用初等有趣的方法部分证明了下列游戏(Bulgariansolitaire):将n(n+1)/2个棋子穿成任意多串,每串上有任意多个棋子,并排放在一起,然后作如下操作:将每串棋子的最上面一颗棋子拿下穿成一个新串,放在原来棋子串的后面,对新的棋子串作同样的操作,则经有限次操作后(包括以后的操作后),只有n串上面有棋子,并且此时排序后各串上棋子的数目分别是1,2,3,…,n.
This article partly solves a game (Bulgarian solitaire). Given n(n+1)/2 chessmen, if they aredivided into arbitrary heaps and each heap with arbitrary chessmen, then the numbers of chessmen in each heap must bel,2,...,n after a series of the following operation, pick a chessman from each heap and compose a new heap.