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Let $ \sigma (n) = \sum_{k|n}^{}{k} $. I need to solve $\sigma(n)=12$. Probably the following might be of use: if $n={p_1}^{a_1}{p_2}^{a_2}...{p_s}^{a_s}$ then $\sigma(n)=\frac{{p_1}^{a_1+1}-1}{p_1-1}\frac{{p_2}^{a_2+1}-1}{p_2-1}...\frac{{p_s}^{a_s+1}-1}{p_s-1}$.

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    $\begingroup$ What is the RHS $6$ or $12?$ $\endgroup$ – lab bhattacharjee Jun 13 '13 at 17:44
  • $\begingroup$ It is 12. I edited it. $\endgroup$ – mr.stealyourgirl Jun 13 '13 at 17:51
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If RHS $=12,$ as $1+p+p^2=7$ or $\ge 1+3+3^3=13,n$ must be square-free

So, $\sigma(n)=\prod(1+p_i)$ where $p_i$s are distinct prime divisors $(\ge2)$ of $n$

Also, $\sigma(n)\ge 1+n \ \ \ \ (1)$

The equality occurs if $n$ is prime i.e, here $n+1=12\implies n=11$ which is prime.

$(1)\implies n\le 11$

So, other the values of $n$ must be non-prime & square-free

As $2\le n\le 11,$ the values can be $6,10$

$\sigma(6)=\sigma(2\cdot3)=(1+2)(1+3)=12$

$\sigma(10)=\sigma(2\cdot5)=(1+2)(1+5)=18$


Alternatively,

As $n$ is square-free, the values of $1+p_i$s are $3,4,6,12$ as $\ge2$

If $1+p_1=3,1+p_2=4\implies n=p_1\cdot p_2=2\cdot 3=6$

If $1+p_1=6,1+p_2=2\implies p_2=1<2$ so this case does not arise

If $1+p_1=12, n=p_1=11$

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  • $\begingroup$ $p_i \le n$ as well. $\endgroup$ – Zackkenyon Jun 13 '13 at 17:53
  • $\begingroup$ @Zackkenyon, I think, that's too obvious to be mentioned $\endgroup$ – lab bhattacharjee Jun 13 '13 at 17:54
  • $\begingroup$ I sort of agree, but if it were obvious, then this person would need no help solving this particular case... $\endgroup$ – Zackkenyon Jun 13 '13 at 17:55
  • $\begingroup$ @Zackkenyon, you mean he could have played with the primes $<12$. But before that, we need to establish the square-free ness $\endgroup$ – lab bhattacharjee Jun 13 '13 at 17:57
  • $\begingroup$ no, I mean he could simply have tested the sum of the divisors of all of the numbers less than 12. $\endgroup$ – Zackkenyon Jun 13 '13 at 18:00

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