Another of those “drop-a-digit” Le Monde mathematical puzzle:

Find all integers n with 3 or 4 digits an single interior zero digit, such that removing that zero digit produces a divider of x.

As in puzzle #904, I made use of the digin R function:

digin=function(n){
  as.numeric(strsplit(as.character(n),"")[[1]])}

and simply checked all integers up to 10⁶:

plura=divid=NULL
for (i in 101:10^6){
 dive=rev(digin(i))
 if ((min(dive[1],rev(dive)[1])>0)&
    (sum((dive[-c(1,length(dive))]==0))==1)){
   dive=dive[dive>0]
   dive=sum(dive*10^(0:(length(dive)-1)))
 if (i==((i%/%dive)*dive)){
   plura=c(plura,i)
   divid=c(divid,dive)}}}

which leads to the output

> plura
1] 105 108 405 2025 6075 10125 30375 50625 70875
> plura/divid
[1] 7 6 9 9 9 9 9 9 9

leading to the conclusion there is no solution beyond 70875. (Allowing for more than a single zero within the inner digits sees many more solutions.)