http://en.wikipedia.org/wiki/Fermat's_little_theorem

Take a number any number greater than 0. Call it Q. Take a prime number and prime number P greater than 0 and that does not divide Q (i.e. must be a remainder when Q/P). You are guaranteed that [Q^(P-1)]-1 is divisible by P(i.e. there is no remainder). Simply amazing, and if you dont think so, try to comeup with a theorem like that someday. Its more useful than your toothbrush.

Take a number any number greater than 0. Call it Q. Take a prime number and prime number P greater than 0 and that does not divide Q (i.e. must be a remainder when Q/P). You are guaranteed that [Q^(P-1)]-1 is divisible by P(i.e. there is no remainder). Simply amazing, and if you dont think so, try to comeup with a theorem like that someday. Its more useful than your toothbrush.