An 8 digit number is divisible by 99. If the digits are shuffled, then the number is always divisible by __________. 

Given that the 8-digit number is divisible by 99, it must be divisible by both 9 and 11. When the digits of a number are shuffled, the sum of its digits remains the same, and so does the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions. ⇒ The sum of the digits remains divisible by 9. ⇒ The difference between the sum of the digits in the odd positions and the sum of the digits in the even positions remains divisible by 11. Therefore, the number formed by shuffling the digits will also be divisible by both 9 and 11. ∴ The number is always divisible by 9.