# Euler Problem 26

kinuthiA muchanE muchanek at gmail.com
Mon Jun 30 19:13:58 BST 2008

```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...

```