Let the digit in unit’s place be ‘x’ and the digit in ten’s place be ‘y’.
|
Digit in tens place |
Digit in units place |
Number |
Sum of the digits |
Original number |
y |
x |
10y + x |
y + x |
Number obtained by interchanging the digits |
x |
y |
10x + y |
x + y |
According to the first condition,
A two digit number is 3 more than 4 times the sum of its digits.
10y + x = 4(x + y) + 3
∴ 10y + x = 4x + 4y + 3
∴ x – 4x + 10y – 4y = 3
∴ – 3x + 6y = 3
Dividing both sides by -3,
x – 2y = -1 …(i)
According to the second condition,
if 18 is added to the number, the sum is equal to the number obtained by interchanging the digits.
10y + x + 18= 10x + y
∴ x – 10x + 10y – y = -18
∴ – 9x + 9y = -18
Dividing both sides by – 9,
x – y = 2 ……(ii)
Subtracting equation (ii) from (i),
∴ y = 3
Substituting y = 3 in equation (ii),
x – y = 2
∴ x – 3 = 2
∴ x = 2 + 3 = 5
∴ Original number = 10y + x
= 10(3) + 5
= 30 + 5
= 35
The required number is 35.