Perfect Number ends in 6 or 28 preceded by Odd Digit

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Theorem

Let $n$ be an even 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:

\(\ds n\) \(=\) \(\ds 2^{p - 1} \paren {2^p - 1}\)
\(\ds \) \(=\) \(\ds 2^{4 n} \paren {2 \times 2^{4 n} - 1}\)
\(\ds \) \(\equiv\) \(\ds 16 \paren {2 \times 16 - 1}\) \(\ds \pmod {20}\) Powers of 16 Modulo 20
\(\ds \) \(\equiv\) \(\ds 496\) \(\ds \pmod {20}\)
\(\ds \) \(\equiv\) \(\ds 16\) \(\ds \pmod {20}\)

showing our first claim.

$\Box$


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:

\(\ds n\) \(=\) \(\ds 2^{p - 1} \paren {2^p - 1}\)
\(\ds \) \(=\) \(\ds 4 \times 2^{4 n} \paren {8 \times 2^{4 n} - 1}\)
\(\ds \) \(=\) \(\ds 4 \paren {20 K + 16} \paren {8 \paren {20 K + 16} - 1}\)
\(\ds \) \(=\) \(\ds \paren {80 K + 64} \paren {160 K + 127}\)
\(\ds \) \(=\) \(\ds 12800 K^2 + 20400 K + 8128\)
\(\ds \) \(\equiv\) \(\ds 128\) \(\ds \pmod {200}\)

showing our second claim.

$\blacksquare$


Sources

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