If G is a group not cyclic then its order can be: If G is a group not cyclic then its order can be: 
a)15
b)35 
c)77 
d)120 
e)2011
Well, i know that if G is not cyclic then it is not isomorphic to Zn, but i think it does not help much. 
Any tips? 
 A: Proposition Let $n$ be a positive integer. Then there is only one group of order $n$ if and only if gcd$(n,\varphi(n))=1$. 
Note that such a group must be necessarily cyclic. Except for 120 all other numbers satisfy this criterion. $S_5$ has order $120$ and is not cyclic.
A: Let $G$ be a group from order $pq$ which $p,q$ are distinct prime numbers and $p\lt q$ and more $q-1$ not divisible to $p$ then, $G$ is cyclic group. And we Know every group from order of a prime number, is cyclic. So, the only possible case is $120$.
A: Actually the term G is non-cyclic helps pretty much.It says that G is not isomorphic to Z/n.But one knows that $$ \mathbb Z_{pq}\cong\mathbb Z_p\times \mathbb Z_q, $$when p and q are coprime.Using this arguement you can easily reach at your ans which should be option D.
A: If this is a multiple choice question with one answer, then you can find it easily.
There is a nonabelian dihedral group of order $2n$ for ever $n\ge 3$
Or you can notice that if $p^2$ is a factor of the order of your group, then you can create a group of requisite order as a direct product including the non-cyclic component $\mathbb Z_p \times \mathbb Z_p$ (e.g. the non-cyclic group of order $4$)
Or you could know something about the symmetric group $S_5$ of order $120$
Any way, you know one answer. Are the others possible? Well the Sylow theorems deal with the first three (but there is a non-cyclic group of order $21$ because $7\equiv 1 \bmod 3$). And $2011$ is prime.
Moral: you can't really do group theory without having a good understanding of some standard and basic examples of groups.
