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

(A) Let the ten's digit be $x$ and the unit's digit be $y$. The number is $10x + y$.
When the digits are reversed,the new number is $10y + x$.
According to the problem,$(10x + y) + (10y + x) = 66$.
$11x + 11y = 66$,which simplifies to $x + y = 6$ $...(1)$.
Given that the digits differ by $2$,we have two cases:
Case $1$: $x - y = 2$ $...(2)$.
Adding $(1)$ and $(2)$: $2x = 8 \implies x = 4$. Substituting $x=4$ in $(1)$,$y = 2$. The number is $42$.
Case $2$: $y - x = 2$ $...(3)$.
Adding $(1)$ and $(3)$: $2y = 8 \implies y = 4$. Substituting $y=4$ in $(1)$,$x = 2$. The number is $24$.
Thus,there are two such numbers: $42$ and $24$.