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DJMcMayhem
DJMcMayhem
3:16 PM
 
El'endia Starman
El'endia Starman
@DJMcMayhem Ermagherrrrrd...
 
DJMcMayhem
DJMcMayhem
3:45 PM
That's what I thought
 
 
2 hours later…
DJMcMayhem
DJMcMayhem
5:38 PM
So inspired by a recent CMC in TNB, I have a question that maybe you math nerds/geniuses could help with. 3 and 9 have interesting modular properties in base 10. The sum of the digits of any factor of 3 or 9 will always be another factor of 3 or 9. Now it's pretty obvious why this works for 9, because in base b=10, 9 is b-1. And this works for other bases (multiples of 4 in base 5 sum to factors of 4 for example).
But why does this work for 3? Is it because it's the root of b-1 or because it's the root of b-1 or because it's a factor?
 
Cows quack
Cows quack
$\sum_i a_i×b^i (\mod n)$ if we want that to equal $\sum_i a_i (\mod n)$ (so that the position of digits don't matter), then $b^i ≡ 1 (\mod n)$
when i=1, $b≡1 (\mod n)$ so $b-1 ≡ 0 (\mod n)$
we could test this for base 16 and n=3,5
 
Nathan Merrill
Nathan Merrill
@DJMcMayhem because it's a factor
 
Dennis
Dennis
The difference will be zero mod d whenever d is a divisor of b-1, which is a divisor of b^i-1.
 
Cows quack
Cows quack
5:59 PM
why do you subtract the digitsum from the number?
what is that supposed to represent?
 
Dennis
Dennis
a-b is zero mod d iff a and b are congruent mod d.
 
Cows quack
Cows quack
ah I see
 
Dennis
Dennis
So the above proves that a positive integer is congruent mod d to its digit sum in base b whenever d divides b-1.
 

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