27 Apr
2010
27 Apr
'10
12:40 a.m.
On Tue, Apr 27, 2010 at 9:27 AM, Hans Havermann <pxp@rogers.com> wrote:
A large number n = 689*10^((5889*10^((578*10^196-218)/81)-165)/81)-1 and I would like to know the value of n mod 81. Is this doable?
Yes, n mod 81 = 31. Since 10^81 mod 729 = 1 and ((578*10^196-218)/81) mod 81 = 34, we have (5889*10^((578*10^196-218)/81)-165) mod 729 = (5889*10^34-165) mod 729 = 567 = 7*81 So ((5889*10^((578*10^196-218)/81)-165)/81) mod 9 = 7. Since 10^9 mod 81 = 1, we have (689*10^((5889*10^((578*10^196-218)/81)-165)/81)-1) mod 81 = (689*10^7-1) mod 81 = 31. Warut