Notice that $10$ is a triangular number. Thus, the desired $10$ numbers are of the form $p_1 p_2, p_1 p_3,p_1 p_4,p_1 p_5, p_2 p_3, p_2 p_4, p_2 p_5, p_3 p_4, p_3 p_5, p_4 p_5$ Thus, the answer is $\boxed{5}$

Indeed

I might be mistaken, but I think I have an example for $3$ primes. (clearly $2$ primes is not possible) Take: $2\mid a_{i}/\forall i\in[1; 9]$ $3\mid a_{j}/\forall j\in[2; 10]$ $5\mid a_{1}, 5\mid a_{10}$ And for them to be distinct just take different exponents. (Please correct me if I'm wrong)

Let's prove that the smallest number of distict prime factors is $3$. If there were only one prime factor, all the $10$ numbers should be multiples of it and then they wouldn't be all prime to each other, contradiction. If there were only two distinct prime factors, say $p$ and $q$, there should be a number of the form $p^{x}$ or $q^{x}$, otherwise all the numbers would be multiples of $pq$, contradiction. WLOG suppose the number is of the form $p^{x}$. Since the other numbers aren't prime to $p^{x}$, they all must be multiples of $p$, but then they wouldn't be all prime to each other, contradiction. Now let's give an example for $3$ primes. Let $p$, $q$ and $r$ be distinct prime numbers. The example of ten numbers $pq$, $qr$, $rp$, $p^{2}q$, $q^{2}r$, $r^{2}p$, $pq^{2}$, $qr^{2}$, $rp^{2}$, $pqr$ works. With $p = 2$, $q = 3$ and $r = 5$, we get $6$, $15$, $10$, $12$, $45$, $50$, $18$, $75$, $20$, $30$.