Fives everywhere

Let the positive integer is $n$ .It can be expressed as product of $5$ primes.So $n=p_1 p_2 p_3 p_4 p_5$ .

No of proper divisor is $5$ .So ,total divisors $5+1=6$ .Each power of primes is $1$ Then total number if facotrs will be $2^5=32$ .

$6=2 \times 3=1 \times 6$ .pwers of $2-1$ and $3-1$ will not be allowed because then it cannot be expressed of $5$ primes.So power $(6-1)$ will be accepted which also be expressed as product of 5 primes.Now $n=p^5$ .The smallest n will be $n=2^5=32$ which also satisfies third condition.