Edit on March 2, 2018: I just noticed that this is almost identical to a question asked on MO by David Harden in 2011, and that I had even given an (incomplete) answer to that one. I would delete the question, but I think this can't be done when an answer has been accepted ( I did try).

The title says it all really. I know that in some "small" simple groups such as ${\rm PSL}(2,p)$ it is known that if the order is divisible by $60,$ then $A_{5}$ does occur as a subgroup ( this is because sufficient control of the character table is available to exhibit an involution and an element of order $3$ whose product has order $5$). However, at the moment, I don't see a general strategy for attacking this question ( or similar ones), though of course it seems likely that some use of the classification of finite simple groups would be needed.

Later edit: It is a little late for motivation, and I did not realise it when I asked the question, but it seems to follow from the affirmative answer outlined by Derek Holt that a finite group $G$ is $5$-solvable (ie has all composition factors of order $5$ or of order prime to $5$) if and only if every subgroup of $G$ has a Hall $\{2,5\}$-subgroup. Recall that a Hall subgroup of $G$ is one whose order and index are relatively prime.

Here is an outline of the difficult half of the proof: suppose then that every subgroup of $G$ has a Hall $\{2,5\}$-subgroup, but that $G$ is not $5$-solvable, and $G$ is chosen to be of minimal order subject to this. Note that every proper subgroup of $G$ is $5$-solvable. I claim that $G/F(G)$ is non-Abelian simple. For let $M$ be a maximal normal subgroup of $G$, and suppose that $M \neq 1$. Let $p$ be a prime divisor of $|M|$ and let $P$ be a Sylow $p$-subgroup of $M$. Then $G = MN_{G}(P)$ by a Frattini argument. If $N_{G}(P) < G$ then it easily follows that both $M$ and $G/M$ are $5$-solvable, and then $G$ is, contrary to assumption. Hence $P \lhd G$. Since $p$ as arbitrary, $M$ is nilpotent. Hence $M = F(G)$ is the unique maximal normal subgroup of $G$, and $G/M$ is non-Abelian simple. Now it easy to check that every subgroup of $G/M$ has a Hall $\{2,5\}$-subgroup, since $G$ has that property.

Now, however, $G/M$ can't have order divisible by $60$, using the answer to this question (otherwise, it has a subgroup isomorphic to $A_{5}$, but $A_{5}$ has no subgroup of order $20$). However, $G/M$ must have order divisible by $5$, otherwise $G$ would be $5$-solvable. Also, by Feit-Thompson, $G/M$ has order divisible by $4$. Hence $G/M$ has order prime to $3$, so that $G/M$ is a Suzuki group. However, no simple Suzuki group has a Hall $\{2,5\}$-subgroup, a contradiction.