In the described game, can the rubies be divided into 105 piles of one?

Question

Problem

In a pirate ship, there is a chest with 3 sacks containing 5, 49, and 51 rubies respectively. The treasurer of the pirate ship is bored and decides to play a game with the following rules:

• He can merge any two piles together into one pile, and
• he can divide a pile with an even number of rubies into two piles of equal size.

He makes one move every day, and he will finish the game when he has divided the rubies into 105 piles of one. Is it possible for him to finish the game?

Solution attempt

It seems to me that it's not possible to finish the game, and here is the argument that I could come up with:

Assume, for the sake of reaching a contradiction, that a state with 105 piles of 1 can be reached.

The start state has no piles of 1, so the piles of 1 must have been obtained from other piles. From the rules, the only way of obtaining a pile of 1 from other piles is by splitting a pile of 2 into two piles of 1. So, each two piles of 1 must have originated from splitting a pile of 2. The resulting number of piles of 1 generated this way is even, because every pile of 2 generates 2 piles of 1. However, there is an odd number (105) of piles of 1, so this state is impossible to from the given start state using the defined rules.

Is this correct, or at least on the right track?

Answer 1

As lulu mentioned in a comment, your proof is incorrect.

The reason for this is that you assume that the only way the number of piles of $1$ can be modified is by dividing a pile of $2$ into two piles of $1$, and saying that since this preserves parity $105$ piles is unreachable.

However, what you have failed to take into account is that the number of piles of $1$ can also decrease: you can merge a pile of $1$ with another pile not of size $1$, decreasing the number of piles of size $1$ by $1$, or even merge two piles of size $1$ (although there would not be any good reason to do so).

With regards to the actual solution, I would recommend the following tactic: think about what your first move can be, and what that results in. Obviously it has to be a pile merge. I'll give you a head start: if you start by merging the $5$ and $51$, then all pile sizes are divisible by $7$, and this does not change through either merges or splits. Can you finish from here?

Answer 2

All initial piles are odd, so your first move must be one of three mergings:

• If you merge $5 + 49$, your two resulting piles ($54$ and $51$) are each divisible by $3$. Your two legal operations retain this property for each pile, so you can never finish.
• If you instead merge $5 + 51$, your two resulting piles ($56$ and $49$) are each divisible by $7$. Your two legal operations retain this property for each pile, so you can never finish.
• If you instead merge $49 + 51$, your two resulting piles ($5$ and $100$) are each divisible by $5$. Your two legal operations retain this property for each pile, so you can never finish.

Hence: You can never finish.