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]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
