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
50 balloons and 34 stickers
25 balloons and 17 stickers
4 balloons and 4 stickers
17 balloons and 25 stickers
To find the greatest number of bags that can be made with an equal number of balloons and stickers, we need to find the greatest common divisor (GCD) of 100 and 68.
The prime factorization of 100 is 2^2 * 5^2
The prime factorization of 68 is 2^2 * 17
To find the GCD, we take the highest power of each common prime factor:
GCD = 2^2 = 4
Therefore, the greatest number of bags that can be made is 4.
To calculate the number of balloons and stickers in each bag, we divide the total number of balloons and stickers by the number of bags:
Number of balloons in each bag = 100 / 4 = 25
Number of stickers in each bag = 68 / 4 = 17
So the correct answer is: 25 balloons and 17 stickers.