If a is odd, how do I prove that 3a is also odd? I know that if something is odd then $\exists k \in \mathbb{Z}: a = 2k + 1$. But what I get is: 

$n = 2k + 1$ 

$3n = 3(2k+1)$ 

$3n = 6k + 3 $

But i can't factor 6k + 3 to give me 2k + 1 ! 

Any suggestions?
If you factor out a 3, $(6k+3)$ $=$ $3(2k +1) = 2(3k + 1) + 1.$ This is now an odd number. You are here in your post, try thinking about taking an odd number and multiplying it by 3. You get an odd number in any example you try. This is because an odd integer multiplied by an odd integer is always odd, the proof of this is another good exercise.
If you factor out a 3, $(6k+3)$ $=$ $3(2k +1) = 2(3k + 1) + 1.$ This is now an odd number. You are here in your post, try thinking about taking an odd number and multiplying it by 3. You get an odd number in any example you try. This is because an odd integer multiplied by an odd integer is always odd, the proof of this is another good exercise.