If b=12a, where a and b are positive integers, then b can be multiple of which if the following integers.
A) 4
B) 6
C) 8
D) 12
E) 24
E) 48
\rm{b} = 2^2 \times 3^1 \times \rm{a} \ \text{ where a is from } 1.... +\infty
Now , plug in +ve integer values for a
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Answer are 4,6 and 12.
As per explanation they have taken b=k*x and b=12a (given)
Now they have taken k*x =12a
And then calculate the value for x =12a/k
We need to find x
Case 1: k=a ; x=12 
Case 2: k=2a ; x=6
Case 3: k=3a ; x = 4
Case 4: k= 4a ; x=3
Case 5 : k= 6a ; x= 2
Case 6 : k= 12a ; x =1
That’s how they have explained this which is beyond my “KEN”.
Is this a GregMat question ?
No it’s not. I was doing it from some third party and found this.
Even at first I was not able to digest the explanation. That’s why I put it here to check if it’s explained correctly or not.