Quote:
Originally Posted by science_man_88
I know (a*b) mod x = (a mod x * b mod x)mod x, I've thought about addition and a few others as well looks like it holds for other operations.

Quote:
Originally Posted by 3.14159
Operations using modular arithmetic;
(a mod x + b mod x) mod x = (a + b) mod x.
(a mod x * b mod x) mod x = (a * b) mod x.

Yes. But it's sometimes useful to reduce only partially: (a  Ax)(b  Bx) mod x = a * b mod x. This is used in some efficient algorithms where fully reducing at each step would be more costly.