Posted by Denis Borris on November 03, 2002 at 12:01:25:
In Reply to: divisible by 19 posted by Joel on November 03, 2002 at 10:11:32:
: I'm trying to prove that 2^2^n + 3^2^n + 5^2^n is divisible by 19 for all positive integers n (by induction).
: Easy enough to show this is true for n=1 and n=2, etc., as a basis step.
if n=3, then above = 864 and 2/19
if n=6, ...and 14/19
if n=9, ...and 3/19
then the above repeats for n=12/15/18, 21/24/27 .....
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