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hartnn
Fermat's Little Theorem. a^{p-1}=(corresponds to)1mod p using this, we have remainder when 3^100 is divided by 101 as 1. Is this correct ? Can i have few more examples where this theorem is applied but not so directly... @mukushla
@siddhantsharan , @sauravshakya
yes, that was direct use of the theorem....
i think we just use it directly
any other interesting topic or theorem in modular arithmetic ?
Wilson's theorem is interesting
(n-1)!=-1(mod n) how do we use it ?
If \(p\) is a prime, then \((p-1)!+1\) is a multiple of \(p\), that is\[(p-1)! \equiv -1 \ \ \text{mod} \ p\]
i will be glad if someone posts a link for a good reference in this topic...
FLT is also a^p = a (mod p) Show that the two formulations are equivalent.
divide by a on both sides? maybe.....
It's an iff type, so both ways......
idk....maybe multiply both sides by a to get a^p = a (mod p) but is it correct to do so ?
On the right track, sort of...
a^p = a mod p, then if a not = 0 mod p then can cancel by a to get a^(p-1) = 1 mod p That's the first part...
same for other part, right ?
Yes, more or less.....
FLT is also used in some combinatorial problems....
u have some practice/examples with those ?
OK, you can try this one: Say you have enough beads in n colours, how many different necklaces consisting of p beads can be made, where p is prime?
I guess I should tell you that you will need to think a bit in order to work it out, it is not straightforward....
Or I can tell you the answer and you can try to work it out.....
let me try....after few minutes, u give answer...
I think u need more than a few minutes....:-)
(n^p-n)/2p + (n^((p+1)/2) -n)/2 + n
It's an integer so the first term still incorporates FLT...
Do you want links only about FLT/Wilsons or congruences/modular stuff as well....
modular stuff as well...
Similar this http://www.math.sc.edu/~filaseta/gradcourses/Math780notes.pdf Or more detailed....
thanks, i'll go through it......and will ask u for any doubts.