Okay. Here is some code to determine the number of digits in the
period of repetition of the decimal expansion of 1/n, where n > 0:
int period(int n);
{
int i=1,j=0;
while( n % 2 == 0 )
n /= 2;
while( n % 5 == 0 )
n /= 5;
do
{
i = (10 * i) % n;
+
hmm. :-o
On Fri, Jul 2, 2010 at 5:57 PM, Dave wrote:
> Yes. With a period of 16:
> 1/17 = 0.0588235294117647 0588235294117647 0588235294117647 ...
>
> Dave
>
> On Jul 2, 5:22 am, jalaj jaiswal wrote:
> > @dave
> > is 1/17 recurring...??
> > @abhirup
> > now convert float to string ..only pa
Yes. With a period of 16:
1/17 = 0.0588235294117647 0588235294117647 0588235294117647 ...
Dave
On Jul 2, 5:22 am, jalaj jaiswal wrote:
> @dave
> is 1/17 recurring...??
> @abhirup
> now convert float to string ..only part after decimal
>
> now let the string be .346346346.
> take an auxilarry
Yes. What if the recurring number is more than 20 digits?
On Fri, Jul 2, 2010 at 9:33 AM, Dave wrote:
> Does it work for 1/17, 2/17, 3/17, etc.?
>
> Dave
>
> On Jul 1, 5:23 pm, jalaj jaiswal wrote:
> > we are given with Numerator and Denominator. After division we might get
> a
> > recurring d
Does it work for 1/17, 2/17, 3/17, etc.?
Dave
On Jul 1, 5:23 pm, jalaj jaiswal wrote:
> we are given with Numerator and Denominator. After division we might get a
> recurring decimal points float as the answer.
> For example 23.34563456 ...
> return 3456 i.e the recurring part
>
> i did it by c
