Proving grammar only generate strings that is multiple of 3 Hello I have an exercise for homework and I was hopping to get some hints in order to solve it.

num-&gt; 11 | 10 num' 01 | num 0 | num num
num'-&gt; 00 num' | 1 num' | ε

I need to prove that my grammar produces only strings that is multiply of 3.
What i know is:

11 is 3 so no problem there
adding 0 to the end of the num is irrelevant because it will be still dividable by 3.
same with num num
I know that is something about the 10 num' 01 pattern. (1001 is divided by 3)

I was thinking to try to prove it with sums (this is how i solve a previous exercise). However i am not sure that i can solve it that way.
Hint: num' generates (1) numbers of even length which are divisible by 3 and (2) numbers of odd length which equal 1 mod 3. For the rule num &#8594; 10 num' 01, consider the parity of the length of the number generated by num'.