In partition theory, a counting number n is expressed as the sum of positive integers, without regard to order.

If n is a triangular number it has a special partition, namely 1+2+3+4+ ...

Examples of triangular numbers are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, and so on.

If we select a triangular number of cards from a deck of 52, the largest number we'd have is 49 45.

Let's go with 49 45cards [any 49 45cards, it doesn't matter] and play a little game.

Deal the cards into some number of piles, each pile having some number of cards.

Doesn't matter how many piles - could be a single pile, could be 49 45 piles. Your choice.

What you've done is create a partition of the number 49 45.

Play proceeds as follows:

.

Remove 1 card from each pile.

Place these cards together and use them to form a new pile.

.

For example, if you began with 49 45 piles of one card each, after the first play you'd have a single pile of 49 45 cards. After two plays you'd have a pile of 48 cards and a pile of 1 card. And so on. Piles are created and sometimes piles are used up.

Continue repeating steps 1 and 2.

The puzzle is to find the answer to these questions:

Does the game ever get stuck in an endless loop?
i.e. Does a partition ever revert to a previous partition? .

Does the game ever reach a stable state?
i.e. Is a partition ever reached that is invariant to steps 1 and 2? .

In answering these questions remember that, in a partition, the order does not matter.

So in a game of 3 cards, piles of 1 1 and 1 card and a single pile of 3 cards are different partitions.

But piles of 1 and 2 cards and piles of 2 and 1 cards are the same partition.

Shuffle up, deal and play the game, or just think about it.

## Question

## bonanova 84

In partition theory, a counting number

is expressed as the sum of positive integers, without regard to order.nIf

is a triangular number it has a special partition, namely 1+2+3+4+ ...nExamples of triangular numbers are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, and so on.

If we select a triangular number of cards from a deck of 52, the largest number we'd have is 49 45.

Let's go with 49 45cards [any 49 45cards, it doesn't matter] and play a little game.

Deal the cards into some number of piles, each pile having some number of cards.

Doesn't matter how many piles - could be a single pile, could be 49 45 piles. Your choice.

What you've done is create a partition of the number 49 45.

Play proceeds as follows:.

- Remove 1 card from each pile.
- Place these cards together and use them to form a new pile.

.For example, if you began with 49 45 piles of one card each, after the first play you'd have a single pile of 49 45 cards. After two plays you'd have a pile of 48 cards and a pile of 1 card. And so on. Piles are created and sometimes piles are used up.

Continue repeating steps 1 and 2.

The puzzle is to find the answer to these questions:- Does the game ever get stuck in an endless loop?
- Does the game ever reach a stable state?

In answering these questions remember that, in a partition, the order does not matter.i.e. Does a partition ever revert to a previous partition?

.

i.e. Is a partition ever reached that is invariant to steps 1 and 2?

.

So in a game of 3 cards, piles of 1 1 and 1 card and a single pile of 3 cards are different partitions.

But piles of 1 and 2 cards and piles of 2 and 1 cards are the same partition.

Shuffle up, deal and play the game, or just think about it.

Either way, enjoy!

Edited by bonanovachange 49 to 45

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