About The Sum of Positive Divisors of $n$

Question

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?

Answer 1

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 $23$.

Answer 2

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, $\sum_{k=1}^{\infty}\left ( \left \lfloor \frac{n}{k} \right \rfloor-\left \lfloor \frac{n-1}{k} \right \rfloor \right )=2$