Question

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?

Answer 1

Final Answer : 72

Steps:
1) Let the ten's and unit's digit be 'x' respectively .
Then, the number is 10x +y.

2) According to the question,
Two digit number /Sum of digits = 8
=>
$\frac{10x + y}{x +y } = 8 \\ = > 10x + y = 8x + 8y \\ = > 2x = 7y - - - - (1)$

3) Ten's digit is diminished by three times the unit digit gives 1.
$= > x - 3y = 1 \\ = > 2x - 6y = 2 - - (2)$

4) Using equation (1) & (2) ,
$7y - 6y = 2 \\ = > y = 2$
And,
$2x = 7 \times 2 \\ = > x = 7$
Therefore, the number is
$\boxed {10(7) + 2 = 72}$

Answer 2

Let the digit at units place be y and 10s place be x
So,
the number can be written as,
10x+y
Now, as per conditions given
(10x+y)/(x+y)=8
10x+y = 8x+8y
2x=7y
So,
x=(7y)/2

Now by other condition when tens digit is diminished 3 times units place, the remainder is 1
So,
x-3y=1
But we know that x= (7y)/2 so,
(7y)/2 - 3y = 1
(7y-6y)/2 = 1
y=2

So,
x= (7y)/2
x=(7*2)/2
x=7
So, the number is 72