Re: last few digits of a^(a^(a^(a....z times) a and z integers < 1000
- From: tryitn1@xxxxxxxxx
- Date: Sat, 6 Sep 2008 13:05:40 -0700 (PDT)
On Sep 6, 10:39 pm, "christian.bau" <christian....@xxxxxxxxxxxxxxxxxx>
wrote:
On Sep 6, 3:21 pm, tryi...@xxxxxxxxx wrote:
What i'm thinking was find the cycle length.But the cycle lengths are
huge (because for 1000 i found some as 20) if we have to find it for
10^10 .
The cycle length for (a^k) mod n is phi (n).
"The cycle length for (a^k) mod n is phi (n)".How did you get this ?
What is this theorem and preconditions. Can you send me links
please ?
6^k mod 10 ,phi(10) =4 , 6^k mod 10 is always 6.same for 5 .
For an easy calculation
of phi (n) check the Wikipedia article; for example phi
(10,000,000,000) = 4,000,000,000. If you continue the calculation the
numbers go down very quickly. And there is a very simple algorithm
that lets you find (a^k) mod n for lets say 10 digit k using less than
70 multiplications.
.
- References:
- last few digits of tetration
- From: tryitn1
- last few digits of a^(a^(a^(a....z times) a and z integers < 1000
- From: tryitn1
- Re: last few digits of a^(a^(a^(a....z times) a and z integers < 1000
- From: amzoti
- Re: last few digits of a^(a^(a^(a....z times) a and z integers < 1000
- From: tryitn1
- Re: last few digits of a^(a^(a^(a....z times) a and z integers < 1000
- From: christian.bau
- last few digits of tetration
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