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),"")[[1]])}

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)}

which leads to the output

> sol
 [1] 11 12 15 21 22 24 25 31 32 33 35 36
[13] 41 42 44 45 48 51 52 55 61 62 63 64
[25] 65 66 71 72 75 77 81 82 84 85 88 91
[37] 92 93 95 96 99 125 225 312 315 325 375 425
[49] 525 612 615 624 625 675 725 735 825 832 912
[61] 915 925 936 945 975 1125 2125 3125 3375 4125
[70] 5125 5625
[72] 6125 6375 7125 8125 9125 9225 9375 53125
[80] 91125 95625

leading to the conclusion there is no solution beyond 95625.