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A problem to end all shift-a-digit problems (Posted on 2007-03-26) Difficulty: 3 of 5
Prove that there exists no natural number such that shifting its first digit to the end, multiplies it by 5, 6, or 8.

See The Solution Submitted by e.g.    
Rating: 4.0000 (1 votes)

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re: Something of a solution (one example) | Comment 2 of 3 |
(In reply to Something of a solution. by Leming)

6129 = 17510
1629 = 3510
6129 / 5 = 1629

7141 = 28810
1741= 4810
7141 / 6 = 1741

9171 = 64010
1971= 8010
9171 / 8 = 1971

Edited on March 27, 2007, 1:14 am
  Posted by Dej Mar on 2007-03-27 01:03:34

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