# the sum of two numbers is 384.What are the possible number if their is HCF 48?

*print*Print*list*Cite

Expert Answers

Student Comments

pramodpandey | Student

Let number be x and y ( positive integer).

So by given condition

`x+y=384` (i)

since HCF of x and y is 48

therefore

`x xx y=m(48)^2` (ii) m is an integer.

`x={m(48)^2}/y` (iii)

substitute x from (iii) in (i)

`{m(48)^2}/y+y=384`

`m(48)^2+y^2=384y` (iv)

let `y=48l`

`therefore ` (iv) reduces to

`m(48)^2+(48l)^2-384xx48l=0`

`l^2-8l+m=0` (v)

solving (v) for l and m ,we may get many solutions.

**m=16 ,l= 4 then y=192, x=192**

**m=15 ,l=3 then y=144 x=240**

**m=12 ,l= 2 then y= 96 ,x= 288**

**m=7 ,l=1 then y=48, x=336**