Question

The units digit of a two-digit number is $3$ and seven times the sum of the digits is the numbers itself. Find the number.

Answer 1

Say, unit digit of two-digit number $=x$
tens digit of two-digit number $=y$
So, number will be represented as $=10y+x$
From question,
unit digit of two-digit number, $x=3$

ATQ,
$7$ times the sum of digits of the number $(10y+x)=10y+x$
$7(x+y)=10y+x...\left(1\right)$
$\Rightarrow 7(3+y)=10y+3[\because x=3]$
$\Rightarrow 21+7y=10y+3$
$\Rightarrow -3y=-18$
$\Rightarrow \frac{-3y}{-3}=\frac{-18}{-3}$
$\Rightarrow y=6$

So, the number $=10y+x=10\times 6+3=63$

Therefore, the required number is $63$.