Problem 759

Solution
Let $n=p q r$, where $p, q$ and $r$ are distinct primes.
Its factors are $1, p, q, r, p q, p r, q r$ and $p q r$.
The mean of its factors is $\frac{1+p+q+r+p q+p r+q r+p q r}{8}$.
This can be factorised into $\frac{(1+p)(1+q)(1+r)}{8}$.
For this to not be an integer the number of factors of 2 in the numerator must be at most 2 .
This means at least one bracket must be odd, so one of the primes must be even. However, the only even prime is 2 , so one prime must be 2 and the other brackets are both even.

Therefore, neither of the other brackets can have a second factor of 2 and so be a multiple of 4 , so the primes cannot be 1 less than a multiple of 4 . The smallest two examples of this are 5 and 13.

The smallest value for $n$ is therefore $2 \times 5 \times 13=130$.