Problem source

A \textit{proper factor} of a positive integer $N$ is an integer $M$, with $M\ne 1$ and $M\ne N$, which divides $N$ without remainder.
Show that $12$ has $4$ proper factors and $16$ has $3$.
Suppose that $N$ has the prime factorisation
\[N=p_{1}^{m_{1}}p_{2}^{m_{2}}\dots p_{r}^{m_{r}},\]
where  $p_{1}, p_{2}, \dots, p_{r}$ are distinct primes and $m_{1}, m_{2}, \dots, m_{r}$ are positive integers.
How many proper factors does $N$ have and why?
Find:
\begin{questionparts} 
\item the smallest positive integer which has
precisely 12 proper factors;
\item the smallest positive integer which has
at least 12 proper factors.
\end{questionparts}

Solution source

$12$ has factors $1,2,3,4,6,12$ of which $4$ are neither $1$ nor $12$.

$16$ has factors $1,2,4,8,16$ of which $3$ are neither $1$ nor $16$.

If $N = p_1^{m_1} \cdots p_r^{m_r}$ then $N$ has $(m_1+1)\cdots(m_r+1)$ factors since we can have between $0 \leq k \leq m_i$ of the $i$th prime factor, whcih is $m_i+1$ possibilities. We then need to subtract two for the proper factors, ie $(m_1+1)\cdots(m_r+1) - 2$.

\begin{questionparts}
\item We need $(m_1 + 1) \cdots (m_r + 1) - 2 = 12$ ie $(m_1+1) \cdots (m_r +1) = 14$. Since $14 = 2 \times 7$ we should have two prime factors, ie of the form $p_1^6p_2^1$. The smallest number of this form is $2^6 \cdot 3 = 192$

\item We wish to make $(m_1+1)\cdots(m_r+1) \geq 14$ with $2^{m_1} \cdot 3^{m_2} \cdots p_r^{m_r}$ as small as possible and $m_1 \geq m_2 \geq \cdots \geq m_r$.

With $1$ prime, the best we can do is $2^{13}$.
With $2$ primes the best we can do is $2^p \cdot 3^q$ with $(p+1)(q+1) \geq 14$. 

\begin{array}{c|c} 
q & p & N \\ \hline
0 & 13 & 2^{13} \\
1 & 6 & 2^6 \cdot 3^1 \\
2 & 5 & 2^5 \cdot 3^2 \\
3 & 3.& 2^3 \cdot 3^3
\end{array}

So the best here is $2^6 \cdot 3$

With $3$ primes, the best we can do is $2^p 3^q 5^r$ with $p \geq q \geq r$ and $(p+1)(q+1)(r+1) \geq 14$

\begin{array}{c|c|c|c} 
r & q & p & N \\ \hline
1 & 1 & 3 & 2^3 \cdot 3 \cdot 5 \\
1 & 2 & 2& 2^2 \cdot 3^2 \cdot 5 \\
2 & 2& 2 & 2^2 \cdot 3^2 \cdot 5^2
\end{array}

With four primes we can achieve it immediately with $2\cdot3\cdot5\cdot 7$ and more primes clearly wont help, so our best options is $120$ 

\end{questionparts}