Discrete Math - Logic and Proofs

Is there alternative proof method for my proof here?

or all integers a and b, if a is even and b is a multiple of 3, then ab is a multiple of 6.

I have proved this with contradiction

Given a is even it can be represented with 2k. (k is an integer)

b is multiple of 3 therefore it can be represented as 3p (p is an integer)

Ab = 2k* 3p = 6 kp

6kp/6=kp (it is multiple of 6)

What other method can be done and please explain. Thanks!