# Euler Problem 26

kinuthiA muchanE muchanek at gmail.com
Mon Jun 30 21:03:37 BST 2008

```On Mon, 2008-06-30 at 22:04 +0300, Tony White wrote:
> Hi
>
> My guess, your pattern is limiting you to uniquely the digits 0-9, so

...and that is the whole idea IMHO, actually "input-ting" nothing would
match my regular expression. There are only ten non-recurring digits in
the world ie zero to nine, :-).
> that's where it stops. Consider - (1011011101234101111109) could also
> be a valid sequence. ie, digits may be used more than once.

My dear Tony, so what is the question? I thought they wanted the
reciprocal of a number, that will give you the digits "123456789" in a
non-repeating order. And 1011011101234101111109 would not match the
regular expression I had! 1's are and 0's occur more than once!

Mugumo, I think there is always  a mathematical angle to everything! And
I am okay...

Asante kwa kuweza kupata wasaa wa kunijibu, natumai tutaendelea
kuwasiliana.

Kinuthia...

>
> Tony
>
>
> 2008/6/30 kinuthiA muchanE <muchanek at gmail.com>:
> > This email is from a ubuntu-ke subscriber to other subscribers inluding you:)
> > Habari wote,
> >
> >
> > I am trying to solve Problem Number 26
> > (http://projecteuler.net/index.php?section=problems&id=26) on Project
> > Euler but apparently the answer I am submitting is wrong.
> >
> > Here is the problem:
> >
> >
> > A unit fraction contains 1 in the numerator. The decimal representation
> > of the unit fractions with denominators 2 to 10 are given:
> >
> >        1/2
> >        =
> >        0.5
> >        1/3
> >        =
> >        0.(3)
> >        1/4
> >        =
> >        0.25
> >        1/5
> >        =
> >        0.2
> >        1/6
> >        =
> >        0.1(6)
> >        1/7
> >        =
> >        0.(142857)
> >        1/8
> >        =
> >        0.125
> >        1/9
> >        =
> >        0.(1)
> >        1/10
> >        =
> >        0.1
> >
> > Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It
> > can be seen that 1/7 has a 6-digit recurring cycle.
> >
> > Find the value of d < 1000 for which 1/d contains the longest recurring
> > cycle in its decimal fraction part.
> >
> > I am giving the answer 38, because 1/38 = 0.0263157894. It seems I have
> > misunderstood the question or I cant solve it! Here is the code(in
> > Python) that I
> > came up with:
> >
> > def aux(num):
> >        import re
> >        pattern = re.compile(r"^0?1?2?3?4?5?6?7?8?9?\$")
> >
> >        frac ="%.9f"%(1.0/num)
> >        fracSlice = frac[2:]                    # get the decimal
> > fractional part, ie remove
> > '0.'
> >
> >        fracList = list(fracSlice)              #convert string to a
> > list
> >        fracList.sort()                                 # I need to
> > sort , because I will be searching by
> > increasing order
> >
> >        testFrac  = "".join(fracList)   # convert list back to a string,
> > phew!
> >        if re.match(pattern,testFrac):  # if the pattern matches, the
> > number is
> > our candidate
> >                print (num,fracSlice)
> >
> >
> > for b in xrange(1,1000):
> >        aux(b)
> >
> > Er... er, that does not exactly work as expected but it narrows the
> > search to only 3 candidates because of the inclusion of the zero:
> >
> >  (28, '035714286')
> >  (38, '026315789')
> >  (81, '012345679')
> >
> > For 28, the digit, in the fractional part, after 8 is 5, so 5 is
> > repeated and as for, 81 the next digit after 7 is 0, so again 0 occurs
> > twice. But for 38, the next digit after 9 is 4, and because it has NOT
> > occurred before, I assume 38 is the correct answer... and I am wrong!
> >
> > I suspect I have completely misunderstood the question.
> >
> > Any ideas?
> > Thanks!
> >
> > Kinuthia...
> >
> >
> > --
> > Ubuntu-ke mailing list
> > Ubuntu-ke at lists.ubuntu.com
> > https://lists.ubuntu.com/mailman/listinfo/ubuntu-ke
> >
>
>
>

```