# The HCF of two numbers is 4, and the LCM of the two numbers is 60. How would I work this out?

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### 2 Answers

Denote the numbers by x and y. By the conditions 4 is a factor of x and x is a factor of 60. So x=4m for some natural m and 60=x*n for some natural n. Therefore

60=x*n=4*m*n, so 15=m*n and m is a factor of 15. There are not so many factors of 15: 1, 3, 5 and 15. This means that x=4m, i.e. x can be 4, 12, 20 or 60, and the same we can state about y.

Now simply try all variants:

x=4, y=4 -- no, LCM is 4, not 60.

x=4, y=12 -- no, LCM=12.

x=4, y=20 -- no, LCM=20.

x=4, y=60 -- yes, HCF=4 and LCM=60.

x=12, y=12 -- no, LCM=12.

x=12, y=20 -- yes!

x=12, y=60 -- no, HCF=12.

x=20, y=20 -- no, LCM=20.

x=20, y=60 -- no, HCF=20.

x=60, y=60 -- no, LCM=60.

The answer ( two variants): 4 and 60, 12 and 20. (60, 4 and 20, 12 are the same pairs).

Let the two numbers be a & b

Now, Product of the two numbers = product of the H.C.F & L.C.M of two numbers

Thus, a*b = 4*60

or, a*b = 240............(1)

Now, Since the H.C.F of a & b is 4

Thus, these numbers can be represented as :-

a = 4m & b = 4n........(2)

Using (2) in (1) we get

4a*4b = 240

or, 16m*n = 240

or, m*n = 15...........(3)

Now, the possible combinations of m & n can be (1,15) ; (15,1) ; (3,5) & (5,3)

Thus, a = 12 & b = 20 or vice versa

or, a = 4 & b = 60 or vice versa