The sum of two numbers is 135 and their H.C.F. is 27. If their L.C.M. is 162, the numbers are:
81, 54
H.C.F. of two numbers is 27.
So let the numbers be 27a and 27b
Now, 27a + 27b = 135
⇒ a+ b = 5 .........(i)
Also, product of two numbers = H.C.F. of the numbers × L.C.M. of two numbers
27a × 27b = 27 × 162.
⇒ ab = 6 ............(ii)
Now using the identity:
(a−b)2=(a+b)2−4ab
(a−b)2=52−(4×6)
⇒ a - b = 1 ..........(iii)
Solving (i) and (iii) we get
a = 3, b = 2
So, numbers are 27 × 3 = 81 and 27 × 2 = 54.