# Euler Problem 26

Tony White tony.mzungu at gmail.com
Tue Jul 1 01:06:35 BST 2008

```Excellently put Andrew - a much better example than mine ;)

2008/7/1 Andrew Mathenge <mathenge at gmail.com>:
> This email is from a ubuntu-ke subscriber to other subscribers inluding you:)
> Actually, I think that understanding the question is the easiest part.
> 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....
>
> 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.
>
> 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
>
> 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
>>
>
> --
> Ubuntu-ke mailing list
> Ubuntu-ke at lists.ubuntu.com
> https://lists.ubuntu.com/mailman/listinfo/ubuntu-ke
>

--
Tony White
SpeechNet Technologies Ltd

```