>>12575499Yes. I'll give a proof for a 3-digit number, but this logic can be extended to any number of digits pretty intuitively:
A 3-digit number `abc` can be written in the form:
100a + 10b + c
subtracting the sum of its digits gives:
100a + 10b + c - (a + b + c)
= 99a + 9b
= 9(11a + b)
i.e. some number that has 9 as a factor
And, as you likely know from tricks for testing the divisibility of a number, any number thats digits add up to 9 is divisible by 9, and the converse.
? it is true for all natural numbers
Proof of this latter statement is left as an exercise to the OP.
