Question

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

Answer 1

Let the tens digit be x
Let the units digit be y

The required number is 10x+y

The sum of the digits is x+y

Given, (10x+y)/(x+y) = 8
=> 10x+y = 8(x+y)
=> 10x+y = 8x+8y
=> 10x-8x = 8y-y
=> 2x = 7y
=> 2x-7y = 0 ------------(1)

Given, x-3y = 1
=> 2x-6y=2 -------------(2)

Solving (1) and (2), we get x=7 and y=2

The answer is 72

Answer 2

Solution :-

Let the tens digit and one digit be x and y respectively.

Then,
According to the question -

(10x + y)/(x + y) = 8

10x + y = 8x + 8y

10x - 8x - 8y + y = 0

2x - 7y = 0 ...............(1)

Given that the ten's digit is diminished by three times the unit's digit, the remainder is 1.

⇒ x - 3y = 1 ..............(2)

Multiplying the equation (2) by 2, we get.

2x - 6y = 2 ..........(3)

Now, subtracting (3) from (1), we get.

    2x - 7y = 0 
    2x - 6y = 2
   -    +       -
___________
         - y = - 2
___________

- y = - 2

y = 2

Putting y = 2 in (1), we get.

x - 3y = 1

x - (3*2) = 1

x - 6 = 1

x = 1 + 6

x = 7

So, tens digit is 7 and ones digit is 2 and the number is 72.

Answer.