When a two-digit number is divided by the sum of its digits, the quotient is 8. On diminishing the ten’s digit by three times the unit’s digit, the remainder obtained is 1. Find the number.

Let the ten's and unit's digit of required number be x and y respectively.

Number = 10x + y

Given,

On dividing the number by the sum of its digits, the quotient is 8.

$\Rightarrow \Big(\dfrac{10x + y}{x + y}\Big) =8 \\[1em] \Rightarrow 10x + y = 8(x + y) \\[1em] \Rightarrow 10x + y = 8x + 8y \\[1em] \Rightarrow 10x - 8x + y - 8y = 0 \\[1em] \Rightarrow 2x - 7y = 0 \text{ …..(1)}$

Given,

On diminishing the ten’s digit by three times the unit’s digit, the remainder obtained is 1.

⇒ x - 3y = 1

⇒ x = 3y + 1     ………(2)

Substituting the value of x from equation (2) in (1), we get :

⇒ 2(3y + 1) - 7y = 0

⇒ 6y + 2 - 7y = 0

⇒ 2 - y = 0

⇒ y = 2.

Substituting value of y in equation (2), we get :

⇒ x = 3 × 2 + 1

⇒ x = 6 + 1

⇒ x = 7.

Number = 10x + y

= 10 × 7 + 2

= 72.

Hence, the number is 72.