Let the three-digit original number be $xyz,$ then the reversed number will be $zyx.$
We know that,
- $xyz = 100x+10y+z\quad \longrightarrow (1)$
- $zyx = 100z+10y+x\quad \longrightarrow (2)$
Subtract equation $(1)$ from equation $(2).$
$\begin{array}{} zyx = 100z+10y+x \\ xyz = 100x+10y+z \\ \;– \qquad \; – \qquad \;\; – \quad\;\; - \\\hline zyx-xyz = 99z – 99x = 99(z-x) \end{array}$
$\Rightarrow 99(z-x)=198$
$\Rightarrow \boxed{z-x=2}$
$\Rightarrow \boxed{z=x+2}$
The three-digit number is greater than $100.$ So, $x\geq1$
$$\begin{array}{cc} x & z \\\hline 1 & 3 \\ 2 & 4 \\ 3 & 5 \\ 4 & 6 \\ 5 & 7 \\ 6 & 8 \\ 7 & 9 \\ \end{array}$$
So, $x$ can takes value from $1$ to $7$ and correspondingly $z$ can take values from $3$ to $9.$
Hence, $7$ combinations are possible. Also, $y$ can take values from $0$ to $9$, a total of $10$ possible values.
$\therefore$ Total possible numbers $= 7 \times 10 = 70.$
Correct Answer: $70$