Dave's logic is working fine.

On Mon, Jun 7, 2010 at 1:35 PM, Raj N <[email protected]> wrote:

> @Anand: The code you've sent is not correct. It doesn't work for numbers
> 15,12 etc.
>
>
> On Mon, Jun 7, 2010 at 2:20 AM, Minotauraus <[email protected]> wrote:
>
>> @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().
>> >
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