$16=2^4$ has five divisors. The next number to $16$ that has five divisors is $2^4\times 3=48$. So, then the least possible common difference could be $32$. Thankfully, that turns out to be one of the common differences that prove the first part. Indeed, any general term of the sequence is $16+k\times 32=2^4(1+2k)$ whose highest power of $2$ is $4$. So, $5$ divides the number of divisors.

The problem statement seems redundant since by finding a sequence with smallest common difference, one proves that such a sequence exist. The problem could have meant that an infinite number of these sequences exist which also holds since for any common difference $d \equiv 0 \pmod{32}$ each term in th sequence is s.t. $2^4$ is the highest power of $2$ that divides it.

On the other hand, why would they ask about finite arithmetic progression?