@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 <[email protected]> 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 <[email protected]> wrote:
>
>
>
> > @WgpShashank: Yours is an O(log n) solution. Mine is O(1).
>
> > Dave
>
> > On Dec 5, 6:21 am, WgpShashank <[email protected]> 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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