jmc

We need to find the smallest integer, $k,$ that has exactly $10$ factors. $10=5\cdot 2=10\cdot 1,$ so $k$ must be in one of two forms:

$\bullet$ (1) $k=p_1^4\cdot p_2^1$ for distinct primes $p_1$ and $p_2.$ The smallest such $k$ is attained when $p_1=2$ and $p_2=3,$ which gives $k=2^4\cdot 3=48.$

$\bullet$ (2) $k=p^9$ for some prime $p.$ The smallest such $k$ is attained when $p=2,$ which gives $k=2^9>48.$

Thus, the least positive integer with exactly $10$ factors is $\boxed{48}.$