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We call the the ordinary counting numbers (1, 2, 3, and so on), of which there are infinitely many, the set of natural numbers. Let us suppose that I have a magic basket with enough ping-pong balls in it so that each natural number labels exactly one ping-pong ball. (Clearly, this basket is bigger on the inside than it is on the outside – like Mary Poppins’s carpet bag – but never mind about that!)
If I had an infinite amount of time to work in, I could, if I wished, take two other magic baskets and fill them with the balls from the first basket, so that when I was finished my first basket would be empty, and my two new baskets would each contain infinitely many balls. To accomplish this I would begin by labelling my two new baskets “odd” and “even”, respectively, and I would then proceed by always taking the lowest-numbered ball remaining in the first basket and putting it in the odd basket if it was labelled with an odd-number, and putting it in the even basket otherwise. After moving infinitely many balls, my first basket would be empty, and the odd and even baskets would each contain infinitely many ping-pong balls.

Supposing I wanted to fill four baskets instead of only two? I could do it almost exactly the same way, only this time the four new baskets would be labelled “1”, “2”, “3”, and “0”, and each time I took a ball from the first basket I would put it in whichever new basket corresponded to the remainder I got when dividing the number printed on it by 4. Thus, for example, when I took ball number 57 from the first basket, I would put it in the basket labelled “3”, since 57 leaves a remainder of 3 when divided by 4.

Clearly, starting with one magic basket with infinitely many ping-pong balls in it, I can fill any finite number of baskets with infinitely many ping-pong balls each.
Today's challenge: Can you think of a way to fill infinitely many baskets with infinitely many ping-pong balls, starting from just one magic basket?


Solution to yesterday's challenge

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