Question

The sum of a two-digit number and the number formed by reversing the order of digits is $66$. If the two digits differ by $2$, find the number. How many such numbers are there?

Answer 1

Let the two-digit be $x$ and $y$ $.$

Two-digit number $=10x+y.$

Two-digit number formed by interchanging the digits$=10y+x$

$\therefore (10x+y)+(10y+x)=66$$\Rightarrow x+y=6⟶\left(i\right)$

$11x+11y=66$$\Rightarrow x-y=2⟶\left(ii\right)$

Substracting $\left(i\right)$ and $\left(ii\right),$

$2x=8$

$\Rightarrow x=4$

putting $x$ in $\left(i\right),$

$4+y=6$

$y=2$

Two-digit number$=10x+y=42$ or$,$ $10y+x=24$

$\therefore$ $2$ such numbers are possible$.$