# Homework Help: Simple but annoying characteristic of congruences

1. Feb 25, 2013

### trap101

I've been having this small problem with modulo arithmetic that's just irking me and this is the perfect example to get clarifivation

find the remainder for 75 congruent to x (mod 21)

here's my issue....so I'd set this up to solve and see if I could find this relationship

7$\equiv$ x (mod 21)........now 7 is a multiple of 21, but of course 21 doesn't divide 7 unless we take into account the negative numbers which would produce 7 $\equiv$ -12 (mod 21)...........is this the right way to look at that?

the next thing I did and which made it easier was look at 72 $\equiv$ 7 (mod 21) .....now I have 75 $\equiv$ 7 (mod 21) when all is said and done....but the fact that I know 7 has a relationship with 21 is sitting uneasy with me...

Last edited: Feb 25, 2013
2. Feb 25, 2013

### Dick

Don't know what's sitting uneasy with you. Sure, if 7^2=7 (mod 21) then 7^5=7 (mod 21).

3. Feb 25, 2013

### trap101

Well it has to do with an RSA encoding question, in order to encode my message which in this case is 7, I have to send the remainder when I raise 7 to the power of 5.........which ended up being 7. Probably I'm just over thinking it