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 the same number of stickers?(1 point) Responses 25 balloons and 17 stickers 25 balloons and 17 stickers 4 balloons and 4 stickers 4 balloons and 4 stickers 17 balloons and 25 stickers 17 balloons and 25 stickers 50 balloons and 34 stickers

Answer 1

To find the greatest number of bags possible where each bag contains the same number of balloons and stickers, we need to determine the greatest common divisor (GCD) of the two quantities: 100 balloons and 68 stickers.

1. Find the GCD of 100 and 68:

   • The prime factorization of 100 is \(2^2 \cdot 5^2\).
   • The prime factorization of 68 is \(2^2 \cdot 17\).
   • The common factors are \(2^2\), so the GCD is \(4\).

2. Determine the number of bags:

   • The total number of bags possible is the GCD, which is 4.

3. Calculate the number of balloons and stickers per bag:

   • For balloons: \(100 \div 4 = 25\)
   • For stickers: \(68 \div 4 = 17\)

Thus, each bag can contain 25 balloons and 17 stickers.

The correct response is: 25 balloons and 17 stickers.