Consecutive number

For $n(n+1)$ to have exactly 10 positive divisors, it must equal to $pq^4$ , where $p$ and $q$ are primes. That is $n(n+1) = pq^4$ . We note that $n(n+1)$ is always even, then $pq^4$ must also be even. Since 2 is the only even prime, either $p$ or $q$ is 2. If we assume $n+1=p$ and $n=q^4$ , then $p=n+1=q^4+1$ . If we assume $q=2$ , then $p=2^4+1=17$ , which is a prime.

$\implies$ Yes, it is possible that the number can have exactly 10 positive divisors.

Here is an example: $n=16$ .

It is easy to see that $n$ or $n+1$ has to be a prime number, and the other number has to be fourth power of an integer.