Re: modular arithmetic
- From: israel@xxxxxxxxxxx (Robert Israel)
- Date: 27 Feb 2006 00:09:03 GMT
In article <378699.1140991738735.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
Robert <robertduncan2005@xxxxxxxxxxx> wrote:
Hi, I am working on a problem and taking the same steps similar to that
of an example. However, I need help to understand something and would
like some help.
Example:
find the remainder of 3^25 divided by 36.
so mod is 36 therefore
3^4=3^2*3^2=9*9=81=9(mod36)
3^8=(3^4)^2=9^2=9(mod36)
3^16=(3^8)^2=9^2+9(mod36)
3^25=3*3^8*3^16=3*9*9(mod36)
=3*9=27(mod36)
I need help understanding 3^25. I can see that 3 is obviously 3 and that
3^16 is 9(mod36) but why is 3^8 equal to 9?
3^8 = (3^4)^2. You already know 3^4 = 9 (mod 36), so
(3^4)^2 = 9^2, and again you already know from the previous line that
this is 9 (mod 36)
Also, can someone explain
the last line to me too?
25 = 1 + 8 + 16. So 3^25 = 3^1 * 3^8 * 3^16. And you already know ...
The problem that I am working on is to find the remainder of
23^(25^57)by 49. Can someone please tell me the steps in how to approach
this question. Do I start with 25^57 with mod(49)?
No, but mod 42 might help.
Alternatively:
Start with 23^5 mod 49, then raise this to the power 5 to get 23^25 mod
49. Raise this to the 25th power mod 49 in a similar way. Repeat until
you get back to 23...
Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
- References:
- modular arithmetic
- From: Robert
- modular arithmetic
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