On 29/07/05, Kevin <kevin_olbrich / yahoo.com> wrote:
> this is odd since you don't need to loop to find the answer
> 
> maxnum = 10_000
> 
> n = 2 ** (Math.log(maxnum) / Math.log(2)).floor
> 
> puts n + " is the largest power of 2 less than " + maxnum
> 
> 
> 

I'd think it was a educative example?


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