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