Question

A two digit positive number is such that product of its digit is 6.If 9 is added to the number the digit interchange thier places, find the number

Answer 1

Step-by-step explanation:

two digit number xy

x*y=6

x-y+9=yx

put x=6/y

6/y-y+9=y*6/y

6-y²+9y/y=6

6+9y-y²=6y

y²-9y-6+6y=0

y²-3y-6=0

Answer 2

Answer:23

Step-by-step explanation:

Let the 2 digit number be xy.

Therefore the number so formed will be (10x+y).

By question,

xy=6......................(1)

(10x+y)+9=(10y+x)............(2)

10x+y+9=10y+x

Or,9x-9y+9=0

Or,x-y+1=0........(3)

Using equation (1) in (3) we get

(6/y)-y+1=0

Or,6-y^2+y=0

Or,y^2-y-6=0

Or,y^2-3y+2y-6=0

Or,y(y-3)+2(y-3)=0

Therefore y=3,-2

Given two digit number is positive so y= -2 value is neglected.

Therefore y=3

Now with y=3 we get

xy=6

Or,x×3=6

Or x=2.

Therefore, required two digit number is (xy)=23