no simple group of order $945$ I need to show that there are no simple groups of order $945$.

I've tried the regular method using the Sylow theorems.

$$|G|=945=3^3\cdot5\cdot7 $$

If $G$ is simple then there should be 7 Sylow-3 groups ; 21 Sylow-5 groups and 15  Sylow-7 groups. Even if they would all intersect trivially, there will still be no contradiction. 

Any ideas?
If as you say there are 7 Sylow 3-subgroups, then the normalizer $N$ of one of these subgroups has index 7.  If $G$ is simple, then there is an injective permutation representation $G \hookrightarrow S_7$, and so $|G| = 945$ must divide $7!$, but this is not the case.
I think every simple group has even order.