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: "$496\mapsto 4+9+6=19\mapsto 1+9=10$".
 A: For any odd $k$, $(2^k-1)2^{k-1}$ is congruent to $1 \bmod 9$, and all even perfect numbers are of this form (although in one case, $6$, $k$ is even). Unless it happens that, in forming the ultimate digit sum for some perfect number, an intermediate total hits on a higher power of $10$, say $100$, it's likely that $10$ will be reached in the digit-sum sequence.
Since the order of $2\bmod 9$ is $6$, you have essentially demonstrated the truth of the above proposition by testing $k=3,5,7$ already. 
A: It is certainly true that even perfect numbers $>6$ are all congruent to $1\pmod 9$. Thus if you keep on adding the digits together until you get a single digit, that digit will always be $1$.
To see this, note that, as per Euclid, even  perfect numbers have the form $n_p=2^{p-1}(2^p-1)$ for some prime $p$ such that $2^p-1$ is also prime.
Excluding $p=2$ (which gives us $6$) we may assume $p$ is odd. 
We note that $n_3=28$ which is indeed $1\pmod 9$.  Form now on we'll assume that $p>3$.
It is easy to verify that $2$ has order $6 \pmod 9$.  $p$ odd and $>3$ implies that $\equiv \pm 1\pmod 6$ and it is now a straightforward exercise to confirm that in both cases $n_p\equiv 1 \pmod 9$.
A: 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.
