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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?

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    $\begingroup$ I think brute search is optimal, if a bit tedious. You'll find the answer fairly quickly. $\endgroup$ – lulu Dec 9 '18 at 13:35
  • $\begingroup$ Oh, thanks, my method sounds good though. $\endgroup$ – Maged Saeed Dec 9 '18 at 13:35
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    $\begingroup$ 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. $\endgroup$ – lulu Dec 9 '18 at 13:36
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    $\begingroup$ Correct again. $\,$ $\endgroup$ – lulu Dec 9 '18 at 13:39
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    $\begingroup$ (+1) for the posted solution, good work. $\endgroup$ – lulu Dec 9 '18 at 13:49
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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 $23$.
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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, $\sum_{k=1}^{\infty}\left ( \left \lfloor \frac{n}{k} \right \rfloor-\left \lfloor \frac{n-1}{k} \right \rfloor \right )=2$

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

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