Question

The sum of a two digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.

Answer 1

Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is `10 y + x`.

The two digits of the number are differing by 3. Thus, we have ` x - y = +-3`

After interchanging the digits, the number becomes ` 10 x + y`.

The sum of the numbers obtained by interchanging the digits and the original number is 99. Thus, we have

` ( 10 x + y)+ (10 y + x)=99`

` ⇒ 10 x + y + 10 y + x = 99`

`⇒ 11 x + 11y = 99 `

` ⇒ 11( x + y )= 99`

` ⇒ x + y 99/11`

` ⇒ x + y =9`

So, we have two systems of simultaneous equations

` x- y = 3`

` x + y = 9`

` x - y = -3`

` x + y = 9`

Here x and y are unknowns. We have to solve the above systems of equations for xand y.

(i) First, we solve the system

` x - y = 3`

` x + y = 9`

Adding the two equations, we have

` ( x - y) + ( x + y) = 3+ 9`

` ⇒ x - y + x + y = 12`

` ⇒ 2x = 12`

` ⇒ x = 12/2`

` ⇒ x = 6`

Substituting the value of x in the first equation, we have 

` 5 - y = 3`

` ⇒ y = 6-3`

` ⇒ y = 3`

Hence, the number is`10 xx 3 + 6 = 36`.

(ii) Now, we solve the system

` x - y = -3`

`x+y=9`

Adding the two equations, we have

`(x -y)+(x+y)= -3 +9`

` ⇒ x - y + x+ y = 6`

`⇒ 2x = 6`

` ⇒ x = 6/2`

` ⇒ x = 3`

Substituting the value of x in the first equation, we have

` 3-y = -3`

`⇒ y = 3 + 3 `

` ⇒ y = 6`

Hence, the number is `10 xx6 + 3 = 63`.

Note that there are two such numbers.