Cat-Questions-Solutions

Question

LCM of three numbers is equal to 1080; HCF of the three numbers is equal to 2. How many such triplets are possible?

Explanatory Answer

HCF = 2. Let the numbers be 2x, 2y and 2z. HCF of (x, y, z) = 1.
LCM = 2xyz => xyz = 540
540 = 22 * 33 * 5

x = 22, y = 33, z = 5
x = 22 * 33, y = 5, z = 1
x = 22 * 5, y = 33, z = 1
x = 33 * 5, y = 22, z = 1
x = 22 * 33 * 5, y = 1, z = 1

These are the basic combinations. The first 4 yield 6 ordered triplets each, the last one yields 3 ordered triplets.

So, there are 5 unordered triplets and 27 ordered triplets

Correct Answer : 27