Question

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

4 balloons and 4 stickers
4 balloons and 4 stickers

25 balloons and 17 stickers
25 balloons and 17 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 while ensuring each bag has 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.

First, we calculate the GCD of 100 and 68.

1. The prime factorization of 100 is \( 2^2 \times 5^2 \).
2. The prime factorization of 68 is \( 2^2 \times 17 \).

The common factors are \( 2^2 \), which means the GCD is \( 4 \).

Now, we can determine how many balloons and stickers can go into each bag:

- For balloons: \( \frac{100}{4} = 25 \)
- For stickers: \( \frac{68}{4} = 17 \)

Thus, the maximum number of bags you can make is 4, and each bag can contain 25 balloons and 17 stickers.

So the answer is **25 balloons and 17 stickers.**