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

Let $xyz$ is a number.

From the question, we have to check for $xyz-zyx$

$xyz-zyx=(100x+10y+z)-(100z+10y+x)=99x-99z=99(x-z)$

Which is always divisible by $99$, that is by both $9$ and $11$, because $99=9\times 11$

Hence the difference of three and the number obtained by putting a digits in reverse order is always divisible by $9$ and $11$.