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?

after each step, the bag has 3 fewer candies than before.

So, starting with 50 candies, after 16 steps, there are just 2 candies left in the bag.

During those 16 steps, 64 candies have been removed. If you are allowed to remove the last two, then that makes 66.

It's a bit disappointing to note that you've been posting this problem for the last three days, and never once offered any of your own ideas. Heck, in that time, you could have done the experiment yourself with real candies!

To find the maximum number of candies this variety of R&Ns will yield, we need to determine the number of candies that can be removed so that an additional candy appears 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 exactly 4 candies, 1 additional candy appears in the bag. This means that the net change in the number of candies is -4 + 1 = -3.

We can represent this with the equation: x - 4 + 1 = x - 3.

Simplifying the equation, we get: x + 1 = x - 3.

If we subtract x from both sides, we get: 1 = -3.

This is not possible, which means our assumption that the initial number of candies is x is incorrect.

Therefore, there is no maximum number of candies that this variety of R&Ns will yield based on the given information.