## Le Monde puzzle [#904] An arithmetics Le Monde mathematical puzzle:

Find all plural integers, namely positive integers such that (a) none of their digits is zero and (b) removing their leftmost digit produces a dividing plural integer (with the convention that one digit integers are all plural).

An easy arithmetic puzzle, with no real need for an R code since it is straightforward to deduce the solutions. Still, to keep up with tradition, here it is!

First, I found this function on Stack Overflow to turn an integer into its digits:

```digin=function(n){
as.numeric(strsplit(as.character(n),"")[])}
```

then I simply checked all integers up to 10⁶:

```plura=NULL
for (i in 11:10^6){
dive=rev(digin(i)[-1])
if (min(dive)>0){
dive=sum(dive*10^(0:(length(dive)-1)))
if (i==((i%/%dive)*dive))
plura=c(plura,i)}}
```

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)}
```

```> sol
 11 12 15 21 22 24 25 31 32 33 35 36
 41 42 44 45 48 51 52 55 61 62 63 64
 65 66 71 72 75 77 81 82 84 85 88 91
 92 93 95 96 99 125 225 312 315 325 375 425
 525 612 615 624 625 675 725 735 825 832 912
 915 925 936 945 975 1125 2125 3125 3375 4125
 5125 5625
 6125 6375 7125 8125 9125 9225 9375 53125
 91125 95625
```

leading to the conclusion there is no solution beyond 95625.

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