discrete mathematics - divisibility proof

For all integers $a$, there exists an integer $b$ such that $3$ divides $a+b$ and $3$ divides $2a+b$.

I think it is false and the negation will be: There exists an integer $a$ such that for all integers $b$, $3$ does not divide $a+b$ or $3$ doesn't divide $2a+b$.
How can I choose a to prove the negation?

Answer

Let $a=1$, if $3$ divides $1+b$ and $3$ divides $2+b$.

Then $3$ must divide $2+b-(1+b)=1$ which is a contradiction.