LCM (a, b) = 23×53
LCM (b, c) = 24×53
LCM (c, a) = 24×53
Therefore prime factorization shows that only 2 and 5 are involved in combination of numbers.
Let a = 2p×5q
b = 2r×5s
c = 2t×5u
Calculation for powers of 2:
Above LCM's shows that 24 came from c thus t = 4. Now either p or r will take 3 and other will take 0, 1, 2, 3.
Total options for (p,r) = (0,3) (1,3) (2,3) (3,3) (3,0) (3,1) (3,2) = 7
Calculation for powers of 5:
Maximum power of 5 is 3. So any two of q, s, u have maximum power 3. and other will take 0, 1, 2, 3.
Total options for (q,s,u) = (0,3,3) (1,3,3) (2,3,3) (3,0,3) (3,1,3) (3,2,3) (3,3,0) (3,3,1) (3,3,2) (3,3,3) = 10
Total combinations = 7 x 10 = 70