A number '$Z$' contains all the digits from $1$ to $9$ exactly once. $Z$ is divisible by $99$. What will be the number on its hundreds place (i.e. its third-to-last digit)?
$99=9\cdot11$ so the addition of numbers $1,2,3,4,5,6,7,8,9$ is $45$ which is divisible by $9$.
To be divisible by $11$, the sum of the odd places of a number subtracted by the even places of a number must be a multiple of $11$.
How do I proceed?