1990 USAMO 第 3 题

Suppose that necklace $\, A \,$ has 14 beads and necklace $\, B \,$ has 19. Prove that for any odd integer $n \geq 1$, there is a way to number each of the 33 beads with an integer from the sequence

$$\{ n, n+1, n+2, \dots, n+32 \}$$

so that each integer is used once, and adjacent beads correspond to relatively prime integers. (Here a "necklace" is viewed as a circle in which each bead is adjacent to two other beads.)

假设项链 $\, A \,$ 有 14 颗珠子，项链 $\, B \,$ 有 19 颗。证明对于任何奇数整数 $n \geq 1$，有一种方法可以用序列中的整数对 33 颗珠子中的每一颗进行编号

$$

\{ n, n+1, n+2, \点, n+32 \}

$$

这样每个整数都使用一次，相邻的珠子对应互质的整数。 (这里的“项链”被视为一个圆圈，其中每个珠子都与其他两个珠子相邻。)