2026/037) GCD of 2472,1284 and a third number n is 12.If their LCM is $2^3* 3^2*5*103 * 107$.

 Because this is problem of CGD and LCM it makes sense to find prime factors of all numbers. 

Because GCD is 12 my approach is to mention it as product of 12 and other prime factors. let n = 12k

2472 = 12 * 206 = 12 * 2 * 103

1284 = 12 * 107

n =  = 12 *k

LCM = 12 * 2 * 3 * 5 * 103 * 107

Let us see that is  

After the 12 there is additional 2 and that comes from 2472 ( so there can be 0 or 1 2 in k)

There is an additional 3 and it has to come from k as it does not come from other numbers

There is an additional 5 and it has to come from k as it does not come from other numbers

There is a 103 in 2472 not in 1284 putting 0 or 1 103 shall not change GCD or LCM

There is a 107 in 1284 not in 2742 putting 0 or 1 107 shall not change GCD or LCM

So $n = 12k = 12 * 3 *5 *2^a * 103^b * 107^c = 180 * 2^a *103^b * 107^c$ where each of a,b,c is 0 or 1