The sum of the digits of a two-digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.

Let the tens and the units digits of the required number be x and y, respectively. 

Required number = (10x + y) 

x + y = 12 ……….(i) 

Number obtained on reversing its digits = (10y + x) 

∴ (10y + x) - (10x + y) = 18 

⇒10y + x – 10x – y = 18 

⇒9y – 9x = 18 

⇒y – x = 2 ……..(ii) 

On adding (i) and (ii), we get: 

2y = 14 

⇒y = 7 

On substituting y = 7 in (i) we get 

x + 7 = 12 

⇒ x = (12 - 7) = 5 

Number = (10x + y) = 10 × 5 + 7 = 50 + 7 = 57 

Hence, the required number is 57.

Let us consider,

One’s digit of a two digit number = x and

Ten’s digit = y

So, the number is x + 10y

By interchanging the digits,

One’s digit = y and

Ten’s digit = x

Number is y + 10x

As per the statement,

x + y = 12 ………. (1)

y + 10x = x + 10y + 18

y + 10x – x – 10y = 18

x – y = 2 …(2)

Adding (1) and (2), we have

2x = 14 or x = 7

On subtracting (1) from (2),

2y = 10 or y = 5

Answer:

Number = 7 + 10 x 5 = 57