1
$\begingroup$

The Wiki page on Perfect numbers says:

[A]dding the digits of any even perfect number (except $6$), then adding the digits of the resulting number, and repeating this process until a single digit (called the digital root) is obtained, always produces the number $1$. For example, the digital root of $8128$ is $1$, because $8 + 1 + 2 + 8 = 19$, $1 + 9 = 10$, and $1 + 0 = 1$. This works ... with all numbers of the form $2^{m−1}(2^m−1)$ for odd integer (not necessarily prime) $m$.

How does that work?

Ok, this means for example perfect numbers (except $6$) never have a factor of $3$, but does this help...?

$\endgroup$
  • 4
    $\begingroup$ The digital root of a number $n$ is the remainder of $n$ divided by 9. $\endgroup$ – Jakube Sep 16 '13 at 20:54
5
$\begingroup$

Let $e = m-1$ be even. Then $2^e$ can be congruent to one of $1, 4, 7$ modulo $9$. Then $2^{e+1}-1$ is congruent to $2\cdot 1-1 = 1,\, 2\cdot 4 - 1 = 7,\, 2\cdot 7 - 1 = 13 \equiv 4$ modulo $9$, hence $2^e(2^{e+1}-1)$ is congruent to $1\cdot 1$ or $4\cdot 7 = 28 \equiv 1$ modulo $9$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.