The correct answer is 8.

Step-by-Step Explanation:

We are given that a natural number $N$ can be expressed as the sum of three distinct factors $p$, $q$, and $r$ of $N$:

$N=p+q+r$

Without loss of generality, assume $p<q<r<N$. Dividing both sides of the equation by $N$, we get:

$\frac{p}{N}+\frac{q}{N}+\frac{r}{N}=1$

Since $p$, $q$, and $r$ are factors of $N$, their reciprocals relative to $N$ must be integers. Let:

$a=\frac{N}{r}$, $b=\frac{N}{q}$, $c=\frac{N}{p}$

Where $1<a<b<c$ are distinct positive integers. Substituting these into our equation yields:

Finding integer solutions for :

1. If $a\ge 3$, then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$, which is impossible. Thus, we must have $a=2$.

2. Substituting $a=2$ gives:

$\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$

3. If $b=3$, then $\frac{1}{c}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\Rightarrow c=6$.

4. If $b\ge 4$, then $\frac{1}{b}+\frac{1}{c}<\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$, which yields no distinct integer solutions.

Therefore, the unique set of values is .

Finding the numbers $N$ below 50:

This implies that the factors are $r=\frac{N}{2}$, $q=\frac{N}{3}$, and $p=\frac{N}{6}$. For these factors to be integers, $N$ must be a multiple of $6$.

Now, let's list all multiples of $6$ strictly less than $50$:

6, 12, 18, 24, 30, 36, 42, 48

Counting these values gives a total of 8 numbers below 50 that satisfy the condition.