To test for divisibility by three (or
nine), we often sum the digits and test the resulting sum. If the resulting
sum has several digits, then sum again. Tartaglia (1506-1559) pointed out that if
we repeatedly sum the digits of an even perfect
number (other than six), we always get one
[Dickson19]. This was later proved by H.
Novarese (1887) and regularly reappears--see for example, [Brooke1960] [Sondow1961] and [Gardner68].
If you sum the digits of any even perfect number (except 6), then
sum the digits of the resulting number, and repeat this process until
you get a single digit, that digit will be one.
28 ¬10 ¬ 1,
496 ¬ 19 ¬ 10 ¬ 1, and
8128 ¬ 19 ¬10 ¬ 1
Let s(n) be the sum of the digits of n. It is easy to
see that s(n) = n (mod 9). So to prove the theorem, we
need only show that perfect numbers are congruent to one modulo nine.
If n is a perfect number, then n has the form 2p-1(2p-1)
where p is prime (see theorem
one). So p is either 2, 3, or is congruent to 1 or 5 modulo
6. Note that we have excluded the case p=2 (n=6). Finally,
modulo nine, the powers of 2 repeat with period 6 (that is, 26
= 1 (mod 9)), so modulo nine n is congruent to one of the three
numbers 21-1(21-1), 23-1(23-1),
or 25-1(25-1), which are all 1 (mod 9).