20210220, 09:12  #1 
Feb 2021
1_{2} Posts 
Digit sums, of any value?
Hello,
I have not worked this out mathematically, I am simply studying primes in my spare time, but now as I look at the table of all of the Perfect numbers found it looks as though the digit sums of the digit sums add up to 10. Except for 6 and 28 this holds for the first 8 Perfect numbers: 28 the first digit sum = 2+8=10 496=4+9+6=19, 1+9=10 8128=8+1+2+8=19, 1+9=10 33550336=3+3+5+5+0+3+3+6=28, 2+8=10 Next digit sum=64, 6+4=10 Next digit sum=55, 5+5=10 Next digit sum=73, 7+3=10 Is there a rule to explain it? Could it be used to find more Perfect numbers? /Sincerely, Chris65 
20210220, 09:39  #2 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
1011111101101_{2} Posts 
And the digit sum of 10 is 1.
BTW: Repeated digit sum is just another way of computing the remainder when divided by 9. Or more succinctly: digit_sum^{n}(x) = x mod 9 
20210220, 09:48  #3 
Dec 2012
The Netherlands
11×151 Posts 

20210220, 09:53  #4  
"Robert Gerbicz"
Oct 2005
Hungary
2663_{8} Posts 
Quote:


20210221, 15:23  #5 
Nov 2016
101100000011_{2} Posts 
Code:
n possible values for even perfect numbers mod n 1 0 2 0 3 0 (only for 6), 1 4 0, 2 (only for 6) 5 1, 3 6 0 (only for 6), 4 7 0 (only for 28), 1, 6 8 0, 4 (only for 28), 6 (only for 6) 9 1, 6 (only for 6) 10 6, 8 11 1, 4, 6, 10 12 4, 6 (only for 6) 13 1, 2, 3, 6 (only for 6), 8 14 0 (only for 28), 6, 8 15 1, 6 (only for 6), 13 16 0, 6 (only for 6), 12 (only for 28) 17 1, 2, 3, 6 (only for 6), 11 18 6 (only for 6), 10 19 1, 2, 3, 6 (only for 6), 7, 9 (only for 28), 10, 15 20 6 (only for 6), 8, 16 21 1, 6 (only for 6), 7 (only for 28), 13 22 4, 6, 10, 12 23 1, 3, 5, 6, 9, 13, 15, 20 24 4 (only for 28), 6 (only for 6), 16 25 1, 3, 6, 11, 16, 21 (only for 496) 26 2, 6 (only for 6), 8, 14, 16 27 1, 6 (only for 6), 10 28 0 (only for 28), 6 (only for 6), 8, 20 29 1, 3, 4, 6, 7, 8, 16, 18, 26, 28 30 6 (only for 6), 16, 28 31 0 (only for 496), 1, 6, 27, 28 32 0, 6 (only for 6), 16 (only for 496), 28 (only for 28) 
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