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

Question

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

hartnn · Answer

@siddhantsharan , @sauravshakya

hartnn · Answer

yes, that was direct use of the theorem....

anonymous · Answer

i think we just use it directly

hartnn · Answer

any other interesting topic or theorem in modular arithmetic ?

anonymous · Answer

Wilson's theorem is interesting

hartnn · Answer

(n-1)!=-1(mod n)
how do we use it ?

anonymous · Answer

If $p$  is a prime,  then  $(p-1)!+1$ is a multiple of  $p$, that is$$(p-1)! \equiv -1 \ \ \text{mod} \ p$$

hartnn · Answer

i will be glad if someone posts a link for a good reference in this topic...

anonymous · Answer

FLT is also a^p = a (mod p)

Show that the two formulations are equivalent.

hartnn · Answer

divide by a on both sides?
maybe.....

anonymous · Answer

It's an iff type, so both ways......

hartnn · Answer

idk....maybe multiply both sides by a to get  a^p = a (mod p)
but is it correct to do so ?

anonymous · Answer

On the right track, sort of...

anonymous · Answer

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...

hartnn · Answer

same for other part, right ?

anonymous · Answer

Yes, more or less.....

anonymous · Answer

FLT is also  used in some combinatorial problems....

hartnn · Answer

u have some practice/examples with those ?

anonymous · Answer

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?

hartnn · Answer

(mod(n/p))!

anonymous · Answer

I guess I should tell you that you will need to think a bit in order to work it out, it is not straightforward....

anonymous · Answer

Or I can tell you the answer and you can try to work it out.....

hartnn · Answer

let me try....after few minutes, u give answer...

anonymous · Answer

I think u need more than a few minutes....:-)

hartnn · Answer

ok, whats the answer ?

anonymous · Answer

(n^p-n)/2p  + (n^((p+1)/2) -n)/2 + n

hartnn · Answer

OMG!

anonymous · Answer

It's an integer so the first term still incorporates FLT...

anonymous · Answer

Do you want links only about FLT/Wilsons or congruences/modular stuff as well....

hartnn · Answer

modular stuff as well...

anonymous · Answer

Similar this http://www.math.sc.edu/~filaseta/gradcourses/Math780notes.pdf Or more detailed....

hartnn · Answer

thanks, i'll go through it......and will ask u for any doubts.