Question

The HCF of two number is one sixth of the LCM of those number. I If one of the number is 15, Find the other?

Answer 1

Let the L.C.M of two numbers 'a' & 'b' be 'X'
H.C.F(a,b) = 1/6 of L.C.M = X/6
a = 15
We know,
L.C.M*H.C.F = Product of numbers
X * X/6 = 15*b
X^2 = 90*b = 9*10*b
X = root(9*10*b)
X = 3*root(10*b)
['b' has to be such a value that 'X' becomes a natural number. so 'b' can be 10, 40, 90 etc...Smallest value starts from '10']
If 'b' = 10, X = 30
a=15, b = 10
L.C.M (15,10) = 30
H.C.F (15,10) = 5, which is 1/6 of L.C.M
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[check: if b=40, X = 60 = L.C.M. As per condition H.C.F should be 10. a=15, b= 40. L.C.M (15,40)= 120, H.C.F(15,40) = 5.Hence condition not satisfied]
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So, the other number = 10
Hope it helps.