这题是个什么思路？

独木桥

Let the number be 10x+y, we have (10^n)*y+x=2(10x+y), 19x=(10^n-2)y. so 10^n-2 is divisible by 19.  17 is the smallest number satisfying this condition. all those numbers are of the form n=17+18k.   Divide 1 by 19, we get an infinite sequence of digits with period 18. Begin with 1,2,3 or 4, cut a segment of length  n=17+18k, we get the number. www.ddhw.com

独木桥

1/19=0.052631578947368421052631578947368421.......www.ddhw.com Starting at any place with digit 1,2,3 or 4, cut a segment of this sequence of length 17+18k(k=1,2,3,...), we get the number. All those numbers can be obtained by this way.   原贴： 文章来源: 独木桥® 于 2005-3-15 14:50:59 标题：回复：这题是个什么思路？www.ddhw.com Let the number be 10x+y, we have (10^n)*y+x=2(10x+y), 19x=(10^n-2)y. so 10^n-2 is divisible by 19.  17 is the smallest number satisfying this condition. all those numbers are of the form n=17+18k.   Divide 1 by 19, we get an infinite sequence of digits with period 18. Begin with 1,2,3 or 4, cut a segment of length  n=17+18k, we get the number.   www.ddhw.com