A carnival game gives variety bags as prizes. The game operators 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 the same number of stickers
To find the maximum number of bags, we need to find the common factors of 100 and 68. The prime factorization of 100 is $2^2\cdot5^2$, and the prime factorization of 68 is $2^2\cdot17$. The common factors of 100 and 68 are the exponents of the shared prime factors: $$\begin{array}{cccccccc} 2^2\cdot5 & = & 20 & \text{ factors} & 1, & 2, & 4, & 5, & 10, & 20 \\ 2^2 & = & 4 & \text{ factors} & 1, & 2, & 4 \\ 2^2\cdot17 & = & 68 & \text{ factors} & 1, & 2, & 4, & 17, & 34, & 68 \end{array}$$ The greatest number of bags is given by the greatest common factor of 100 and 68, which is $\boxed{4}$. Thus, there are $\boxed{25}$ balloons and $\boxed{17}$ stickers in each bag.
