Question

The sum of the digits of a two digit number is 8. If the number formed by reversing the digits exceeds the original number by 18 find the original number

Answer 1

Let the 10s digit be 'x' and 1s digit be 'y'
ATQ
x + y = 8 eq.1
The no. will be
10x + y
After reversing the digits the number will be
10y + x
ATQ
10x + y + 18 = 10y + x
9x + 18 = 9y
x + 2 = y eq.2

Putting value of 'y' from eq.2 in eq.1
x + (x+2) = 8
x + x + 2 = 8
2x + 2 = 8
x + 1 = 4
x = 3
Putting this value in eq.1
3 + y = 8
y = 5
So the original number
(10x + y) = 10(3) + 5
= 30 + 5
= 35