Perfect numbers adding up to 10 In aliquot sequences, am I right in saying that all perfect numbers when you add up the digits the result always add up to $10$ ( apart from the perfect no $6$?) if so, is this just a coincidence? 
Example: "$496\mapsto 4+9+6=19\mapsto 1+9=10$".

An even number is perfect if and only if it is of the form $N=2^n\cdot(2^{n+1}-1)$ and $2^{n+1}-1$ is prime (Euler).
This implies that $n+1$ is also prime (although this is not a sufficient condition for $2^{n+1}-1$ to be prime).
On the other hand, if $S(n)$ is the sum of the digits of $n$, then $S(n)\equiv n\pmod 9$. So we need to see if $2^n\cdot (2^{n+1}-1)\equiv 1\pmod 9$.
We have that $2$ is a primitive root modulo $9$, that is, the first power of $2$ that is $1$ in $\Bbb Z_9$ is $2^6$. Now write $n=6q+r$. Since $n+1$ must be prime, if we assume that $n+1$ is not $2$ or $3$, $r+1$ must be $1$ or $5$, so $r$ is $0$ or $4$. Putting all together we get
$$N\equiv2^r(2^{r+1}-1)=2^{2r+1}-2^r\equiv1\pmod 9$$
in both cases.
