@Anand: Thanks for the code. I knew you could do it by bit shifting. :-) On Jun 5, 10:21 pm, Anand <[email protected]> wrote: > Here is a code for it.http://codepad.org/umkh3pjf > > > > On Sat, Jun 5, 2010 at 7:14 PM, Minotauraus <[email protected]> wrote: > > Subtract 3 from the number until either you get 0 or a negative > > number. If you get 0, its divisible, else not. > > You can probably do this by bit shifting too. > > > On Jun 5, 11:45 am, divya <[email protected]> wrote: > > > Find if a number is divisible my 3, without using %,/ or *. You can > > > use atoi(). > > > -- > > You received this message because you are subscribed to the Google Groups > > "Algorithm Geeks" group. > > To post to this group, send email to [email protected]. > > To unsubscribe from this group, send email to > > [email protected]<algogeeks%2bunsubscr...@googlegroups > > .com> > > . > > For more options, visit this group at > >http://groups.google.com/group/algogeeks?hl=en.
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