Common Factor Question

Problem:
Why isn't there an "earlier multiple" for the \(N=3\) case?
i.) \(3\) is a prime
ii.) \(3\) and \(10\) have no common factors
iii.) \(10\) is not prime
iv.) \(3\) is not a perfect squareExplanation:
For there to be an "earlier multiple" of \(3\) that makes a multiple of \(10,\) we would need a number less than \(30\) that is divisible by both \(3\) and \(10.\)
Unfortunately, \(3\) and \(10\) are relatively prime, so their least common multiple is \(30;\) hence, there is no such "earlier multiple." Thus, the correct answer is ii.).Don't \(3\) and \(10\) have a common factor of \(1?\)

Actually, that's a good question. Logically, you are right, but \(1\) is a special number that is a factor of all of integer numbers, since any number is divisible by \(1.\) So any two random integer numbers have a common factor of \(1.\) That is why when mathematicians say that "two numbers doesn't have common factors", they mean that those two numbers don't have any common factors except for \(1.\)