Two friends were bored and decided to play a game... a mathematical game with a paper bag!

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John and Mary have a bag full of integer numbers. In fact, the bag has \(10^{10^{10}}\) integers, each written on a plastic card, and the sum of all the \(10^{10^{10}}\) integers in the bag is \(0\). In turns, Mary and John are going to play with the bag by doing the following:

- Picking two cards from the bag with numbers \(a\) and \(b\) and removing them from the bag;
- Inserting a new card in the bag with the number \(a^3 + b^3\).

Is there any initial number configuration and/or set of moves for which it is possible that, after \(10^{10^{10}} - 1\) moves, the only card in the bag has the number \(73\)?

Give it some thought... and most important of all, try it for real! Let me know how it went in the comment section below ;)

**Hint**: the answer is "no". Can you show why?

**Hint**: look for an invariant of the game! That is, find a property of the game that does *not* change when Mary and John play it.

(If you spot a mistake in my solution *please* let me know in the comments below.)

No, there is no initial configuration nor set of moves that allows us to reach the goal. Note how \(x \equiv x^3 \mod 2\). That is, \(x\) has the same parity as \(x^3\). Let us assume that, at a given point, all the integers in the bag add up to \(S\). We show that the parity of the sum of all integers in the bag doesn't change when we remove the cards \(a,b\) and then add the card \(a^3+b^3\):

\[ x \equiv x^3 \mod 2 \implies S \equiv S - a - b + a^3 + b^3 \iff S + a + b \equiv S + a^3 + b^3 \mod 2\]

Thus we can't end up only with \(73\) in the bag, as everything in the bag should add up to an even number and \(73\) is odd.

Bonus question: find a solution that would still work for \(2, 74, 308\) (and for an infinity of even numbers), even though those are even.

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