Ah I see, in which case could you not generalise your solution for all
integers?
By taking into account the size of words on the computer for example?



On 5 December 2011 15:09, Dave <dave_and_da...@juno.com> wrote:

> @Carl: Yes, as coded, my algorithm is for 32-bit integers. But the
> original poster asked for a solution using bit manipulation, and
> modulus and division are arithmetic operations, not bit operations.
>
> Dave
>
> On Dec 5, 8:56 am, Carl Barton <odysseus.ulys...@gmail.com> wrote:
> > @Dave Yours only works for a certain subset of all possible powers or 3
> > doesn't it? So WgpShashank's would be more general?
> >
> > On 5 December 2011 14:30, Dave <dave_and_da...@juno.com> wrote:
> >
> >
> >
> > > @WgpShashank: Yours is an O(log n) solution. Mine is O(1).
> >
> > > Dave
> >
> > > On Dec 5, 6:21 am, WgpShashank <shashank7andr...@gmail.com> wrote:
> > > > @SAMMM  have a look
> >
> > > > * *solution is to keep dividing the number by 3, i.e, do n = n/3
> > > > iteratively. In any iteration, if n%3 becomes non-zero and n is not 1
> > > then
> > > > n is not a power of 3, otherwise n is a power of 3
> >
> > > > check it out ?http://codepad.org/863ptoBE
> >
> > > > Thanks
> > > > Shashank
> > > > Computer Science
> > > > BIT Mesrahttp://
> > >www.facebook.com/wgpshashankhttp://shashank7s.blogspot.com/
> >
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