# About The Sum of Positive Divisors of $n$

The question says:

Find the smallest positive integer $$n$$ so that $$\sigma(x)=n$$ has no solution, exactly two solutions, exactly three solutions.

I could not come up with a good way to solve this question other that trial and error. But I am questioning this method. Is there any better ideas?

• I think brute search is optimal, if a bit tedious. You'll find the answer fairly quickly. – lulu Dec 9 '18 at 13:35
• Oh, thanks, my method sounds good though. – Maged Saeed Dec 9 '18 at 13:35
• Absolutely, yes. And you are correct about $2,12$. The third part won't take you as long as you fear. Keep in mind $\sigma_1(n)≥n+1$ with equality only for primes. That makes it easy to truncate your search. – lulu Dec 9 '18 at 13:36
• Correct again. $\,$ – lulu Dec 9 '18 at 13:39
• (+1) for the posted solution, good work. – lulu Dec 9 '18 at 13:49

## 2 Answers

By trial and error, I have found that the solutions are:

• $$\sigma(x) = 2$$ has no solution.
• $$\sigma(x) = 12$$ has exactly two solutions that are $$6$$ and $$11$$.
• $$\sigma(x) = 24$$ has exactly three solutions that are $$14,15$$ and $$23$$.

The number of divisors of a natural number $$n$$ is given by $$\sigma(n) = \sum_{k=1}^{\infty}\left ( \left \lfloor \frac{n}{k} \right \rfloor-\left \lfloor \frac{n-1}{k} \right \rfloor \right )$$.

This may be useful when expanding the summation.

Note that, when $$n$$ is a prime number, $$\sum_{k=1}^{\infty}\left ( \left \lfloor \frac{n}{k} \right \rfloor-\left \lfloor \frac{n-1}{k} \right \rfloor \right )=2$$

• how did you come up with this formula? – Maged Saeed Dec 12 '18 at 2:03