Iterated sums of the digits of a perfect number converge to 1  (From the Prime Pages' list of proofs) Home Search Site Largest The 5000 Top 20 Finding How Many? Mersenne Glossary Prime Curios! Prime Lists FAQ e-mail list Titans Submit primes 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]. 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 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). Another prime page by Chris K. Caldwell