod 4$ $J \equiv 1 \mod 4$

四条线的和相加后可知中心数字只能为 $J=1,5,9$ 。先看 $J=1$ 的情况。

当 $J=1$ 时，四对相对顶点必须配成相同的和；四对可分配给四条直线，有 $4!$ 种，每对内部可交换，有 $2^4$ 种，其中 $2+9\iff 9+2$ 表示同一对的两种顺序。 $2+9 = 3+8 = 4+7 = 5+6$

对 $J=5$ 和 $J=9$ 也同理，因此总数为 $3\cdot 4! \cdot 2^4 = 1152$

所以正确答案是 C。

Let $S$ be defined as: $\begin{aligned} S &= A+J+E \\ &=B+J+F \\ &=C+J+G \\ &=D+J+H \end{aligned}$ $\begin{aligned} 4S &= A+B+C+D+E \\ &\quad+F+G+H+4J \end{aligned}$ $4S = 45+3J$ $45+3J \equiv 0 \mod 4$ $3J \equiv 3 \mod 4$ $J \equiv 1 \mod 4$

This means that $J=1,5,9.$ From here, let's assume $J=1.$ We will see that the other cases are similar enough to omit.

If $J=1,$ then we know that the pairs of numbers that satisfy the equality above are: $2+9 = 3+8 = 4+7 = 5+6$ There are $4!$ ways to distribute the pairs over the four groups, and then $2^4$ ways for these groups to swap elements (i.e. $2+9\iff 9+2$ ).

Now, if we look at the $J=5$ and $J=9$ cases, we see a similar pattern in the number of groupings and swaps. As such, we have: $3\cdot 4! \cdot 2^4 = 1152$ possibilities.

Thus, the correct answer is C .

2013 AMC 10B 第 22 题

答案：C

难度评级：2060

解答：

其他年份的第 22 题