Euler Problem 26
kinuthiA muchanE
muchanek at gmail.com
Wed Jul 2 11:46:39 BST 2008
Habari wajamaa,
I got it but I am not enthusiastic about it because it was largely trial
and error. I have never done so much long division with a pen and paper
trying to look for a pattern in my life! What I was sure about was that
it had to be a prime.
Using the python's decimal module I set the decimal precision to 1000
places! Then using my eyes and maybe a bit of luck, I started with 997,
991, and then 983... bingo!
00101729399796541200406917599186164801627670396744659206510681586978636826042726347914547304170905391658189216683621566632756866734486266531027466937945066124109867751780264496439471007121057985757884028484231943031536113936927772126144455747711088504577822990844354018311291963377416073245167853509664292980671414038657171922685656154628687690742624618514750762970498474059003051881993896236012207527975584944048830
11190233977619532044760935910478128179043743641912512716174974567650050864699898270600203458799593082400813835198372329603255340793489318413021363173957273652085452695829094608341810783316378433367243133265513733468972533062054933875890132248219735503560528992878942014242115971515768056968463886063072227873855544252288911495422177009155645981688708036622583926754832146490335707019328585961342828077314343845371312309257375381485249237029501525940996948118006103763987792472024415055951169888097660223804679552390640895218718209562563580874872838250254323499491353
982 digits!
Andrew, I do not know if you would like to see the code, if you can call
it that! How did you guys do it?
Kinuthia...
On Tue, 2008-07-01 at 18:06 +0300, Miano Njoka wrote:
> I was wrong the first time, got it after a couple of tries. good
> luck!!! :)
>
> On Jul 1, 2008, at 4:37 PM, kinuthiA muchanE wrote:
>
> > Miano,
> >
> > Just got back to the computer, right now I am struggling with "for"
> > loops :-/
> > I do not know what the answer is but check here
> > http://projecteuler.net/index.php?section=problems&id=26
> >
> > Kinuthia...
> > On Tue, 2008-07-01 at 14:34 +0300, Miano Njoka wrote:
> >> I've written a solution using C, I got d=831, one cycle is 69 digits
> >> long. What is the answer?
> >>
> >>
> >> On Jul 1, 2008, at 12:02 PM, Andrew Mathenge wrote:
> >>
> >>> This email is from a ubuntu-ke subscriber to other subscribers
> >>> inluding you:)
> >>> Only glad I could be of assistance. Let me know if you find a python
> >>> solution to the problem. I wrote a solution in perl, but I'm
> >>> interested in learning python.
> >>>
> >>> Good luck!
> >>>
> >>> Andrew.
> >>>
> >>> On Tue, Jul 1, 2008 at 2:50 AM, kinuthiA muchanE
> >>> <muchanek at gmail.com> wrote:
> >>>>
> >>>> 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
> >>>>>>>>
> >>>>>>>
> >>>>>>>
> >>>>>>>
> >>>>>>
> >>>>>>
> >>>>>> --
> >>>>>> Ubuntu-ke mailing list
> >>>>>> Ubuntu-ke at lists.ubuntu.com
> >>>>>> https://lists.ubuntu.com/mailman/listinfo/ubuntu-ke
> >>>>>>
> >>>>
> >>>>
> >>>
> >>> --
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> >>
> >
>
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