Iterated sums of the digits of a perfect number converge to 1

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).
Printed from the PrimePages <> © Chris Caldwell.