Well, there's also 15, but you can rule that out, because...?
Suppose that R is a commutative ring and |R|=30. If I is an ideal of R and |I|=10, prove that I is a maximal ideal.
Im pretty sure I understand why its maximal, its because if there was an ideal that properly contained I then the order would be greater than 10 but the order of all proper sub-rings of R would be positive divisors which stop at 10. ok my problem is proving this elegantly on paper... any thoughts. thanks