jmc

Let $k$ be the number of tiles. There are two cases: If $k$ has twenty divisors, then we can divide them into ten pairs, which gives us 10 ways to write $k$ as the product of two positive integers. Alternatively, if $k$ has 19 divisors, then $k$ is a square. So other than the square case, there are $(19-1)/2=9$ ways to write $k$ as the product of two positive integers, which gives us a total of $9+1=10$ ways.

If the prime factorization of $k$ is $p_1^{e_1}{p}_2^{e_2}\cdots p_n^{e_n},$ then the number of divisors of $k$ is $$(e_1+1)(e_2+1)\cdots (e_n+1).$$Note that $e_i\ge 1$ for each $i,$ so each factor $e_i+1$ is at least 2.

If $k$ has 19 divisors, then $k$ must be of the form $p^{18},$ where $p$ is prime. The smallest number of this form is $2^{18}=262144.$

Otherwise, $k$ has 20 divisors. We want to write 20 as the product of factors, each of which are least 2. Here are all the ways: $$20=2\cdot 10=4\cdot 5=2\cdot 2\cdot 5.$$Thus, we have the following cases:

(i). $k=p^{19}$ for some prime $p.$ The smallest such $k$ is attained when $p=2,$ which gives $k=2^{19}.$

(ii). $k=p{q}^9$ for distinct primes $p$ and $q.$ The smallest such $k$ is attained when $p=3$ and $q=2$ which gives $k=2^9\cdot 3.$

(iii). $k=p^3{q}^4$ for distinct primes $p$ and $q.$ The smallest such $k$ is attained when $p=3$ and $q=2,$ which gives $k=2^4\cdot 3^3=432.$

(iv). $k=pq{r}^4$ for distinct primes $p,$ $q,$ and $r.$ The smallest such $k$ is attained when $p=3,$ $q=5,$ and $r=2,$ which gives $k=2^4\cdot 3\cdot 5=240.$

Therefore, the least number of tiles Emma could have is $\boxed{240}$ tiles.