Another variety of R&Ns has a different set of magical (yet mathematical) properties. Remove exactly 4 candies and 1 additional candy appears in the bag. According to the package, a bag contains 50 candies. What is the maximum number of canies that this variety of R&Ns will yield?
To find the maximum number of candies that this variety of R&Ns will yield, we need to determine the minimum number of candies that can appear in the bag.
Let's assume that the initial number of candies in the bag is x.
According to the given information, if we remove 4 candies, 1 additional candy appears in the bag. This means that after removing 4 candies, there will be x - 4 candies in the bag.
To find the maximum number of candies, we want to minimize the number of times we remove 4 candies and maximize the number of times we gain 1 additional candy.
If we keep removing 4 candies until we can no longer do so, the number of candies left in the bag should be 0, 1, 2, or 3.
However, the problem states that the bag initially contains 50 candies, which means that there must be a total of 50 - (x - 4) candies removed.
So, we need to find the closest multiple of 4 that is smaller than or equal to 50 - (x - 4).
To maximize the number of candies, we need to find the largest value of x that satisfies this condition.
Starting from x = 50, we subtract 4 at each step until we find a multiple of 4.
50 - 4 = 46 (not a multiple of 4)
46 - 4 = 42 (a multiple of 4)
Therefore, the maximum number of candies that this variety of R&Ns will yield is 42.