Can you find a number which, when multiplied by 3, gives an answer equivalent to moving its last digit to the front? (see image on right)
Puzzle created and kindly supplied by Michael Jørgensen. The number 142857 has the nice property that multiplying it by 3 is equivalent to simply moving the first digit to the end of the number (see image on left)
Can you find a number which, when multiplied by 3, gives an answer equivalent to moving its last digit to the front? (see image on right)
11 Comments
Phil Ellinger
3/5/2014 01:21:44 am
2413793103448275862068965517x3
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Paul
3/5/2014 01:32:03 am
Thanks Phil. That was the quickest response I've ever had to my puzzle of the month. I've just posted it! Super fast :)
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mehtamatics
3/5/2014 02:12:19 am
I am not going to repeat Phil's answer which I obtained in a similar fashion. Maybe a little less guesswork and bit more trial and error (and more error!) including some digital roots. But I did want to mention that 285714 shares the same property as 142857 - since 3*leading digit<10.
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Phil Ellinger
3/5/2014 02:58:53 am
Yes, 1/7,2/7,3/7,4/7,5/7,6/7 are all recurring decimals with cyclic arrangements of the same 6 digits, 142857, repeated. So the pairs 1/7 & 3/7 and 2/7 & 6/7 both have that property, and consequently so do the integers formed by taking just the first 6 decimal places. I haven't checked but it's seems likely 1/29 has somewhat similar behaviour, so quite possibly there could be other solutions starting at other points in the sequence.
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Phil Ellinger
3/5/2014 03:11:16 am
My "lucky guess" at splitting the sequence at "72" wasn't that lucky after all. I could have split in lots of places to get:
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Jim Walker
3/5/2014 10:44:33 am
1428567x3 does not equal 4285671. I'm not sure I understand the parameters of the problem... Can some digits change?
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Brian
3/5/2014 11:23:12 am
Jim actually checked the question. Good on you!
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Phil
3/5/2014 04:14:57 pm
Well spotted! I never checked. I guess the "6" was a typo. Remove that and it works for the resulting 6 digits.
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Paul
4/5/2014 12:26:55 am
Thanks to the eagle-eyed Jim for spotting the deliberate error :)
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4/5/2014 11:10:49 am
Paul didn't say it had to be base 10. Turns out that base 10 makes it very hard to find a solution. However, base 11 numbers solve the puzzle with ease. Example:
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David Tonn
7/5/2014 10:12:52 am
Format:
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