Question

The sum of digits of a two digit number is $7$. If the digits are reversed and the resulting number is decreased by $2$, twice the original number is obtained. Find the original number

Answer 1

Let the original number be $\overline{yx}$; i.e., $10y+x$. We know $x+y=7$. The number obtained by reversing the digits in $\overline{xy}$, i.e., $10x+y$. The second condition gives $10x+y-2=2(10y+x)$. Thus we have two equations:
$x+y=7....\left(1\right)$
$8x-19y=2....\left(2\right)$
Multiply the equation $\left(1\right)$ by $19$ and get
$19x+19y=133$.
Adding this to $\left(2\right)$, we obtain $27x=135$. This gives $x=5$. Hence $y=7-x=7-5=2$.
The required number is $25$.