A carnival game gives variety bags as prizes. The game operator uses 100 balloons and 68 stickers to put into a certain number of bags. How many balloons and how many stickers can go in each bag if they make the greatest number of bags possible so that each bag has the same number of balloons and stickers?
To find the maximum number of bags, we need to find the greatest common divisor (GCD) of 100 and 68.
The prime factorization of 100 is $2^2\cdot5^2$.
The prime factorization of 68 is $2^2\cdot17$.
Therefore, the GCD of 100 and 68 is $2^2=4$.
So, each bag can contain 4 balloons and 4 stickers.
Therefore, the answer is $\boxed{4}$.