## Le Monde puzzle [#909]

**A**nother of those “drop-a-digit” Le Monde mathematical puzzle:

Find allintegers 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)}}}

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.)

May 1, 2015 at 5:10 pm

Doesn’t 100 technically qualify?

May 1, 2015 at 5:38 pm

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

May 1, 2015 at 5:39 pm

Which was not mentioned in the original post, sorry!

May 1, 2015 at 11:01 am

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

May 2, 2015 at 5:58 pm

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