The problem statement from Slovenia reads as follows (link here): Quote: You start with three boxes, each of which contain some number of marbles. At any step, you can double the number of marbles in some box by removing the marbles required to double it from some other box. Show that after a finite number of steps of this form, we can force some box to be empty. I spent this morning trying to tackle this problem and came up with this solution.

I have only seen the second part of this problem now, but I believe the answer is no. Edit: Well the solution here is pretty easy. Take any configuration where the sum is odd. Then the final configuration must be 0 0 (odd number), but that's impossible to reach.

A simple starting configuration, with forced moves, is $(1,1,1)$, from which we may only derive $(0,2,1)$, then $(0,1,2)$, then $(0,2,1)$ again, and we cycle between these last two states.

Yongyi781 wrote: Furthermore, we will always have enough marbles in the third box because the first box will hold at most $i_1+i_2$ marbles at the end, so we will have to take at most $i_2\leq i_3$ marbles from the third box. No, consider (13,37,42) --> (26,37,29) --> (52, 11, 29), but 52>13+37=50. In fact, what actually happens is that the sum in the second decreases by at least $i_2-i_1$, whereas the sum in the first increase by no more than $2i_2-i_1$, so the net increase in the sum of the first 2 is at most $i_2$.

This is exactly like USAMTS number 5...

http://www.usamts.org/Solutions/Solutions_23_1.pdf

A really nice that I saw as an example in the book here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=721&p=3573948#p3573948 Example 7 of Chapter 1

Peter cannot transfer all his money into one account. Consider the case when the total amount of money is odd, the final move to transfer it to the final account clearly cannot be double, contradiction. WLOG let the three accounts in the bank be "$Evan$", "$Sheldon$" and "$Fish$" WLOG suppose that $Fish \geq Sheldon \geq Evan$ we claim that we can always operate such that $min{ Evan, Sheldon, Fish}$ Decreases (monovariant). Write $Sheldon$ in terms of $Evan$. $Sheldon = Evan( X_1 +2X_2 +....... 2^kX_k) + r $ such that $Evan < r$ and for all $X_k$, $X_k= 0 or 1$. Note that Sheldon can give Evan* power of two to Evan at every turn( working up from $1,2,4,...2^k$). Say Sheldon has $X_k=0$, then Fish can give $Evan(2^kX_k)$ to Evan. This will ensure that Sheldon will become reduced to r. Note now, $ min{ Evan, Sheldon, Fish}=r < Evan$ Now repeat this process until it terminates: When the min is 0, sp we are done.

Solved with nukelauncher and islander7. Solution (a) Without loss of generality the accounts contain \(a<b<c\) dollars. I will show we can perform some moves to obtain an account with strictly fewer than \(a\) dollars. Write \(b=ab'+r_1\) and \(c=ac'+r_2\), with \(0\le r_1,r_2<a\). Let \(b'=(\overline{1b_{k-1}b_{k-2}\ldots b_0})_2\) in binary. Iterating over \(i=0,1,\ldots,k-1\), perform the following: If \(b_i=1\) then transfer money from the second account to the first account, doubling \(a\) and removing the \(b_i\) bit from \(b'\). If \(b_i=0\) then transfer money from the third account to the first account, doubling \(a\). At the end the first account contains \(2^ka\) dollars and the second account contains \(2^ka+r_1\) dollars, so transferring money from the second account to the first account leaves the second account with \(r_1<a\) liters. Solution (b) Peter can't always transfer his money into one account: if we start with at least two nonzero balances and the sum of the three balances is odd, then observe that at the end of every move, the recipient balance must be even. But the final position includes a single nonzero balance that is odd.

Can someone please explain the line with example? I didn't understand this line, $$\textbf{"He is only allowed to transfer money from one account to another so that the amount of money in the latter is doubled."}$$

@above This means that if $a \geq b$, then the bank accounts $(a,b)$ can become $(a-b,2b)$.

Apologies for double post. One detail not to miss is that we have to justify why the process leaves the largest bank account non-negative. This is true because the last move is from the second account, making the second account reduce more than the third account ($2^n > 2^{n-1} + 2^{n-2} + \cdots + 1$). But the second account remains non-negative, so the third account will also be non-negative.