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? Responses 50 balloons and 34 stickers 50 balloons and 34 stickers - no response given 25 balloons and 17 stickers 25 balloons and 17 stickers - no response given 4 balloons and 4 stickers 4 balloons and 4 stickers - no response given 17 balloons and 25 stickers

1 answer

To solve this problem, we need to determine the greatest common divisor (GCD) of the number of balloons (100) and the number of stickers (68). The GCD will tell us how many bags we can make while ensuring each bag has the same number of balloons and the same number of stickers.

  1. Find the GCD of 100 and 68:

    • The prime factorization of 100: \(100 = 2^2 \times 5^2\)
    • The prime factorization of 68: \(68 = 2^2 \times 17\)

    The GCD is found by taking the lowest power of each common prime factor.

    • The common prime factor is \(2\), and the lowest power is \(2^2\).
    • So, the GCD(100, 68) = \(2^2 = 4\).
  2. Determine the number of bags:

    • The greatest number of bags possible is \( \frac{100}{4} = 25\) bags for balloons and \( \frac{68}{4} = 17\) bags for stickers.
  3. Conclusion: Each bag can contain:

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

Therefore, the maximum number of bags that can be made is 25, and each bag can hold 4 balloons and 17 stickers.

None of the options you provided match this conclusion exactly. Please verify if there's a mistake in the options. The correct option should be 4 balloons and 17 stickers.