Perfect Number ends in 6 or 28 preceded by Odd Digit

Theorem
Let $n$ be a perfect number.

Then $n$ ends either in $6$ or $28$ preceded by an odd digit.

Proof
By the Theorem of Even Perfect Numbers:
 * $n = 2^{p - 1} \paren {2^p - 1}$

where $p$ is prime.

With the exception of $6 = 2^1 \paren {2^2 - 1}$ and $28 = 2^2 \paren {2^3 - 1}$:
 * $p$ is odd and $p > 4$.

We claim that:
 * $n$ ends in $\phantom 0 6$ preceded by an odd digit if $p \equiv 1 \pmod 4$
 * $n$ ends in $28$ preceded by an odd digit if $p \equiv 3 \pmod 4$

These statements are equivalent to:
 * $n \equiv \phantom 0 16 \pmod {\phantom 0 20}$ if $p \equiv 1 \pmod 4$
 * $n \equiv 128 \pmod {200}$ if $p \equiv 3 \pmod 4$

By Powers of 16 Modulo 20, we have:
 * $2^{4 n} = 16^n \equiv 16 \pmod {20}$ for $n \ge 1$

Case $1$: $p \equiv 1 \pmod 4$
Write $p = 4 n + 1$.

Then:

showing our first claim.

Case $2$: $p \equiv 3 \pmod 4$
Write $p = 4 n + 3$.

By Powers of 16 Modulo 20, we can write $2^{4 n} = 20 K + 16$ for some $K \in \Z$.

Then:

showing our second claim.