[Tutor] Fwd: Problem Euler 26

W W srilyk at gmail.com
Tue Jul 1 11:48:56 CEST 2008

Just a tidbit:

A neat function my friend came up with last year to figure out the
length of a whole number (now converted to python for your viewing

from math import log10 as log
from math import floor

def findNumberLength(number):
    number = float(number)
    x = log(number)
    x = floor(x)
    x += 1
    x = int(x)
    return x

The only catch is that this only works for a whole (non-decimal) number:
>>>print findNumberLength(123)
>>>print findNumberLength(1234.5234)

Though you could easily split your number in two:

def splitNum(mynum):
    mynum = str(mynum)
    mynum = mynum.split('.')
    for x in range(2):
        mynum[x] = float(mynum[x])
    return mynum

and then run the operation on both. A simple error check would
eliminate, say, working this on 123.0 (which would give you 4 digits,
and I suppose if you were looking for significant figures that would

if (mynum % 1) == 0:
    print "This number is X.0"

Although, now that I think about it, if you were really lazy, you
could just convert the int to a str and run len on it. Heh... Can't do
that (easily) in c++.

*Shrugs* oh well.

On Tue, Jul 1, 2008 at 1:12 AM, Andre Engels <andreengels at gmail.com> wrote:
> On Sun, Jun 29, 2008 at 10:34 PM, kinuthiA muchanE <muchanek at gmail.com> wrote:
>> 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.
> Yes, it seems you have... I will try to rephrase:
> For each fraction x/y (y>x to get it below 1), its representation as a
> decimal fraction consists of two parts: first some arbitrary series of
> numbers, then a series of numbers that is repeated over and over
> again. For example, 1/55 = 0.018181818181818..., or in short 0.0(18) -
> the arbitrary series here is 0, the repeating part 18. You are asked
> to get the largest repeating part.
> Your solution finds the longest time before a digit is repeated
> instead, which is incorrect because the same digit can be used more
> than once in the repeating part (and also because it might occur once
> or more in the non-repeating part) - for example, the 1/38 that you
> mentioned equals:
> 0.0(263157894736842105)
> A hint for solving this problem: use methods similar to the once you
> use for a long division with pen and paper.
> --
> Andre Engels, andreengels at gmail.com
> ICQ: 6260644 -- Skype: a_engels
> _______________________________________________
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