Q:

Prove that the sum of two multiples of 3 still a multiple of 3

If a and b are multiples of 3 and if I add a+b the result will be a multiple of 3.
How can I prove that is true?
First of all I know that multiples of 3 are : 3,6,9,12,15 etc.
I tried to find a counterexample but it did not work because for example 3+6=9, 9+12=21 and so on.
Also I know that Numbers are divisible by 3 if the sum of all the individual digits is divisible by 3.
I tried to use that and I started with:
let a and b be multiples of 3 , then 3a+3b=3n, but I am not sure how to continue,can anyone help me?
Thank you.

A:

If $a$ and $b$ are multiples of $3$ then there are integers $x,y$ such that $a=3x$ and $b=3y$ (by definition). Then
$a+b=3x+3y=3(x+y)$
The result follows.

A:

You need to start with a definition of "multiple".
Let $a$ be an integer. We call $a$ a multiple of 3 if and only if there is a $k\in \mathbb{Z}$ such that $a=3k$.
Let $a$ and $b$ be multiples of 3. Therefore there are $k, l \in \mathbb{Z}$ such that $a=3k$, $b=3l$. We have: $a+b=3(k+l)$. The sum of two integers is a integer, so $a+b$ does conform to our definition of multiple of 3, since $a+b=3(k+l)$.

A:

"let a and b be multiples of 3 , then 3a+3b=3n"
Not quite.  If $a$ is a multiple of $3$ then $a = 3n$ for some integer $n$.  We can skip the variable $a$ altogether and say "a multiple of $3$ is of the form $3n$ for some integer $n$.
So if $3n$ is a multiple of $3$, and $3m$ is a multiple of $3$ then you want to prove that:
To prove:  $3n + 3m$ is a multiple of 3.
And as $3n + 3m = 3(n+m)$ then $3n + 3m$ is a multiple of 3. (As it is $3$ times the integer (n + m).)
....
To be neat and to have style points:
An integer $n$ is a multiple of $3$ if  $n = 3k$ for some integer $k$.
So if $a$ and $b$ are two multiples of 3, then $a = 3n$ and $b = 3m$ for some integers, $n,m$.  Then $a + b = 3n + 3m = 3(n+m)$.  As $n + m$ is an integer, $a + b = 3(n + m)$  is a multiple of 3. 

