Some divisors of positive integers really hate each other and they don't like to share one or more common digits.

Those integers are called Hostile Divisor Numbers (HDN)


Number 9566 has 4 divisors: 1, 2, 4783 and 9566
(as you can see, no two of them share the same digit).
Thus, 9566 is a Hostile Divisor Number

Number 9567 is NOT HDN because its divisors (1, 3, 9, 1063, 3189, 9567) share some common digits.

Here are the first few HDN



The above list goes on and your task is to find the nth HDN


A positive integer n from 1 to 4000


The nth HDN

Test Cases

here are some 1-indexed test cases.
Please state which indexing system you use in your answer to avoid confusion.

input -> output     
 1        1     
 10       23       
 101      853     
 1012     26053     
 3098     66686      
 4000     85009      

This is , so the lowest score in bytes wins.


Good news! I submitted my sequence to OEIS and...
Hostile Divisor Numbers are now OEIS A307636

  • 1
    \$\begingroup\$ I think square numbers would be the least hostile of numbers. \$\endgroup\$ May 3, 2019 at 17:59
  • 3
    \$\begingroup\$ @JoeFrambach That I do not understand. There are perfect-square HDN. For a somewhat large example, 94699599289, the square of 307733, has divisors [1, 307733, 94699599289] which shows it is a HDN. Seems hostile to me. \$\endgroup\$ May 5, 2019 at 17:21
  • \$\begingroup\$ @JeppeStigNielsen For a much smaller example, why not just 49? Factors [1, 7, 49] qualifies as hostile... Or, well, 4: [1, 2, 4]... \$\endgroup\$ May 6, 2019 at 13:27
  • \$\begingroup\$ @DarrelHoffman Not to mention, the square number 1 with divisor list [1]. (Maybe large HDN are more interesting?) \$\endgroup\$ May 6, 2019 at 18:09
  • \$\begingroup\$ I interpreted 49 as having divisors [7, 7], which not only share digits but are the same digits. 49 has factors [1, 7, 49] \$\endgroup\$ May 6, 2019 at 19:59

20 Answers 20


05AB1E, 12 10 bytes


-2 bytes thanks to @Emigna.


Try it online or verify most test cases (last two test cases are omitted, since they time out).


µ           # Loop while the counter_variable is not equal to the (implicit) input yet:
 N          #  Push the 0-based index of the loop to the stack
  NÑ        #  Get the divisors of the 0-based index as well
            #   i.e. N=9566 → [1,2,4783,9566]
            #   i.e. N=9567 → [1,3,9,1063,3189,9567]
    €Ù      #  Uniquify the digits of each divisor
            #   → ["1","2","4783","956"]
            #   → ["1","3","9","1063","3189","9567"]
      S     #  Convert it to a flattened list of digits
            #   → ["1","2","4","7","8","3","9","5","6"]
            #   → ["1","3","9","1","0","6","3","3","1","8","9","9","5","6","7"]
       D    #  Duplicate this list
        Ù   #  Unique the digits
            #   → ["1","2","4","7","8","3","9","5","6"]
            #   → ["1","3","9","0","6","8","5","7"]
         Q  #  And check if it is still equal to the duplicated list
            #   → 1 (truthy)
            #   → 0 (falsey)
            #  And if it's truthy: implicitly increase the counter_variable by 1
            # (After the loop: implicitly output the top of the stack,
            #  which is the pushed index)
  • 2
    \$\begingroup\$ You beat me to it this time. I had µNNÑ€ÙSDÙQ for 10. \$\endgroup\$
    – Emigna
    May 3, 2019 at 14:05
  • 2
    \$\begingroup\$ @Emigna Ah, I was just working on an alternative with µ, so you save me the trouble. ;) \$\endgroup\$ May 3, 2019 at 14:06
  • \$\begingroup\$ this is poetically eloquent \$\endgroup\$
    – don bright
    May 8, 2019 at 23:45

Python 2, 104 bytes

while n: 
 while i:m=set(`i`*(x%i<1));c*=d-m==d;d|=m;i-=1
print x

Try it online!



JavaScript (ES6), 78 bytes



Try it online!

Faster version, 79 bytes

n=>{for(k=0;n;n-=!d)for(s=d=++k+'';k%--d||d*!s.match(`[${s+=d,d}]`););return k}

Try it online!


Given an integer \$k>0\$, we build the string \$s\$ as the concatenation of all divisors of \$k\$.

Because \$k\$ is always a divisor of itself, \$s\$ is initialized to \$k\$ (coerced to a string) and the first divisor that we try is \$d=k-1\$.

For each divisor \$d\$ of \$k\$, we test whether any digit of \$d\$ can be found in \$s\$ by turning \$d\$ into a character set in a regular expression.


  • \$s=\text{"}956647832\text{"}\$, \$d=1\$"956647832".match(/[1]/) is falsy
  • \$s=\text{"}9567\text{"}\$, \$d=3189\$"9567".match(/[3189]/) is truthy


This is the version without eval(), for readability

n => {                   // n = input
  for(                   // for() loop:
    k = 0;               //   start with k = 0
    n;                   //   go on until n = 0
    n -= !d              //   decrement n if the last iteration resulted in d = 0
  )                      //
    for(                 //   for() loop:
      s =                //     start by incrementing k and
      d = ++k + '';      //     setting both s and d to k, coerced to a string
      k % --d ||         //     decrement d; always go on if d is not a divisor of k
      d *                //     stop if d = 0
      !s.match(          //     stop if any digit of d can be found in s
        `[${s += d, d}]` //     append d to s
      );                 //
    );                   //   implicit end of inner for() loop
                         // implicit end of outer for() loop
  return k               // return k
}                        //

Jelly, 10 bytes


Try it online!

-1 byte thanks to ErikTheOutgolfer

Takes input from STDIN, which is unusual for Jelly but normal where nfind is used.

ÆDQ€FQƑµ#Ṫ  Main link
         Ṫ  Get the last element of
        #   The first <input> elements that pass the filter:
ÆD          Get the divisors
  Q€        Uniquify each (implicitly converts a number to its digits)
    F       Flatten the list
     QƑ     Does that list equal itself when deduplicated?


  • \$\begingroup\$ is this 2-indexed? It's ok with me but please state it for others \$\endgroup\$
    – ZaMoC
    May 3, 2019 at 14:20
  • \$\begingroup\$ It's whatever your test cases were, so 1 \$\endgroup\$
    – hyper-neutrino
    May 3, 2019 at 14:21
  • 3
    \$\begingroup\$ No it isn't. 101 returns 839. and 102 -> 853. It works fine but it is 2-indexed \$\endgroup\$
    – ZaMoC
    May 3, 2019 at 14:23
  • 1
    \$\begingroup\$ @J42161217 wait what? i guess when i moved the nfind it changed the indexing lol \$\endgroup\$
    – hyper-neutrino
    May 3, 2019 at 14:33
  • 1
    \$\begingroup\$ ⁼Q$ is the same as . \$\endgroup\$ May 3, 2019 at 14:35

Perl 6, 53 bytes

{(grep {/(.).*$0/R!~~[~] grep $_%%*,1..$_},^∞)[$_]}

Try it online!


/(.).*$0/ matches any number with a repeated digit.

grep $_ %% *, 1 .. $_ returns a list of all divisors of the number $_ currently being checked for membership in the list.

[~] concatenates all of those digits together, and then R!~~ matches the string on the right against the pattern on the left. (~~ is the usual match operator, !~~ is the negation of that operator, and R is a metaoperator that swaps the arguments of !~~.)

  • \$\begingroup\$ 50 bytes \$\endgroup\$
    – Jo King
    May 8, 2019 at 21:43
  • \$\begingroup\$ Ah wait, this checks for digits within the same divisor. The regex should be /(\d).*\s.*$0/, where you don't concatenate the divisors \$\endgroup\$
    – Jo King
    Sep 12, 2020 at 11:11

Python 2 (PyPy), 117 114 bytes

Uses 1-indexing

while k:n+=1;k-=1-any(set(`a`)&set(`b`)for a in r(1,n+1)for b in r(1,a)if n%a<1>n%b)
print n

Try it online!


Wolfram Language 103 bytes

Uses 1-indexing. I'm surprised it required so much code.

  • \$\begingroup\$ Can you please add a TIO link so that everybody can check your answer? \$\endgroup\$
    – ZaMoC
    May 3, 2019 at 16:40
  • \$\begingroup\$ 95 bytes: (n=t=1;While[t<=#,If[!Or@@IntersectingQ@@@Subsets[IntegerDigits@Divisors@n,{2}],t++];n++];n-1)& I am not planning to post an answer so I will leave this here \$\endgroup\$
    – ZaMoC
    May 3, 2019 at 16:45
  • \$\begingroup\$ @J42161217, I've been trying to get the code to work in TIO without success. There must be some trick I'm missing. \$\endgroup\$
    – DavidC
    May 3, 2019 at 20:55
  • \$\begingroup\$ @J42161217, Your code seems to work but takes 3 times the runtime. You can submit it as your own. (Maybe I'll learn how to implement TIO from your example.) \$\endgroup\$
    – DavidC
    May 3, 2019 at 20:57
  • 1
    \$\begingroup\$ 72 bytes using For instead of While, E!=##& to test for repeated digits \$\endgroup\$
    – att
    May 3, 2019 at 22:23

PowerShell, 112 bytes

for($a=$args[0];$a-gt0){$z=,0*10;1..++$n|?{!($n%$_)}|%{"$_"|% t*y|sort -u|%{$z[+"$_"]++}};$a-=!($z|?{$_-ge2})}$n

Try it online!

Takes 1-indexed input $args[0], stores that into $a, loops until that hits 0. Each iteration, we zero-out a ten-element array $z (used to hold our digit counts). Then we construct our list of divisors with 1..++$n|?{!($n%$_)}. For each divisor, we cast it to a string "$_", cast it toCharArray, and sort those digits with the -unique flag (because we don't care if a divisor itself has duplicate digits). We then increment the appropriate digit count in $z. Then, we decrement $a only if $z contains 0s and 1s (i.e., we've found an HDN). If we've finished our for loop, that means we found the appropriate number of HDNs, so we leave $n on the pipeline and output is implicit.

  • \$\begingroup\$ you could save some bytes: $a-=!($z-ge2) instead $a-=!($z|?{$_-ge2}) \$\endgroup\$
    – mazzy
    May 4, 2019 at 6:40
  • \$\begingroup\$ a bit golfed \$\endgroup\$
    – mazzy
    May 4, 2019 at 8:13

Python 3, 115 bytes


f=lambda n,x=1,s="",l="",d=1:n and(d>x+1and f(n-1,x+1)or{*s}&{*l}and f(n,x+1)or f(n,x,s+l,(1-x%d)*str(d),d+1))or~-x

Try it online!

This uses a lot of recursion; even with increased recursion limit, it can't do f(30). I think it might be golfable further, and I tried finding something to replace the (1-x%d) with, but couldn't come up with anything (-~-x%d has the wrong precedence). Any bytes that can be shaved off are greatly appreciated.

How it works

# n: HDNs to go
# x: Currently tested number
# s: String of currently seen divisor digits
# l: String of digits of last tried divisor if it was a divisor, empty string otherwise
# d: Currently tested divisor

f=lambda n,x=1,s="",l="",d=1:n and(                    # If there are still numbers to go
                             d>x+1and f(n-1,x+1)or     # If the divisors have been
                                                       #  exhausted, a HDN has been found
                             {*s}&{*l}and f(n,x+1)or   # If there were illegal digits in
                                                       #  the last divisor, x isn't a HDN
                                                       # Else, try the next divisor, and
                                                       #  check this divisor's digits (if
                                                       #  if is one) in the next call
                             )or~-x                    # Else, return the answer

Brachylog (v2), 14 bytes


Try it online!

Function submission; input from the left, output to the right. (The TIO link contains a command-line argument to run a function as though it were a full program.)


"Is this a hostile divisor number?" code:

ℕ       number is ≥0 (required to match the question's definition of "nth solution")
 f      list of all factors of the number
   ᵐ    for each factor
  d       deduplicate its digits
    c   concatenate all the deduplications with each other
     ≠  the resulting number has no repeated digits

This turned out basically the same as @UnrelatedString's, although I wrote it independently.

"nth solution to a " wrapper:

    &      output the successful input to
  {  }ᶠ    the first n solutions of the problem
       ⁽   taking <n, input> as a pair
;A         form a pair of user input and a "no constraints" value
        t  take the last solution (of those first n)

This is one of those cases where the wrapper required to produce the nth output is significantly longer than the code required to test each output in turn :-)

I came up with this wrapper independently of @UnrelatedString's. It's the same length and works on the same principle, but somehow ends up being written rather differently. It does have more potential scope for improvement, as we could add constraints on what values we were looking at for free via replacing the A with some constraint variable, but none of the possible constraint variables save bytes. (If there were a "nonnegative integer" constraint variable, you could replace the A with it, and then save a byte via making the the unnecessary.)

  • \$\begingroup\$ It’s 2-indexed? \$\endgroup\$
    – FrownyFrog
    May 9, 2019 at 10:03

Java 10, 149 139 138 126 125 120 119 bytes

n->{int r=0,i,d;for(;n>0;n-=d){var s="1";for(r+=d=i=1;i++<r;)if(r%i<1){d=s.matches(".*["+i+"].*")?0:d;s+=i;}}return r;}

-10 bytes by using .matches instead of .contains per digit, inspired by @Arnauld's JavaScript answer.
-5 bytes thanks to @ValueInk
-1 byte thanks to @ceilingcat


Try it online.


n->{                 // Method with integer as both parameter and return-type
  int r=0,           //  Result-integer, starting at 0
      i,             //  Index integer
      d;             //  Decrement integer
  for(;n>0;          //  Loop until the input `n` is 0:
      n-=d){         //    After every iteration: decrease `n` by the decrement integer `d`
    var s="1";       //   Create a String `s`, starting at "1"
    for(r+=d=i=1;    //   (Re)set the decrement and index integers to 1,
                     //   and increase the result by 1 as well
        i++<r;)      //   Inner loop `i` in the range [2, r]:
      if(r%i<1){     //    If `r` is divisible by `i`:
                     //     If string `s` contains any digits also found in integer `i`:
           0         //      Set the decrement integer `d` to 0
          :d;        //     Else: leave `d` unchanged
        s+=i;}}      //     And then append `i` to the String `s`
  return r;}         //  After the loops, return the result `r`

Brachylog, 16 bytes


Try it online!

Very slow, and twice as long as it would be if this was a . 1-indexed.

                    The output
               t    is the last
             ᵘ⁾     of a number of unique outputs,
g                   where that number is the input,
 {          }       from the predicate declaring that:
     .              the output
    <               which is greater than
   0                zero
  ∧                 (which is not the empty list)
      f             factorized
        ᵐ           with each factor individually
       d            having duplicate digits removed
          ≠         has no duplicate digits in
         c          the concatenation of the factors
           ∧        (which is not the output).
  • 1
    \$\begingroup\$ If you just read that explanation as a sentence though... \$\endgroup\$
    – FireCubez
    May 4, 2019 at 11:24
  • \$\begingroup\$ I try to write my explanations like plain English, which typically ends up just making them harder to read \$\endgroup\$ May 4, 2019 at 20:25

Wolfram Language (Mathematica), 74 bytes


Try it online!


Japt v2.0a0, 17 bytes

_=â ®sâìUµZ¶â}f1

Try it

Port of this Brachylog answer.

Credit: 4 bytes savings total thanks to Shaggy who also suggested there was a better solution leading to many more bytes :)

Original answer 28 byte approach:

Èâ¬rÈ«è"[{Y}]" ©X+Y}Xs)«U´Ãa

Try it

Port of this JavaScript answer.

  • \$\begingroup\$ 28 bytes \$\endgroup\$
    – Shaggy
    May 7, 2019 at 17:04
  • \$\begingroup\$ Nice - I hadn't used the « shortcut before :) I figure if Shaggy is only improving on my score by a handful of bytes, I must be getting (somewhat) decent at this? \$\endgroup\$
    – dana
    May 7, 2019 at 17:42
  • \$\begingroup\$ It can be done in 20 (maybe less) b7 employing a slightly different method. \$\endgroup\$
    – Shaggy
    May 7, 2019 at 18:43
  • \$\begingroup\$ Hah - I guess I spoke too soon :) yeah, some of the other golfing lang's have much shorter solutions. \$\endgroup\$
    – dana
    May 7, 2019 at 19:29
  • 1
    \$\begingroup\$ 17 bytes \$\endgroup\$
    – Shaggy
    May 8, 2019 at 17:56

Perl 5 -p, 66 bytes

map{1while(join$",map{$\%$_==0&&$_}1..++$\)=~/(\d).* .*\1/}1..$_}{

Try it online!

1 indexed


Husk, 10 bytes


Try it online!

Same method as Jelly.

Husk, 19 bytes


Try it online!

The more manual cehcker.


Icon, 123 bytes

procedure f(n)
while m<n do{
every k%(i:=1 to k)=0&(upto(i,s)&r:=1)|s++:=i
return k

Try it online!

1-indexed. Really slow for big inputs.


Perl 6, 74 bytes

{(grep {!grep *>1,values [(+)] map *.comb.Set,grep $_%%*,1..$_},1..*)[$_]}

0-indexed. Only the first three cases are listed on TIO since it's too slow to test the rest.

Try it online!

  • 1
    \$\begingroup\$ 57 bytes \$\endgroup\$
    – Jo King
    May 8, 2019 at 6:07

Ruby, 110 97 92 84 bytes

-13 bytes by leveraging @Arnauld's JavaScript regex check.

-5 bytes for swapping out the times loop for a decrementer and a while.

-8 bytes by ditching combination for something more like the other answers.

->n{x=0;n-=1if(s='';1..x+=1).all?{|a|x%a>0||(e=/[#{a}]/!~s;s+=a.to_s;e)}while n>0;x}

Try it online!


J, 87 59 bytes

-28 bytes thanks to FrownFrog

0{(+1,1(-:~.)@;@(~.@":&.>@,i.#~0=i.|])@+{.)@]^:(>{:)^:_&0 0

Try it online!


J, 87 bytes

[:{:({.@](>:@[,],([:(-:~.)[:-.&' '@,/~.@":"0)@((]#~0=|~)1+i.)@[#[)}.@])^:(#@]<1+[)^:_&1

Try it online!


This is atrociously long for J, but I'm not seeing great ways to bring it down.


It helps to introduce a couple helper verbs to see what's happening:

h=. [: (-:~.) [: -.&' '@,/ ~.@":"0
  • d returns a list of all divisors of its argument
  • h tells you such a list is hostile. It stringifies and deduplicates each number ~.@":"0, which returns a square matrix where shorter numbers are padded with spaces. -.&' '@,/ flattens the matrix and removes spaces, and finally (-:~.) tells you if that number has repeats or not.

With those two helpers our overall, ungolfed verb becomes:

[: {: ({.@] (>:@[ , ] , h@d@[ # [) }.@])^:(#@] < 1 + [)^:_&1

Here we maintain a list whose head is our "current candidate" (which starts at 1), and whose tail is all hostile numbers found so far.

We increment the head of the list >:@[ on each iteration, and only append the "current candidate" if it is hostile h@d@[ # [. We keep doing this until our list length reaches 1 + n: ^:(#@] < 1 + [)^:_.

Finally, when we're done, we return the last number of this list [: {: which is the nth hostile number.

  • \$\begingroup\$ 66 \$\endgroup\$
    – FrownyFrog
    May 9, 2019 at 9:21
  • \$\begingroup\$ 62 \$\endgroup\$
    – FrownyFrog
    May 9, 2019 at 9:41
  • \$\begingroup\$ This is great, many thanks. Will go over it and update tonight \$\endgroup\$
    – Jonah
    May 9, 2019 at 12:39
  • \$\begingroup\$ 59 \$\endgroup\$
    – FrownyFrog
    May 9, 2019 at 23:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.