## Le Monde puzzle [#905]

A recursive programming  Le Monde mathematical puzzle:

Given n tokens with 10≤n≤25, Alice and Bob play the following game: the first player draws an integer1≤m≤6 at random. This player can then take 1≤r≤min(2m,n) tokens. The next player is then free to take 1≤s≤min(2r,n-r) tokens. The player taking the last tokens is the winner. There is a winning strategy for Alice if she starts with m=3 and if Bob starts with m=2. Deduce the value of n.

Although I first wrote a brute force version of the following code, a moderate amount of thinking leads to conclude that the person given n remaining token and an adversary choice of m tokens such that 2m≥n always win by taking the n remaining tokens:

```optim=function(n,m){

outcome=(n<2*m+1)
if (n>2*m){
for (i in 1:(2*m))
outcome=max(outcome,1-optim(n-i,i))
}
return(outcome)
}
```

eliminating solutions which dividers are not solutions themselves:

```sol=lowa=plura[plura<100]
for (i in 3:6){
sli=plura[(plura>10^(i-1))&(plura<10^i)]
ace=sli-10^(i-1)*(sli%/%10^(i-1))
lowa=sli[apply(outer(ace,lowa,FUN="=="),
1,max)==1]
lowa=sort(unique(lowa))
sol=c(sol,lowa)}
```

```> subs=rep(0,16)