#### Question

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

#### Solution

Let the digit at the tens place be ‘a’ and at units place be ‘b’.

The two-digit so formed will be 10a + b.

According to the first condition, the product of its digits is 6.

⇒ a x b =6

`=> x = 6/b` ...(1)

According to second condition

10a + b + 9 = 10b + a

`⇒ 9a - 9b = 9

`=> a - b = 1`

`=> a - 6/a= 1` From 1

`=> a^2 - a - 6 = 0`

`=> (a - 3)(a + 2) = 0`

`=> a= -3 or 2`

Since a digit cannot be negative, a = 2.

`=> b = 6/a = 6/2 = 3`

Thus, the required number = 10a + b = 10(2) + 3 = 23

Is there an error in this question or solution?

#### APPEARS IN

Solution A Two Digit Positive Number is Such that the Product of Its Digits is 6. If 9 is Added to the Number, the Digits Interchange Their Places. Find the Number. Concept: Factorising a Polynomial Completely After Obtaining One Factor by Factor Theorem.