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- Newsgroups: rec.puzzles
- Path: sparky!uunet!charon.amdahl.com!pacbell.com!ames!saimiri.primate.wisc.edu!zazen!schaefer.math.wisc.edu!mueller
- From: mueller@schaefer.math.wisc.edu (Carl Douglas Mueller)
- Subject: Number Puzzle
- Message-ID: <1992Nov18.152111.27777@schaefer.math.wisc.edu>
- Reply-To: mueller@math.wisc.edu
- Organization: Univ. of Wisconsin Dept. of Mathematics
- Distribution: na
- Date: Wed, 18 Nov 92 15:21:11 GMT
- Lines: 22
-
- Here's a puzzle that was told to me on the phone last night by a friend.
- He thought it was a pretty hard one. It turns out to be fairly easy to
- solve if you look at it right, but I thought it interesting enough to
- post here.
-
- Find a number ending in 7 such that when this number is multiplied by 7
- the resulting number consists of the same digits in the same order but
- with the 7 moved to the front. That is:
-
- 7*(abc...xyz7) = 7abc...xyz
-
- where the abc...xyz is some string of digits (you'll have to figure out
- how many digits).
-
- After solving this one (and finding the EASY way to solve it) I extended
- the puzzle to numbers other than 7. ie
-
- N*(abc...xyzN) = Nabc...xyz.
-
- I've found the numbers for N = 1 (trivial) up to N=9.
-
- Carl Mueller (mueller@math.wisc.edu)
-