HCF is 9 so the 2 numbers are 9x, 9y where HCF(x,y) = 1 and x and y >0 and without loss of generality let x > y
product of the 2 numbers are 9xy = 270
or $xy = 30\cdots(1)$
sum of the 2 numbers are 9(x+y) = 99 or $x + y = 11\cdots(2)
we have $(x-y)^2 = (x+y)^2 - 4 * xy = 11^2 - 120 = 1$
so $x - y = 1\cdots(3)$
so we get x = 6 and y = 5 and numbers are $54, 45$
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