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(This article was first published on <strong><a href="https://xianblog.wordpress.com/2016/06/14/le-monde-puzzle-965/"> R – Xi'an's Og</a></strong>, and kindly contributed to <a rel="nofollow" href="http://www.r-bloggers.com/">R-bloggers)</a> </div>
<p><strong><img height="185" alt="" width="273" src="https://encrypted-tbn3.gstatic.com/images?q=tbn:ANd9GcRsQd3v7efgSlmsNPKddzl-oQS9fZOF8a68V6KPmf5dnmI_EJ66rw">A</strong> game-related <a title="Le Monde puzzle [website]" target="_blank" href="https://xianblog.wordpress.com/2011/09/03/le-monde-puzzle-website/">Le Monde mathematical puzzle</a>:</p>
<blockquote>
<p><em>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</em><em>?<br>
</em></p>
</blockquote>
<p>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):</p>
<pre>
solvant &lt;- function(piles){
 if ((length(piles)>1)&amp;((max(piles)==2)||(min(piles)==max(piles)))){
  return(piles)}else{
   i=sample(rep(1:length(piles),2),1,prob=rep(piles-min(piles)+.1,2))
   while (piles[i]&lt;3)
    i=sample(rep(1:length(piles),2),1,prob=rep(piles-min(piles)+.1,2))
   split=sample(rep(2:(piles[i]-1),2),1,
        prob=rep(abs(2:(piles[i]-1)-piles[i]/2)+.1,2))
   piles=c(piles[-i],split-1,piles[i]-split)
   solvant(piles)}}

disolvant &lt;- function(piles){
 sol=solvant(piles)
 while (min(sol)&lt;max(sol))
 sol=solvant(piles)
 return(sol)}

resolvant &lt;- function(piles){
 sol=disolvant(piles)
 lasol=sol;maxle=length(sol)
 for (t in 1:piles){
 sol=disolvant(piles)
 if (length(sol)>maxle){
 lasol=sol;maxle=length(sol)}}
 return(lasol)}
</pre>
<p>Filed under: <a target="_blank" href="https://xianblog.wordpress.com/category/kids/">Kids</a>, <a target="_blank" href="https://xianblog.wordpress.com/category/statistics/r-statistics/">R</a> Tagged: <a target="_blank" href="https://xianblog.wordpress.com/tag/le-monde/">Le Monde</a>, <a target="_blank" href="https://xianblog.wordpress.com/tag/mathematical-puzzle/">mathematical puzzle</a>, <a target="_blank" href="https://xianblog.wordpress.com/tag/r/">R</a>, <a target="_blank" href="https://xianblog.wordpress.com/tag/recursive-function/">recursive function</a> <a target="_blank" rel="nofollow" href="http://feeds.wordpress.com/1.0/gocomments/xianblog.wordpress.com/30734/"><img alt="" src="http://feeds.wordpress.com/1.0/comments/xianblog.wordpress.com/30734/" border="0"></a> <img height="1" alt="" width="1" src="https://pixel.wp.com/b.gif?host=xianblog.wordpress.com&blog=5051449&%23038;post=30734&%23038;subd=xianblog&%23038;ref=&%23038;feed=1" border="0"></p>
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