The key to solving this is to design a pattern after each bunny has gone.
After the first bunny has gone: every egg would be out of their corresponding basket.
After the second bunny has gone: every second egg would be placed back into their basket, because they were out of their basket. This means basket 2, 4, 6, and so on, would have their eggs placed back in.
After the third bunny has gone: every third egg would be affected. That is, the third egg would be placed back in, but the sixth egg would be placed back out because it was previously placed back in.
We can generalise this to determine which eggs are out, and which are in, by their factors, because the factors are the causes of each egg's status. For example, egg number 2 has a factor of 1 and itself. Thus, we know that the egg's status remains unchanged. In fact, every number with an even number of factors would have their basket status unchanged.
Thus, the only group of numbers that will never satisfy this are perfect squares. Perfect squares have an odd number of factors, because they will have a factor that, when multiplied by itself, produce the desired number.
Let's take 4, as an example. Since its factors are 1, 2, and 4, then bunny number 1 will take the egg out, bunny 2 will place it back in, and bunny 4 will take the egg out.
Thus, we know that 4 is a solution. Using this, we can further generalise a statement. "Every factor will pair up with another factor, unless the desired number is produced as a perfect square" and we know this is true.
Thus, we can conclude that the only solutions to this question are the perfect squares from 1 to 100: 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.
