This post contains a problem statement and my proposed solution. Please remember that I am only human and that I am (very) flawed. If you spot a mistake in my solution please let me know.
Problem statement:
John and Mary have a bag full of integer numbers. In fact, the bag has $10^{10^{10}} $ numbers, 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:
Solution:
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.
Let me know what you think!
-RGS
Problem statement:
John and Mary have a bag full of integer numbers. In fact, the bag has $10^{10^{10}} $ numbers, 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, let us say the cards have the numbers $a $ and $b $, and removing them from the bag;
- Inserting a new card in the bag with the number $a^3 + b^3$.
Q: 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 $73$?
Solution:
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.
Let me know what you think!
-RGS
Problem #02 - a bag full of numbers
![Problem #02 - a bag full of numbers]()
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November 05, 2017
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