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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 integer.

a slight modification in the R code allows for a faster exploration, based on the fact that solutions add one extra digit to solutions with one less digit:

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

pluri=plura=NULL
#solutions with two digits
for (i in 11:99){
dive=rev(digin(i)[-1])
if (min(dive)>0){
dive=sum(dive*10^(0:(length(dive)-1)))
if (i==((i%/%dive)*dive))
pluri=c(pluri,i)}}
for (n in 2:6){ #number of digits
plura=c(plura,pluri)
pluro=NULL
for (j in pluri){
for (k in (1:9)*10^n){
x=k+j
if (x==(x%/%j)*j)
pluro=c(pluro,x)}
}
pluri=pluro}