Proof/Disproof: ab = 3k for all b ∈ ℤ

[mateomy]

Prove or disprove:

There exists an integer "a" such that $ab\equiv\,0\,(mod 3)$ for every integer "b".

I know I can rewrite the above as $ab=3k$ for some k$\,\in\,\mathbb{Z}$, but other than that I'm not sure where to go. I realize that dividing any of the above will not necessarily result in an integer which contradicts the initial statement, but I'm sort of lost on the wording. Am I on the right path?

Thanks.

 

[scurty]

Try dividing both sides by 3 (so the right side is an integer). Do you see an obvious choice for a so that the left side is an integer or are there no choices?

 

[mateomy]

As long as a is a multiple of 3 it would work. I just don't know how to word that correctly.

 

[algebrat]

Here's how you might start the proof:

Indeed, let a=3, then for any b...