## Fall 2016, problem 32

Written on a blackboard are $n$ nonnegative integers whose greatest common divisor is $1$. A move consists of erasing two numbers $x$ and $y$, where $x\ge y$, on the blackboard and replacing them with the numbers $x-y$ and $2y$. Determine for which original $n$-tuples of numbers on the blackboard it is possible to reach a point, after some number of moves, where $n-1$ of the numbers on the blackboard are zeroes.

### Comments

The necessary condition is in fact also sufficient. Let us firstly point out some remarks.

- We will always assume that the sum is a power of 2.
- In particular the number of odd terms must be even.
- If, after some moves, all terms are even, we can work with the "halved" board. In fact, if a sequence of moves solves the problem corresponding to the halved board, the same sequence applied to the original board solves the true problem.
- We will work by induction on $M$, the maximum of the values appearing on the board. The base case ($M=1$) is quite trivial: if $M=1$, because of remark 2, the non vanishing elements are $2p$ elements all with value 1; replacing any couple $(1,1)$ with a couple $(0,2)$ we end up with only $2(p-1)$ non vanishing elements all with value 2. After applying remark 3 we are back with the original problem with $p$ replaced by $p-1$. Step by step we finish with a single couple $(0,1)$ that becomes $(2,0)$.

In order to prove the inductive step, let us call "strategy" the family of moves defined by:

- if all element are even, we half the board;else:
- we list the odd elements in decreasing order, $O_1\ge O_2\ge\dots,\ge O_{2p}$ (see remark 2). Then we replace $(O_1,O_2)$ by $(O_1-O_2,2O_2)$, $(O_3,O_4)$ by $(O_3-O_4,2O_4)$ and so on. All the newly generated numbers are even, as well as the untouched entries; so that we can apply remark 3.

**CLAIM:** starting from a board with maximum $M$ where there are $NV$ non-vanishing elements, after applying the strategy we end up with a board whit a maximum $M'\le M$; and, if $M'=M$, the new number $NV'$ of non vanishing elements is strictly less than $NV$.

Of course the claim is sufficient to prove the induction step. when $M' \le M$ we are done; if instead $M'=M$ we continue to apply our strategy until $M' \le M$ or $NV'=2$, In this last case the element will have the same parity (remark 2) thus applying one more time the strategy we end up with a smaller value for $M$ or just one number as required.

Finally, let us prove the claim. The maximum cannot incease because $O_{2j+1}-O_{2j+1} \le O_{2j + 1}$ and $2O_{2j + 2}$, after halving, is $O_{2j + 2}$; and the even terms obiously cannot increase. Furthermore: if the maximun $M$ is odd, it must coincide with the value noted $O_1$; we can assume $O_1>O_2$ otherwise the generated element $O_1-O_2$ vanishes; thus the maximum anong the newly generated elements, after halving, is $O_2$ that is less than $O_1=M$; and the even elements have a lesser maximum. If, on the contrary, $M$ is even, the odd elements are less than $M$ as well as the halved newly generated ones; and the halved even elements.

A partial solution is that it is necessary that the sum of the numbers be a power of two. Consider the game in reverse, where moves consist of taking 2x and y and replacing them with x and x+y. If our list of numbers has any common divisors other than 2 before a move, it will still have those divisors afterwards, so if we start with only one nonzero number and eventually get to a list with no common divisors, then the nonzero number was a piwer of 2. Since the total sum of the elements in our list is constant, the result follows.