Question

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

Answer 1

Let the ten digits be $x$ and let the units digits be $y$

Then, the required number is $10x+y$

According to the question sum of digits $=x+y$

$\frac{10x+y}{x+y}=8$

$\Rightarrow 10x+y=8x+8y$

$\Rightarrow 10x-8y=8y-y$

$\Rightarrow 2x=7y$

$\Rightarrow 2x-7y=0----------\left(1\right)$

Given,

$x-3y=1-------------\left(2\right)$

from equation $\left(1\right)$ and $\left(2\right)$

$2x-7y=0$

$(x-3y=1)\times 2$

$2x-7y=02x-6y=2-+----------y=-2$

$y=2$

put in equation $\left(1\right)$

$2x-7y=0$

$2x-\times 2=0$

$2x=14$

$x=7$

Hence the number is $10x+y$

$\Rightarrow 10\times 7+2$

$=72$

Hence, this is the answer.