Rare Rotative Relationship! (Posted on 2004-09-23)

	Submitted by Old Original Oskar!
Rating: 4.7500 (4 votes)
Solution:	(Hide)
Suppose the number is abc...z. Let's write N=0.abc...zabc...z.... Therefore, (N+z)/10=0.zabc...zabc...z.... We want this to equal KN, with K an (unknown) integer, 10>K>1. (If K≥10, KN would have more digits than K.) If (N+z)/10=KN, then N=z/(10K-1). As a≥1, then z≥K. So, if we study z/(10K-1) for 10>K>1, z≥K, and look for the smallest abc...z, we'll have the answer. For K=z=4 we produce N=0.102564102564... so the answer is 102564.

	Subject	Author	Date
	Puzzle Solution With Explanation	K Sengupta	2007-06-07 05:34:24
	Answer	K Sengupta	2007-06-07 05:14:55
	Full Algebraic Solution	Candide	2004-09-28 17:04:01
	The algebraic way	Jer	2004-09-27 13:11:37
	re: Another idea	Old Original Oskar!	2004-09-27 12:20:30
	Another idea	Gamer	2004-09-26 11:01:05
	No Subject	abhi ghosh	2004-09-26 07:28:54
	re(2): Solutions with interpretations	Richard	2004-09-24 00:47:05
	re: Solutions with interpretations	Nosher	2004-09-23 21:47:29
	re: Solutions with interpretations	David Shin	2004-09-23 19:06:04
	The first few	Charlie	2004-09-23 13:27:56
	Solutions with interpretations	Jer	2004-09-23 13:16:29
	Solution	David Shin	2004-09-23 13:07:08