Let a=6m where m is any integer since a is divisible by 6. (a+12)/3=(6m+12)/3=2m+4 which is also an integer. The remainder is 0.
Since n is divisible by 3, n can be written as 3a, where a is any integer and m can be written as 2b, where b is any integer. n*m+12=6ab+12 which is divisible by 2.
If a is divisible by 6 it is also divisible by 3 which when added to 12 which is also divisible by 3 will result in a number divisible by three which means remainder will be 0
if n is divisible by 3, m is divisible by 2 then n.m + 12 will always be divisible by 2 because any number multipled by a number mutiple of 2 will be always divisible by 2 and when 12 added to it will also be divisible by 2.
