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 10y + x.

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

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

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

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

⇒ 10x + y + 10y + x = 99

⇒ 11x + 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 x and 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 

x - y = 3

⇒ y = 6 - 3

⇒ y = 3

Hence, the number is 10 × 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 × 6 + 3 = 63.

Note that there are two such numbers.