Talk:Linear congruence theorem
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Sorry to make something that seems easy a bit more confusing, but the steps taken to solve the system of congruences (the steps are known as the Chinese remainder theorem, by the way) only work if the bases (numbers being used as mod's) are relatively prime. In the example mod 4 and mod 6 are both used. Both these have a HCF of 2, not one, so they are not relatively prime. They can be decomposed into relatively prime congruences, but I'm not sure how to do that (I was searching for a way to do it when I stumbled across this page)
