HCF and LCM

The product of 2 numbers and its HCF is 1080. How many such pair of numbers are possible which satisfy the above condition?




Answer:

let the numbers be p and q.

Let p = ha and q = hb where h = HCF

So pq.HCF = h3(ab) = 1080 = (23)(33)(5).
We need to write 1080 as a product of a perfect cube and another number.

Now let us substitute  the values of h as 1,2,3,4,5,6 and find p and q.

1. h = 1, ab = 1080. We have 4 pairs (1,1080), (8, 135), (27, 40) and (5, 216). 
2. h = 2, (ab) = (33)(5). This can be done in two ways - (2, 270), (54,10)
   number of pairs = 2, 
3. h = 3, (ab) = (23)(5). This can be done in two ways - (3, 120), (24,15)
   number of pairs = 2,
4. h = 4 or 5, the LHS and RHS of equation cannot be same, so h = 4, 5 cannot exist. 


5. h = 6, number of pairs = (6,30)

Hence total pairs of (p,q) = 9,