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if you divide my first digit by my last digit then you can find me for ever

If you multiply me by any number equal to or less than my length (but greater than zero) and I will only cycle.

If you divide any number equal to or less than my length (but greater than zero) by the last digit of me you will still find me for ever in different cycles

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  • $\begingroup$ When you say in the second line "and multiply the rest of me" does this mean excluding the last digit? $\endgroup$
    – Kenshin
    Nov 8, 2014 at 8:57
  • $\begingroup$ Edited that part out.... $\endgroup$
    – skv
    Nov 8, 2014 at 9:02
  • $\begingroup$ should the first line read, if you divide me by my last digit, you can find me for ever? $\endgroup$
    – Kenshin
    Nov 8, 2014 at 9:05
  • $\begingroup$ I think that part is right $\endgroup$
    – skv
    Nov 8, 2014 at 9:06
  • $\begingroup$ -1/-1 would be +1 right you wont find me :) $\endgroup$
    – skv
    Nov 8, 2014 at 9:07

1 Answer 1

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Here you are

$142857$

Because

$\frac17 = 0.\overline{142857}$

And

$142857\times2=285714$
$142857\times3=428571$
$142857\times4=571428$
$142857\times5=714285$
$142857\times6=857142$

And finally:

$\frac27 = 0.\overline{285714}$, but it can be rewritten as $0.2857\overline{142857}$
$\frac37 = 0.\overline{428571}=0.42857\overline{142857}$
$\frac47 = 0.\overline{571428}=0.57\overline{142857}$
$\frac57 = 0.\overline{714285}=0.7\overline{142857}$
$\frac67 = 0.\overline{857142}=0.857\overline{142857}$

The way to find it is actually not that difficult:

After you recognise that "cycle" refers to the repeating digits you need to ask what single-digit denominators produce repeating patterns in decimal. We have 3, 6, 7, 9 as candidates. However: 3, 6, and 9 produce repeating pattern of a single digit for any single-digit numerator, so we know that the last digit must be 7. So we try 1 as the first digit and it works.

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  • $\begingroup$ I edited the question $\endgroup$
    – skv
    Nov 8, 2014 at 9:10
  • $\begingroup$ @skv Oh, i see ))) $\endgroup$
    – v010dya
    Nov 8, 2014 at 9:11
  • $\begingroup$ @skv, is this the answer? $\endgroup$
    – Kenshin
    Nov 8, 2014 at 9:11
  • $\begingroup$ If it is, i'll add the explanation of how i found it. $\endgroup$
    – v010dya
    Nov 8, 2014 at 9:12
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    $\begingroup$ Ok, unless you'll change the question again, my answer now reflects the current version and has the explanation. $\endgroup$
    – v010dya
    Nov 8, 2014 at 9:24

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