When a two digit number is divided by the sum of its digits the quotient is 8. If the ten's digit is diminished by three times the unit's digit, the remainder is 1. What is the number?

Let's consider the digits at ten's place as x and let the digit at unit's place be y.

Number = 10 × x + y = 10x + y,

According to first condition, we have

$\Rightarrow \dfrac{10x + y}{x + y} = 8$

⇒ 10x + y = 8(x + y)

⇒ 10x + y = 8x + 8y

⇒ 10x - 8x + y - 8y = 0

⇒ 2x - 7y = 0 …….(i)

According to second condition, we have

x - 3y = 1 ……(ii)

Multiplying eq. (ii) by 2 we get,

2x - 6y = 2 …….(iii)

Subtracting eq. (i) from (iii) we get,

⇒ 2x - 6y - (2x - 7y) = 2 - 0

⇒ 2x - 2x - 6y + 7y = 2

⇒ y = 2.

Substituting value of y in (ii) we get,

⇒ x - 3(2) = 1

⇒ x - 6 = 1

⇒ x = 7.

Number = 10 × x + y = 10 × 7 + 2 = 72.

Hence, number = 72.