Euler Problem 26

Tue Jul 1 07:50:26 BST 2008

On Mon, 2008-06-30 at 18:33 -0400, Andrew Mathenge wrote:
> Actually, I think that understanding the question is the easiest part.
Not for me it seems, I think it is one of those days for me  when even
the obvious escapes you :-)
> Here's the question as copied from the website:
> 
> Find the value of d < 1000 for which 1/d contains the longest
> recurring cycle in its decimal fraction part.
> 
> Let's take the example that's given where d <= 10.
> 
> When d = 7, the value of 1/d (1/7) is:
> 
> 0.142857142857142857142857142857142857142857....

Ah...okay
> 
> As you can see, (142857) repeats indefinitely. It has a sequence of 6
> digits that repeat. This is longer than any of the others where d <=
> 10.
That was the crux of the matter!
> 
> Therefore, the question asks, for d = 1 to 1000, which value of d has
> the longest sequence of repeating digits in the fractional part?
> 
> Quite rightfully so, you can't limit the number of digits to 9 as in
> your example since there will be a number with a series of repeating
> digits greater than 9.
> 
> For example, 1/17 is:
> 
> 0.0588235294117647058823529411764705882352941176470588235294117647.....
> 
> or
> 
> 0.(0588235294117647)
> 
> The number of repeating digits in this sequence is 16.
> 
> Hope that helps. I'm not a python programmer so I really can't help
> debug your code.
It is now crystal clear, Anrdew, and thank you very much for bailing me
out, do not worry about the code, I just wanted a nudge in the right
direction 	;-)	
> 
> Andrew.
> 
> On Mon, Jun 30, 2008 at 4:03 PM, kinuthiA muchanE <muchanek at gmail.com> wrote:
> > This email is from a ubuntu-ke subscriber to other subscribers inluding you:)
> >
> > 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
> >> >
> >>
> >>
> >>
> >
> >
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