About the property of $m$: if $n < m$ is co-prime to $m$, then $n$ is prime The number $30$ has a curious property:
All numbers co-prime to it, which are between $1$ and $30$ (non-inclusive) are all prime numbers!
I tried searching(limited search, of course) for numbers $\gt 30$ that have this property, but could not find any.
Are there any such numbers $\gt 30$?

Let $n>30$ be such a number. Observe that $p^2<n$ implies $p|n$.
Thus $2|n$, $3|n$, $5|n$ follow directly from $n>25>9>4$.
Thus $n$ is a multiple of $30$ and $n>30$, which implies $n\ge 60>49$ and hence $7|n$. 
Up to now, $n$ is divisible by the four smallest primes.
Lemma. The product of four consecutive primes is greater than the square of the next prime:
$$\tag1p_kp_{k+1}p_{k+2}p_{k+3}> p_{p+4}^2.$$
Proof:
Using Bertrand's postulate, we have $p_{k+2}>\frac{p_{k+3}}2$, and $p_{k+4}\le 2p_{k+2}$, hence $(1)$ follows from
$$p_kp_{k+1}p_{k+2}p_{k+3}>3\cdot 5\cdot \frac{p_{k+3}}{2}p_{k+3}> 4p_{k+3}^2>p_{k+4}^2$$
at least if $p_k\ge 3$. The case $p_k=2$ is verified directly: $2\cdot3\cdot 5\cdot 7>11^2$. $_\square$
Now if $n$ is divisible by  four consecutive primes $p_k,\ldots,p_{k+3}$, we conclude from $(1)$ that $n$ is also divisible by the next prime, because otherwise $p_{k+4}^2<n$ is relatively prime to $n$. Hence $n$ is divisible by the four consecutive primes $p_{k+1},\ldots,p_{k+4}$.
By induction, we conclude that $n$ is divisible by all primes, but that is of course absurd.
Hence there is no such number beyond $30$.
