# Le Monde puzzle [#1075]

December 11, 2018
By

(This article was first published on R – Xi'an's Og, and kindly contributed to R-bloggers)

A new Le Monde mathematical puzzle in the digit category:

Find the largest number such that each of its internal digits is strictly less than the average of its two neighbours. Same question when all digits differ.

For instance, n=96433469 is such a number. When trying pure brute force (with the usual integer2digits function!)

```le=solz=3
while (length(solz)>0){
solz=NULL
for (i in (10^(le+1)-1):(9*10^le+9)){
x=as.numeric(strsplit(as.character(i), "")[[1]])
if (min(x[-c(1,le+1)]<(x[-c(1,2)]+x[-c(le,le+1)])/2)==1){ print(i);solz=c(solz,i); break()}}
le=le+1}
```

this is actually the largest number returned by the R code. There is no solution with 9 digits. Adding an extra condition

```le=solz=3
while (length(solz)>0){
solz=NULL
for (i in (10^(le+1)-1):(9*10^le+9)){
x=as.numeric(strsplit(as.character(i), "")[[1]])
if ((min(x[-c(1,le+1)]<(x[-c(1,2)]+x[-c(le,le+1)])/2)==1)&
(length(unique(x))==le+1)){ print(i);solz=c(solz,i); break()}}
le=le+1}
```

produces n=9520148 (seven digits) as the largest possible integer.

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