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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 (15061559) 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 reappearssee for example, [Brooke1960] [Sondow1961] and [Gardner68].
 Theorem.
 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.
 Examples.

 28 ¬10 ¬ 1,
 496 ¬ 19 ¬ 10 ¬ 1, and
 8128 ¬ 19 ¬10 ¬ 1
 Proof.
 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 2^{p1}(2^{p}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, 2^{6}
= 1 (mod 9)), so modulo nine n is congruent to one of the three
numbers 2^{11}(2^{1}1), 2^{31}(2^{3}1),
or 2^{51}(2^{5}1), which are all 1 (mod 9).
