Posted by Joel on November 03, 2002 at 12:34:08:
In Reply to: Re: divisible by 19 posted by Denis Borris on November 03, 2002 at 12:01:25:
: : 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
NO. 2^2^3 + 3^2^3 + 5^2^3 = 397442 which is evenly divisible by 19.
Read it as 2^(2^n) + 3^(2^n) + 5^(2^n). Otherwise the problem doesn't make sense. See the thread "order of operations" from a few days ago.
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