Question

The middle digit of a number between $100$ and $1000$ is zero, and the sum of the other digits is $11$. If the digits be reversed, the number so formed exceeds the original number by $495$; find it.

Answer 1

Let the number be $X0Y$

Value of the number is $100X+Y$

Value of number when it reversed is $100Y+X$

Then, $100Y+X-(100X+Y)=495$

$99Y-99X=495$

As we know $X+Y=11$ we can replace Y by 11-X

$99(11-X)-99X=495$

$(1089-99X)-99X=495$

$-198X=-594$

$X=3$

Hence $Y=8$

The no. is 308