Here is a code for it. http://codepad.org/umkh3pjf
On Sat, Jun 5, 2010 at 7:14 PM, Minotauraus <anike...@gmail.com> 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 <sweetdivya....@gmail.com> 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 algoge...@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
