Problems & Puzzles: Puzzles Puzzle 35. 1999 and the perfect numbers Tom Moore (5/1/99) made me notice that 1999 is the least prime number such that the sum of its digits is a perfect number (28). “SOD(1999) = 28 = perfect number” A perfect numbers is equal to the sum of all of its divisors, excluding itself. Vgr. 6 =1 + 2 + 3. The first 5 perfect numbers are: 6, 28, 496, 8128 and 33550336. Of course it does not exists any prime whose sum of digits is 6 (because it will be divisible by 3, and then a composite). On the other hand it is already known that the least prime whose sum of digits is 28 is 1999. a) Can you find the following 10 primes (apart from 1999) containing only two different kind of digits and whose SOD is 28? b) Can you find the least primes (or pseudo prime) whose SOD is 496, and 8128? c) Last: Any idea why all (Perfect Number>6)@9=1 (CBRF observation)? Solution Robert T. McQuaid (18/1/99) has solved a) and b): a) 9199, 338383, 383833, 1181881, 1881181, 1881811, 2222929, 2922229, 8118181, 8188111 b) SOD(2999999999999999999999989999999999999999/ 9999999999999999) = 496 The same solution was found by J .R. Howell independently one day after. c) SOD(3*10^903110^858) =8128 Trevor Green (18/1/99) has solved c). Here is his answer: "Of course any even perfect
number P can be written as P = (1/2)(M)(M+1), where M is
a Mersenne prime (2^p  1). Observing that 1/2 = 5 (mod
9), and that 2^p  1 = 1 or 4 or 7 (mod 9), for all odd
p; Then P = 5(M)(M+1) (mod 9), where M = 1 or 4 or
7. In each of these cases, P = 1 (mod 9). Of course
for p = 2 this does not hold, and consequently 6 != 1
(mod 9). However, this must hold for all other perfect
numbers. For odd perfect numbers, who can
say?" J .R. Howell's (19/1/99) explanation is this: "Why is every even perfect number (other than 6) == 1 mod 9?. Solution: Perfect number = (2^(p1)) * (2^p  1). Notice that the second factor (2^p  1) is (two times the first factor) minus 1. Notice also that 2^odd == 2 mod 3 and 2^even == 1 mod 3. If "p" is odd (true for all perfect numbers except 6): 2^(p1) = 3k+1 for some integer k. (Because (p1) is even) (2^p)1 = 2(3k+1)1 = 6k+1.Perfect number = (3k+1)*(6k+1) = 18k^2 + 9k + 1, which is 1 mod 9." 




