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How many three-digit numbers are greater than $100$ and increase by $198$ when the three digits are arranged in the reverse order?
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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: $70$

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