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**A**n algorithmic Le Monde mathematical puzzle with a direct

Alice and Bob play a game by picking alternatively one of the remaining digits between 1 and 10 and putting it in either one of two available stacks, 1 or 2. Their respective gains are the products of the piles (1 for Alice and 2 for Bob).

The problem is manageable by a recursive function

facten=factorial(10) pick=function(play=1,remz=matrix(0,2,5)){ if (sum(remz==0)==1){#finale remz[remz==0]=(1:10)[!(1:10)%in%remz] return(prod(remz[play,])) }else{ gainz=0 if (min(remz[1,])==0){ for (i in (1:10)[!(1:10)%in%remz]){ propz=rbind(c(remz[1,remz[1,]>0],i, rep(0,sum(remz[1,]==0)-1)),remz[2,]) gainz=max(gainz,facten/pick(3-play,remz=propz))}} if (min(remz[2,])==0){ for (i in (1:10)[remz==0]){ propz=rbind(remz[1,],c(remz[2,remz[2,]>0],i, rep(0,sum(remz[2,]==0)-1))) gainz=max(gainz,facten/pick(3-play,remz=propz))}} return(gainz)}}

that shows the optimal gain for Alice is 6480=3x5x6x8x9, versus Bob getting 1680=1x2x4x7x10. The moves ensuring the gain are 2-10-…

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