20090719, 23:25  #12 
Apr 2009
near Chicago
2×11 Posts 

20090719, 23:30  #13 
"Kyle"
Feb 2005
Somewhere near M50..sshh!
2·3·149 Posts 
It makes sense if you acknowledge K to be an integer, and n to be an integer, then 2^n is always an integer, multiplied by another integer (k) will always be an integer.
Last fiddled with by Primeinator on 20090719 at 23:30 
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