## Le Monde puzzle [#909]

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

Find all integers n with 3 or 4 digits, no exterior zero digit, and a 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)}}}
```

```> 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.)

### 5 Responses to “Le Monde puzzle [#909]”

1. Doesn’t 100 technically qualify?

• No, only integers with non-zero extreme digits can be used.

• Which was not mentioned in the original post, sorry!

2. Philip Whittall Says:

So, are the following propositions true if we allow any number of digits …
1> All solutions are divisible by 3.
2> 108 is the only solution not divisible by 5.
3> All solutions > 105 are divisible by 9

Philip

• Judging from the only possible solutions, the answer is yes!

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