Question

The difference of three-digit number and the number obtained by putting the digits in reverse order is always divisible by $9$ and ___________.

Answer 1

Determine the number that divides the difference of a three digits number and its reverse.

A three digits number is expressed as $100x+10y+z$.

So, the number obtained by reversing its digits is $100z+10y+x$.

The difference of the numbers is :

$\begin{array}{rcl}100x+10y+z-\left(100z+10y+x\right)& =& 100x+10y+z-100z-10y-x\\ & \Rightarrow & 100x+10y+z-\left(100z+10y+x\right)=99x-99z\\ & \Rightarrow & 100x+10y+z-\left(100z+10y+x\right)=9\times 11\times \left(x-z\right)\end{array}$

Clearly, the difference is a multiple of $9$ and $11$.

So, the difference is always divisible by $9$ and $11$.