The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » Math Topics » alt.math.undergrad

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Congruences
Replies: 0  

Advanced Search

Back to Topic List Back to Topic List  
Scott Phung

Posts: 32
Registered: 12/6/04
Posted: Nov 7, 1996 12:50 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

Thanks to the two people (and possibly more after this
message) who sheded light on this subject.

I do get the lessons taught here. Looking up my number
thoery book (it's not too good at providing examples),
i see a more useful function called Euler's generization of
Fermat's function.

First, i would like to ask a question concerning the
previous question i just asked.

5^999,999 = x mod7
5^6 = 1 mod 7
so 5^999,999 = 5^3 * (5^6)^166,666 mod7

But is it alright to replace (5^6) with 1,
and still leave the ^166,666 above the 1?
does this work, in theory, if say 5^6 was
replaced with say n (just imagine it)?

5^999,999 = 5^3 * n^166,666 mod7

ok, now for euler's phi function,
It is the one, a^phi(m) = 1 (mod m)
with gcf of (a, m) = 1

But what if a is a factor of m (a | m) or vice versa?
Do you reduce one or the other? Consider this example.

2^32 = x mod6

what value k for the exponent will make
2^k = 1 mod6?

i don't think we can apply euler's function here, since
2 is a factor of 6.

2^2 = 1 mod3
2^0 = 1 mod2


Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.