a rwo digit positive no. is such that the product of its digitis 6.if 9 is added to the no. the digit interchange their places
Answers
Let the ones digit of the number be x and the tens digit of the number be 6/x.
The number = 10 × 6/x + x
= (x^2 + 60)/x
After interchanging,
The number = 10 × x + 6/x
= (10x^2 + 6)/x
It is given that if 9 is added to the no. the digit interchange their places.
Therefore, (x^2 + 60)/x + 9 = (10x^2 + 6)/x
= 9 = (10x^2 + 6)/x - (x^2 + 60)/x
= 9 = (10x^2 + 6 - x^2 - 60)/x
= 9 = (9x^2 - 54)/x
= 9x^2 - 54 = 9x
= 9 ( x^2 - 6 ) = 9x
= x^2 - 6 = x
= x^2 - x - 6 = 0
= x^2 + 2x - 3x - 6 = 0
= x ( x + 2 ) - 3 ( x + 2 ) = 0
= ( x - 3 ) ( x + 2 ) = 0
= x = 3 or x = -2
Now, the number = (x^2 + 60)/x
= (3^2 + 60)/3
= (9 + 60)/3
= 69/3
= 23
or
The number = (x^2 + 60)/x
= (-2^2 + 60)/-2
= (4 + 60)/-2
= 64/-2
= -32
Since the question has mentioned that it is a positive number, the number is 23
Ans. The number is 23