147% Perfect

The numbers are powers of $3$ .

$\underline{PROOF:-}$

Let $N = a^{p}\times{b^{q}}\times{c^{r}}\times{d^{s}}$

Then sum of divisors are $\dfrac{{(a^{p+1} - 1)}\times{(b^{q+1} - 1)}\times{(c^{r+1}-1)}\times{(d^{s+1}-1)}}{{(a-1)}\times{(b-1)}\times{(c-1)}\times{(d-1)}}$ .

So now let N be $x = 3^{n}$ , then the sum of divisors = $\dfrac{3^{n+1}-1}{3-1}$ .

Replacing $x = 3^{n}$ ,

The sum of divisors $= \dfrac{3x-1}{2}$ .

I just started listing them and noticed that the numbers which satisfied these properties were only power of 3, then just summed up - 1+3+9+27+81+243+729=1093

These are simply powers of 3...