Prove Sum of Digits Divisible by 9 iff Num is Div by 9

[RK1992]

Homework Statement

prove that the sum of a number's digits is a multiple of 9 if and only if it is divisible by 9

Homework Equations

none that i can think of

The Attempt at a Solution

say any real integer can be written as:
a_na_n-1...a₂a₁a₀
where a₀ is the units, a₁ is the tens digit and so on.

if we have the number x=a_na_n-1...a₂a₁a₀

we can say that x (mod 9) = 0 $$\Leftrightarrow$$ x divisible by 9

rewriting the number x:
a_na_n-1...a₂a₁a₀ = a_na_n-1...a₂a₁0 + a₀ = a_na_n-1...a₂00 + a₀ + 10a₁

x= 10ⁿa_n + 10^n-1a_n-1 + ... +100a₂ + 10a₁ + a₀

we can then break up the so 10a₁ = 9a₁ + a₁
and likewise the 100a₂ = 99a₂ + a₂

however, working mod 9, 9a₁=99a₂ = 0
so we can rewrite x (mod 9) as:
x (mod 9) = a_n + a_n-1 + ... + a₂ + a₁ + a₀

as we said that:
x (mod 9) = 0 $$\Leftrightarrow$$ x divisible by 9

we can say that:
a_n + a_n-1 + ... + a₂ + a₁ + a₀ = 0 (mod 9)
but 0 (mod 9) = 9k
so if:
x = a_na_n-1...a₂a₁a₀
is divisible by 9
a_n + a_n-1 + ... + a₂ + a₁ + a₀ = 9k


is this sound reasoning?

apologies if I've messed up with subscripts anywhere, i did check but its hurting my eyes :p

thanks

 

[micromass]

This sounds like a good proof!

 

[RK1992]

okay thanks :) i know its not rigourous but i think its watertight enough for the level I am working at

 

[micromass]

I think the proof IS rigourous enough. Much more rigour would be too much, in my opinion