# Digit Root of $2^{m−1}(2^m−1)$ is $1$ for odd $m$. Why?

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...?

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The digital root of a number $n$ is the remainder of $n$ divided by 9. –  user38034 Sep 16 '13 at 20:54
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 $9$.