Relaxation Problem Number 1

Let the number of factors of $n$ be denoted $\sigma_{0}(n)$ . Turns out that for $\displaystyle n = \prod_{i} p_{i}^{a_{i}}$ (Fundamental Theorem of Arithmetic), where $p_{i} < p_{i+1}$ are both primes and $a_{i} = 0, 1, 2, \ldots$ , $\displaystyle \sigma_{0}(n) = \prod_{i} (a_{i}+1)$ .

Three possibilities:

$p \neq q \neq r \longrightarrow \sigma_{0}(pqr) = \sigma_{0}(p^{1}q^{1}r^{1}) = (1+1)(1+1)(1+1) = 8$

$p \neq q = r \longrightarrow \sigma_{0}(pqr) = \sigma_{0}(p^{1}q^{2}) = (1+1)(2+1) = 6$

$p = q = r \longrightarrow \sigma_{0}(pqr) = \sigma_{0}(p^{3}) = 3+1 = 4$

Their sum is then $8+6+4 = \boxed{12}$ .