Question

If the digits of a three digit number are reserved, then the number so obtained is less than the original number by $297$. If the sum of the digits 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 $1$

Let the number be $x+10y+100z$

According to the question,

Upon reversing, the number obtained $=z+10y+100x$

$x+10y+100z-z-10y-100x=297$

$\Rightarrow 99(z-x)=297$

$\Rightarrow z=x+3$

Also,

$x+y+z=8$

Possible cases :

$\left(I\right)x=1,z=4,y=3$

$\left(II\right)x=2,z=5,y=1$

Given that Hundredth's place is maximum possible, therefore $z=5$

$\Rightarrow$ Ten's digit $=1$