If $n$ is an even perfect number $ n> 6$ show that the sum of its digits is $\equiv 1 (\bmod 9)$ If $n$ is an even perfect number $ n> 6$ show that the sum of its digits is $\equiv 1 \mod 9$.
I know perfect numbers are of the form $(2^{p-1})(2^{p}-1)$. I have a few trials that I have done and they check out but as far as a proof I have no idea where to start since I have to use the digits in the number.

Note that if $\mathfrak{d}(n)$ is digit sum of $n$, then $$\mathfrak{d}(n) \equiv n \pmod{9}.$$ Hence we wish to show that the perfect number $n$ is congruent to $1$ modulo $9$. Now note that $$2^0 \equiv 1 \pmod{9}$$$$2^1 \equiv 2 \pmod{9}$$ $$2^2 \equiv 4 \pmod{9}$$ $$2^3 \equiv 8 \pmod{9}$$ $$2^4 \equiv 7 \pmod{9}$$ $$2^{5} \equiv 5 \pmod{9}$$ $$2^{6} \equiv 1 \pmod{9}$$ $$\vdots$$ This exponentiation repeats modulo $6$. As you noted, perfect numbers are of form $2^{p - 1}(2^p - 1)$.
1) $p \equiv 0 \pmod{6}$, then $2^{p - 1}(2^p - 1) \equiv 5 \cdot (1 - 1) \equiv \color{red}0 \pmod{9}$
2) $p \equiv 1 \pmod{6}$, then $2^{p - 1}(2^p - 1) \equiv 1 \cdot (2 - 1) \equiv 1 \pmod{9}$
3) $p \equiv 2 \pmod{6}$, then $2^{p - 1}(2^p - 1) \equiv 2 \cdot (4 - 1) \equiv \color{red}6 \pmod{9}$
4) $p \equiv 3 \pmod{6}$, then $2^{p - 1}(2^p - 1) \equiv 4 \cdot (8 - 1) \equiv 1 \pmod{9}$
5) $p \equiv 4 \pmod{6}$, then $2^{p - 1}(2^p - 1) \equiv 8 \cdot (7 - 1) \equiv \color{red}3 \pmod{9}$
6) $p \equiv 5 \pmod{6}$, then $2^{p - 1}(2^p - 1) \equiv 7 \cdot (5 - 1) \equiv 1 \pmod{9}$
So only cases that could be troublesome are $p \equiv 0,2,4 \pmod{6}$. It is well-known since Euclid that if $2^{p - 1}(2^p - 1)$ is an even perfect number, then $p$ is prime. But then we have $$p = 6k + a \text{, where } a \in \{0,2,4\} \implies p \text{ is composite}$$ which is a contradiction.
