Is one of the easter baskets more skimpy?

The Easter Bunny Has 100 Baskets. He Puts One Egg In Each Basket. He Hires 100 Bunnies; The 1st Bunny Takes Out All Eggs. The 2nd One Puts One Egg In Every 2nd Basket. The 3rd One Puts One Egg In Every 3rd Basket Or Takes Out If One Is Already There. The 4th Bunny Puts An Egg In Every 4th Basket And

• Answer:

  60 baskets have at least one egg in them. Since the 1st bunny takes out all eggs, the only bunnies contributing to the final count are #s 98, 99, & 100. The 98th corresponds to #2, which leaves one in every other basket for a total of 50 baskets with eggs. The 99th corresponds to #3, which alters every 3rd basket. But examining multiples of 3, we see that the sequence is odd, even, odd, even... Since only even numbered baskets have eggs in them, the 99th bunny will put one in, take one out, put one in, take one out, and so on, effectively compromising emptying a basket by filling another empty one... this will continue until basket #99 (since there is no basket #102), when he will proceed to fill a net amount of 1 empty basket #99. So far, 51 baskets have eggs in them, including #{(2,3,4,) (8,9,10) (14,15,16)...}, skipping 3 baskets after every sequence of 3 numbers. We only need to worry about empty baskets because we are finding out how many baskets have at least one egg in them. The 100th bunny will simply encounter an empty basket every 3rd basket (#s 12, 24, 36). Notice that these numbers fall into the middle of each empty set of three numbers. For example, (11,12,13) (23,24,25) (35,36,37). This symmetry implies that the pattern will continue until the last basket. So for all multiples of 12 less than 100, the 100th bunny will fill an empty basket. He will fill 8 empty baskets up to #96. So the total of baskets with at least one egg is = 50 + 1 + 9 =60