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
50 balloons and 34 stickers
17 balloons and 25 stickers
17 balloons and 25 stickers
4 balloons and 4 stickers
4 balloons and 4 stickers
25 balloons and 17 stickers
25 balloons and 17 stickers
To find the answer, we need to find the greatest common divisor (GCD) of 100 and 68. The GCD will give us the largest number that can evenly divide both 100 and 68, which will represent the number of balloons and stickers in each bag.
First, let's factorize 100 and 68:
100 = 2^2 * 5^2
68 = 2^2 * 17
To find the GCD, we take the common factors of both numbers, raised to the power of the lower exponent:
GCD = 2^2 = 4
Therefore, each bag can have 4 balloons and 4 stickers. So, the correct answer is option:
4 balloons and 4 stickers
To find the maximum number of bags that can be made, we need to find the common divisor of 100 balloons and 68 stickers.
The prime factorization of 100 is 2^2 * 5^2
The prime factorization of 68 is 2^2 * 17
To find the common divisor, we take the lowest exponent for each prime factor.
The lowest exponent of 2 is 2.
The lowest exponent of 5 is 0 (since there are no 5s in the prime factorization of 68).
The lowest exponent of 17 is 0 (since there are no 17s in the prime factorization of 100).
So, the common divisor is 2^2 = 4.
Therefore, each bag can have 4 balloons and 4 stickers.