# Le Monde puzzle [#920]

**Xi'an's Og » R**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

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**A** puzzling Le Monde mathematical puzzle (or blame the heat wave):

A pocket calculator with ten keys (0,1,…,9) starts with a random digit n between 0 and 9. A number on the screen can then be modified into another number by two rules:

1. pressing k changes the k-th digit v whenever it exists into (v+1)(v+2) where addition is modulo 10;

2. pressing 0k deletes the (k-1)th and (k+1)th digits if they both exist and are identical (otherwise nothing happens.

Which 9-digit numbers can always be produced whatever the initial digit?

I did not find an easy entry to this puzzle, in particular because it did not state what to do once 9 digits had been reached: would the extra digits disappear? But then, those to the left or to the right? The description also fails to explain how to handle n=000 000 004 versus n=4.

Instead, I tried to look at the numbers with less than 7 digits that could appear, using some extra rules of my own like preventing numbers with more than 9 digits. Rules which resulted in a sure stopping rule when applying both rules above at random:

leplein=rep(0,1e6) for (v in 1:1e6){ x=as.vector(sample(1:9,1)) for (t in 1:1e5){ k=length(x) #as sequence of digits if (k<3){ i=sample(rep(1:k,2),1) x[i]=(x[i]+1)%%10 y=c(x[1:i],(x[i]+1)%%10) if (i<k){ x=c(y,x[(i+1):k])}else{ x=y} }else{ prop1=prop2=NULL difs=(2:(k-1))[abs(x[-(1:2)]-x[-((k-1):k)])==0] if (length(difs)>0) prop1=sample(rep(difs,2),1) if (k<9) prop2=sample(rep(1:k,2),1) if (length(c(prop1,prop2))>1){ if (runif(1)<.5){ x[prop2]=(x[prop2]+1)%%10 y=c(x[1:prop2],(x[prop2]+1)%%10) if (prop2<k){ x=c(y,x[(prop2+1):k])}else{ x=y} }else{ x=x[-c(prop1-1,prop1+1)]} while ((length(x)>1)&(x[1]==0)) x=x[-1]} if (length(c(prop1,prop2))==1){ if (is.null(prop2)){ x=x[-c(prop1-1,prop1+1)] }else{ x[prop2]=(x[prop2]+1)%%10 y=c(x[1:prop2],(x[prop2]+1)%%10) if (prop2<k){ x=c(y,x[(prop2+1):k]) }else{ x=y} x=c(x[1:(prop2-1)], (x[prop2]+1)%%10, (x[prop2]+2)%%10,x[(prop2+1):k])} while ((length(x)>1)&(x[1]==0)) x=x[-1]} if (length(c(prop1,prop2))==0) break() } k=length(x) if (k<7) leplein[sum(x*10^((k-1):0))]= leplein[sum(x*10^((k-1):0))]+1 }}

code that fills an occupancy table for the numbers less than a million over 10⁶ iterations. The solution as shown below (with the number of zero entries over each column) is rather surprising in that it shows an occupancy that is quite regular over a grid. While it does not answer the original question…

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