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So basically, it drills down to performing an infinite amount of tasks, in an finite amount of time, which is the definition of what we refer as a supertask.
We start with task 1, where we add balls 1 to 10, and remove the first, which is ball 1. At task 2, we add balls 11-20, and remove the remaining first, which is ball 2. At task 3, we add balls 21-30, and remove the remaining first, ball 3… and so on and so forth.
Now the question is: how many balls would be in the vase at noon?
Let us look into what happened on the actual the steps. At step 1, we removed ball 1, and then step 2, removed ball 2, step 3, removed ball 3,… so at step n, we would remove ball n, and hence all the balls will be removed from the vase by the end of the tasks. So, the answer to our problem is none.
Apparently, the order in which we add balls, and remove them, affects how many balls are in the vase at noon. Let us look at an alternative scenario.
So again, how many balls would be in the vase at noon?
This time, the answer is different, and it is Infinity. This is actually due to the fact that only balls which are multiples of ten are the ones being removed, so the vase will not contain balls 10, 20, 30,… yet all the other balls will still be there, hence the infinite amount.