Originally Posted by

**math2011** Let $\displaystyle \sigma(n)$ be the sum of all divisors of $\displaystyle n$. See

**sum-of-divisors function **$\displaystyle \sigma_1$ definition in

https://en.wikipedia.org/wiki/Divisor_function.

Solve $\displaystyle \sigma(n) = 48$.

The answers are $\displaystyle 33,35,47$.

What are the strategies for solving this equation?

I can see that since $\displaystyle \sigma(n) = \Pi^s_{k=1} \frac{p_k^{\alpha_k + 1} - 1}{p_k - 1} = 48$, but there are too many factorizations of $\displaystyle 48 = 2^4 \times 3$ to try.

If you are allowed to use computer then this simple Maple program solves your problem :

Code:

with(numtheory):
for n from 1 to 200 do
if sigma(n) = 48 then
print(n);
end if;
end do;