AMC10 2013 B

AMC10 2013 B · Q22

AMC10 2013 B · Q22. It mainly tests Arithmetic misc, Basic counting (rules of product/sum).

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 equal. In how many ways can this be done?

正八边形 ABCDEFGH 的中心为 J。将顶点和中心各关联数字 1 到 9，每个数字用一次，使得直线 AJE、BJF、CJG 和 DJH 上的数字和相等。有多少种方法？

(A) 384 384

(B) 576 576

(D) 1680 1680

(E) 3456 3456

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): The digit $j$ at $J$ contributes to all four sums, and each of the other digits contributes to exactly one sum. Therefore the sum of all four sums is $3j + (1 + 2 + 3 + \cdots + 9) = 45 + 3j$. Because all four sums are equal, this must be a multiple of $4$, so $j = 1, 5,$ or $9$. For each choice of $j$, pair up the remaining digits so that each pair has the same sum. For example, for $j = 1$ the pairs are $2$ and $9$, $3$ and $8$, $4$ and $7$, and $5$ and $6$. Then order the pairs so that they correspond to the vertex pairs $(A, E)$, $(B, F)$, $(C, G)$, $(D, H)$. This results in $2^4 \cdot 4!$ different combinations for each $j$. Thus the requirements can be met in $2^4 \cdot 4! \cdot 3 = 1152$ ways.

答案（C）：位于 $J$ 的数字 $j$ 会对四个和都产生贡献，而其他每个数字恰好只对一个和产生贡献。因此，四个和的总和为 $3j + (1 + 2 + 3 + \cdots + 9) = 45 + 3j$。由于四个和都相等，这个总和必须是 $4$ 的倍数，所以 $j = 1, 5$ 或 $9$。对每一种 $j$ 的选择，将剩余数字两两配对，使每一对的和相同。例如当 $j = 1$ 时，配对为：$2$ 与 $9$，$3$ 与 $8$，$4$ 与 $7$，$5$ 与 $6$。然后将这些数对按顺序对应到顶点对 $(A, E)$、$(B, F)$、$(C, G)$、$(D, H)$。这样对每个 $j$ 会产生 $2^4 \cdot 4!$ 种不同组合。因此满足条件的方法总数为 $2^4 \cdot 4! \cdot 3 = 1152$ 种。

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