Le Monde puzzle [#965]

A game-related Le Monde mathematical puzzle:

Starting with a pile of 10⁴ tokens, Bob plays the following game: at each round, he picks one of the existing piles with at least 3 tokens, takes away one of the tokens in this pile, and separates the remaining ones into two non-empty piles of arbitrary size. Bob stops when all piles have identical size. What is this size and what is the maximal number of piles?

First, Bob can easily reach a decomposition that prevents all piles to be of the same size: for instance, he can start with a pile of 1 and another pile of 2. Looking at the general perspective, an odd number of tokens, n=2k+1, can be partitioned into (1,1,2k-1). Which means that the decomposition (1,1,…,1) involving k+1 ones can always be achieved. For an even number, n=2k, this is not feasible. If the number 2k can be partitioned into equal numbers u, this means that the sequence 2k-(u+1),2k-2(u+1),… ends up with u, hence that there exist m such that 2k-m(u+1)=u or that 2k+1 is a multiple of (u+1). Therefore, the smallest value is made of the smallest factor of 2k+1. Minus one. For 2k=10⁴, this value is equal to 72, while it is 7 for 10³. The decomposition is impossible for 2k=100, since 101 is prime. Here are the R functions used to check this analysis (with small integers, if not 10⁴):

solvant <- function(piles){
 if ((length(piles)>1)&((max(piles)==2)||(min(piles)==max(piles)))){
   while (piles[i]<3)

disolvant <- function(piles){
 while (min(sol)<max(sol))

resolvant <- function(piles){
 for (t in 1:piles){
 if (length(sol)>maxle){

One Response to “Le Monde puzzle [#965]”

  1. Derek Wilson Says:

    Why are we forced to choose the pile at random? Why not always choose and split the pile based on the notion that we create 3 piles of equal size as an objective and repeat?

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