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Find the smallest positive integer $n$ satisfying $\sigma(n) = 3n$.

I've tried for a week and didn't know how to start, please suggest.

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Hints:

  1. Note that such numbers with $\sigma(n)=3n$ are examples of abundant numbers.

  2. There is a formula to compute $\sigma(n)$, given the prime decomposition of $n$.

  3. An upper bound must be $n=120$ since $\sigma(120)=3\cdot 120$. Check that there is no smaller one.

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    To reach the number 120, since we know that 12 is the smallest abundant number, so we have to start from $2^2\cdot 3$ and adjust the exponent or add new prime $ p^k $ until we obtain $\sigma(n)=3n$, is my understanding correct ? – 2017-01-09
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    Yes, that is correct. It is the direct way. Lateron one can write this up more theoretically, of course. – 2017-01-09
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    Thank you very much, Dietrich Burde. – 2017-01-09