Converse of "$3|a \ and \ 3|b \implies 3|(a+b)$"

Question

Here is the implication and my proof :

$3|a, 3|b \implies 3| a+b$

Proof: $a = 3k, b = 3k \implies a + b = 6k = 3(2k)$ and

$\dfrac{3(2k)}3 = 2k$, an integer, which proves the implication.

(should I've used two different variables for instead of k?)

I'm not sure if this is correct but based on this I think that the implication is true. But then I don't know how to prove the converse of this implication which is

$3|a+b\implies 3|a \ and \ 3|b $

I thought maybe I could use the same method but I have trouble finding the term for the first predicate.

Answer 1

if $a=3k+2$ and $b=3j-2$
then $ \qquad 3 \not | a \qquad $ and $ \qquad 3 \not | b$
but
$ \qquad a+b=3(k+j) +(2-2)=3(k+j)$
and
$ \qquad 3 | 3(k+j) \implies 3 | (a+b)$