Question

If the digits of a three-digit number are reversed , then the number so obtained is less than the original number by $297$.If the sum of the digit of the number is $8$ and its hundred's digit has the largest possible value, then the ten's digit of the number is :

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is C

$1$

Step 1: Note the given data:

$$\begin{gathered} \mathrm{Original}\mathrm{number}-\mathrm{New}\mathrm{number}=297 \\ \mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{digit}\mathrm{of}\mathrm{the}\mathrm{number}=8 \end{gathered}$$

Let $\text{'}\mathrm{x}\text{'}$be at hundred's place, $\text{'}\mathrm{y}\text{'}$be at ten's place, $\text{'}\mathrm{z}\text{'}$be at unit's place.

$\therefore$Original three-digit number $=100\mathrm{x}+10\mathrm{y}+\mathrm{z}$

New three-digit number $=100\mathrm{z}+10\mathrm{y}+\mathrm{x}\left[\mathrm{After}\mathrm{reversing}\mathrm{the}\mathrm{digit}\right]$

Step 2: Forming the equation:

$$\begin{gathered} \therefore \left(100\mathrm{x}+10\mathrm{y}+\mathrm{z}\right)-\left(100\mathrm{z}+10\mathrm{y}+\mathrm{x}\right)=297........\left(1\right)\mathbf{[}\mathbf{Given}\mathbf{:}\mathbf{}\mathbf{orginal}\mathbf{}\mathbf{number}\mathbf{-}\mathbf{new}\mathbf{}\mathbf{number}\mathbf{=}\mathbf{297}\mathbf{]} \\ \Rightarrow 100x-x+10y-10y+z-100z=297 \\ \Rightarrow 99x-99z=297 \\ \Rightarrow 99(x-z)=297 \\ \Rightarrow \left(x-z\right)=3..............\left(2\right) \\ x+y+z=8..............\left(3\right)\mathbf{[}\mathbf{}\mathbf{G}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{n}\mathbf{}\mathbf{]} \end{gathered}$$

Step 3: Simplify equations

$$\begin{gathered} \therefore x-z=3\mathbf{[}\mathbf{F}\mathbf{r}\mathbf{o}\mathbf{m}\mathbf{}\mathbf{2}\mathbf{]} \\ \Rightarrow x=z+3 \\ z+3+y+z=8\mathbf{[}\mathbf{}\mathbf{Putting}\mathbf{}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{}\mathbf{v}\mathbf{a}\mathbf{l}\mathbf{u}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{x}\mathbf{}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{e}\mathbf{q}\mathbf{u}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{}\mathbf{3}\mathbf{}\mathbf{]} \\ \Rightarrow 2z+y=5............\left(4\right) \end{gathered}$$

Step 4: Solve the equation$\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{}\mathbf{and}\mathbf{}\mathbf{\left(}\mathbf{4}\mathbf{\right)}$ by a trial-error method in the equation :

$\mathrm{x}$	$\mathrm{y}$	$\mathrm{z}$
$3$	$5$	$0$
$4$	$3$	$1$
$5$	$1$	$2$

$\mathbf{[}\mathbf{x}\mathbf{>}\mathbf{(}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{}\mathbf{}\mathbf{,}\mathbf{}\mathbf{Hundred}\mathbf{\text{'}}\mathbf{s}\mathbf{}\mathbf{digit}\mathbf{}\mathbf{has}\mathbf{}\mathbf{largest}\mathbf{}\mathbf{possible}\mathbf{}\mathbf{value}\mathbf{]}\mathbf{}\mathbf{}\mathbf{}$

The possible number is$\mathbf{}\mathbf{512}$.

Therefore, the ten's digit of the number 512 is $\mathbf{1}\mathbf{.}$

Hence, option C is the correct answer.