Have I found all the numbers less than 50,000 with exactly 11 divisors?

If you're looking at a number with prime factorization $$ p_1^{a_1}p_2^{a_2}\cdots p_n^{a_n} $$ (note, there are no $c$s in a prime factorization), then it has exactly $(a_1+1)\cdot(a_2+1)\cdots(a_n+1)$ divisors.

Since $11$ is prime, the only way it can be written as such a product is just $(10+1)$. So you're looking for numbers of the form $p^{10}$ -- and only that form -- less than $50,000$.

If $n$ has a prime factorization of $p_1^ip_2^jp_3^k$, it has $(i+1)(j+1)(k+1)$ divisors. The numbers with exactly 11 divisors are of the form $p^{10}$ since 11 is prime.

Since $3^{10} > 50,000$, the only answer is $1024$.

To flog a horse consider how many factors $p^k $ have. The factors are:

$1,p,p^2,p^3,.......,p^k$

That's $k+1$ factors.

How many factors does $p^kq^m$ have?

$1,p^2,p^3,......,p^k$

$q,qp,qp^2,qp^3,......,qp^k $

$q^2,q^2p,q^2p^2,q^2p^3......,q^2p^k$

......

......

$q^m,q^mp,q^mp^2,q^mp^3.....,q^mp^k$

That is $(k+1)(m+1)$ total factors.

Now how many factors does $\prod p_i^{k_i} $ have?

Well, each $p_i$ prime has factors for every power from 0 to $k_i $. That's $k_i +1$ factors. The each of those factors can be multiplied by the powers of any other {1,$p_j,p_j^2,...p_j^{k_j} $. The total number of combinations is therefore. $\prod (k_i +1) $.

So if $c = \prod p_i^{k_i} $ then $c $ has $\prod (k_i+1) =11$ factors. as 11 is prime, 11 = $k+1$ so $c=p^{10}$ for some prime $p $.

The only such number in range is $c=2^{10}=1024$ that has 11 factors: 1,2, 4 ,8,16,32,64,128,256,512,1024.