[Question]
  [
I have come across a [question](https://codereview.stackexchange.com/q/107221) on the Code Review site that seems interesting. I think OP is doing it wrong, but cannot be sure... So let's solve it for him! (write a program, not a function/procedure)
## Input (stdin or similar):
An integer `x` in decimal notation. It is greater than 1 and less than 2^31.
## Output (stdout or similar):
An integer `y` in decimal notation. The product `x * y` in decimal representation must contain only digits 0 and 1. It must be the minimal such number greater than 0.
Note: output is not limited - if the minimal `y` is around 10^100, your program must output all of its 100 digits (I don't know whether there is a reasonable limit, like 2^64, on `y` - didn't solve it).
Your program should finish in a reasonable time (1 second? 1 hour? - something like that) for all `x` in range.
### Bonus:
If your program doesn't have a limit on the size of the input (except RAM), and has polynomial complexity, multiply the byte count of your program by `0.8` and round down.
---
Example: Input `2`; output `5`, because 2 \* 5 = 10
Example: Input `21`; output `481`, because 21 \* 481 = 10101
---
Disclaimer: I am not responsible for the question on the Code Review site. In case of any discrepancy, only the description above should be regarded as proper spec.
[**OEIS A079339**](http://oeis.org/A079339)
  
[Answer]
# Python 2, efficient solution, 99
```
n=input()
d={n:0}
k=1
while min(d):[d.setdefault((x+k)%n,d[x]+k)for x in set(d)];k*=10
print d[0]/n
```
Thanks Sp3000 for some golfing tips.
I challenge everybody else to post (in their own answers) how long it takes to get the result for input `72` or `99` :) If those are really fast, try something like `79992` next (still <1 sec here).
**Explanation:**
I thought this wasn't necessary (since the code is fairly readable), but I got a request, so here it goes:
The first idea is that a binary-looking number is a sum of 1 or more different powers of 10. Therefore, we can try to add various powers of 10 in different ways until we get remainder 0.
If we do that naively, it's the same as generating all binary-looking numbers and testing them. But a lot of remainders will be the same. A better way is to record only the smallest number that gave a certain remainder, and successively add greater powers of 10 to numbers we recorded. That's what the program does.
`d` is a dictionary/map where keys are remainders and values are binary-looking numbers with that remainder. The initial `n:0` is a special case: it's supposed to be `0:0` so we can start adding powers to it, but the algorithm stops when finding key 0, so I used `n` instead, which is guaranteed to have the same effect and not interfere with the other values.
Then we start adding powers of 10 (stored in `k`) to all existing numbers and recording the remainders. We add `k` to the remainder: `(x+k)%n` and to the number: `d[x]+k`, and record it only if it's a new remainder: `d.setdefault(…)`, then go to the next power: `k*=10` and repeat until we get key 0: `while min(d)`
At the end, `d[0]` gives the binary-looking number that has remainder 0 mod `n`, so we divide it by `n` to get the solution.
Note: the program can be made more efficient by avoiding large numbers (recording exponents rather than powers of 10, and calculating remainders of powers from previous values), but it's code golf, so...
In fact, here, I wrote a faster version:
```
n=input()
d={n:0}
k=1
b=0
while 0not in d:
 for x in list(d):d.setdefault((x+k)%n,b)
 k=(k*10)%n;b+=1
x=10**d[0]
while x%n:x+=10**d[n-x%n]
print x/n
```
[Answer]
## Python 2, 47 bytes
```
n=a=input()
while'1'<max(str(a)):a+=n
print a/n
```
Tracks the input number `n` and the current multiple `a`. When `a` looks like binary, output the ratio `a/n`. To check that a number is made of `0`'s and `1`'s, we compare the maximum char in its string representation to `'1'`.
Uses `str(a)` instead of ``a`` to avoid longs ending in `L`. Unfortunately, `'L'` is bigger than `'1'`.
[Answer]
# Perl, 27 bytes
```
#!perl -p
1while($_*++$\)=~/[2-9]/}{
```
Counting the shebang as one, input is taken from stdin.
**Sample Usage**
```
$ echo 2 | perl dec-bin.pl
5
$ echo 21 | perl dec-bin.pl
481
$ echo 98 | perl dec-bin.pl
112245
```
---
# Perl, 25 bytes
```
#!perl -p
eval'0b'.++$\*$_||redo}{
```
A two byte improvement by [@skmrx](https://codegolf.stackexchange.com/questions/60942/60981#comment146985_60981).
Rather than checking against a regex, this instead attempts to evaluate the product as a binary literal. Upon failure, it moves on to the next. Typically the `oct` function would be used for this purpose, but it silently trims invalid digits, which isn't useful in this challenge.
---
# Perl, 40 bytes
```
#!perl -p
1while($b=sprintf"%b",++$i)%$_;$_=$b/$_
```
A far more efficient solution. We iterate over binary representations, interpret them as base 10, and then check for divisibility. Runtimes for all values under 100 are negligible.
**Sample Usage**
```
$ echo 72|perl dec-bin.pl
1543209875
$ echo 99|perl dec-bin.pl
1122334455667789
```
[Answer]
# Javascript, 43 bytes
This ended up way shorter than I thought. It basically increments `y` up 1 until `y * (input number) = (binary-looking number)`. Obviously quite inefficient.
```
for(x=prompt(y=0);!+('0b'+x*++y););alert(y)
```
---
# Javascript (more efficient solution), 53 bytes
This one increments `y` in binary until `y / (input number) = (number without a remainder)`. Then it outputs `(number without a remainder)`.
```
for(x=prompt(y=1);(z=y.toString(2))%x;y++);alert(z/x)
```
---
# Javascript (even more efficient solution), 76 bytes
This one combines both of the previous methods described above. It checks increments `y` until either `y * (input number) = (binary-looking number)` (meaning that the output is `y`) OR `y / (input number) = (number without a remainder)` (meaning that the output is `(number without a remainder)`).
```
for(x=prompt(y=a=0);!a;a=+('0b'+x*++y)?y:(z=y.toString(2))%x?0:z/x);alert(a)
```
[Answer]
# Haskell, ~~72~~ ~~70~~ ~~64~~ ~~60~~ 58 bytes
```
main=do x<-readLn;print$[y|y<-[1..],all(<'2')$show$x*y]!!0
```
Edit: @Jan Dvorak helped me saving 4 bytes.
Edit: @BlackCap saved 2 bytes by switching to `do` notation. Thanks!
[Answer]
# Pyth, 9 bytes
```
f!-`*TQ10
```
[Demonstration](https://pyth.herokuapp.com/?code=f%21-%60%2aTQ10&input=21&debug=0)
For each multiple, convert to a string, subtract out the digits in `10` (using Pyth's handy int to str cast in this case) and then logically negate the result, terminating the seach only when the correct multiple is found.
### Bonus solution, 10 bytes:
```
f.xi`*TQ2Z
```
This solution actually checks if the string representation of the number can be treated as a binary number (`i ... 2`) and terminates when an error is not thrown on this attempt.
[Answer]
# Python 2, ~~67~~ ~~65~~ ~~63~~ 60 bytes
```
a=input();b=1
while set(`a*b`)&set('23456789'):b+=1
print b
```
Thanks to [Status](https://codegolf.stackexchange.com/users/34315/status) for 2 bytes and [Shebang](https://codegolf.stackexchange.com/users/38417/shebang) for 5 bytes!
[Answer]
# JavaScript (ES6) 222 ~~250~~
Using arbitrary precision math (operating on strings of decimal digits)
~~This can be golfed a little more~~(done), but I like the fact that it is not limited to JS standard numbers (17 decimal digits of precision) and that it is fast.
Test running the snippet below in an EcmaScript 6 compliant browser. Time is acceptable up to 9998 - don't try 9999 and be patient with 999.
```
// As a complete program with I/O via popup  
for(n=+prompt(a=[0],q=[t=1]);t;){for(c=1,t=i=0;i<a.length;i++)a[i]=a[i]&c?0:a[i]|c?(c=0,t+=q[i],1):c=0;c&&(a[i]=c,t+=q[i]=q[i-1]*10%n);t%=n}a.reverse().map(a=>(z+=[a],d=z/n|0,z%=n,r||d?r+=d:0),r='',z=0);alert([r,a.join``])
// As a testable function
f=n=>{
  for(a=[0],q=[t=1];t;)
  {
    for(c=1,t=i=0;i<a.length;i++)
      a[i]=a[i]&c?0:a[i]|c?(c=0,t+=q[i],1):c=0
    c&&(a[i]=c,t+=q[i]=q[i-1]*10%n);
    t%=n
  }  
  a.reverse().map(a=>(z+=[a],d=z/n|0,z%=n,r||d?r+=d:0),r='',z=0)
  return [r,a.join``]
}
// Test and timing
out = x => O.innerHTML += x + '\n'
setTimeout(_=>{
;[1,2,10, 21, 23, 98, 72, 9, 99, 999]
.forEach((test,i) => { 
  var t0 = ~new Date  
  var result = f(test)
  out('n='+test+' '+result+' time(ms) ' + (t0-~new Date))
})},100)  
```
```
<pre id=O>Timing test cases ...
</pre>
```
**More readable**
This is the first version, with modulus and long division as separated functions.
```
// function M - Modulus with arbitrary precision - a is a string of decimal digits
M = (a, b, q = 1, t = 0, j = a.length) => {
  while (j--) + a[j] ? t += q : 0, q = (q * 10) % b;
  return t % b
}
// function D - Long division with arbitrary precision - a is a string of decimal digits
D = (a, b, r = '', z = 0) => [...a].map(a => (z += a, d = z / b | 0, z %= b, r || d ? r += d : 0)) && r
// Testable function 
f = n => {
  for (i = 0; ++i < 1e7 && (z = M(v = i.toString(2), n)););
  return z ? ['big'] : [D(v, n), v]
}
```
[Answer]
## Perl, 45 bytes
```
do{$a++}while($a*($b||=<>))=~/[2-9]/;print$a;
```
[Answer]
# Python ~~3~~ 2, ~~101~~ 76 Bytes
**-25** bytes thanks to @aditsu
almost as efficient as @aditsu's solution
```
99 -> 0.436 Seconds
72 -> 0.007 Seconds
```
```
b,m,n=1,1,input()
while b%n:
 b=int("{0:b}".format(m))
 m+=1
print b/n
```
Instead of trying to loop through the multiples in increasing order, I'm trying to loop through the products, which I'm generating in 'binary' form.
[Answer]
## Pyth, 10 bytes
```
f<eS`*TQ`2
```
[Run the code.](https://pyth.herokuapp.com/?code=f%3CeS%60%2ATQ%602&input=21&debug=0)
A port of my [Python answer](https://codegolf.stackexchange.com/a/60975/20260), taking from [Maltysen](https://codegolf.stackexchange.com/a/60971/20260) the use of `f` to find the first positive number that meets a condition. 
[Answer]
# PHP, 50 bytes
```
while(preg_match('/[^01]/',$argv[1]*++$y));echo$y;
```
**Some test cases**
```
1 > 1
2 > 5
12 > 925
21 > 481
```
[Answer]
# CJam, ~~19~~ ~~17~~ 16 bytes
```
li:V!{)_V*sAs-}g
```
[Try it online](http://cjam.aditsu.net/#code=li%3AV!%7B)_V*sAs-%7Dg&input=21)
Brute force solution, trying values sequentially until one meeting the condition is found.
The latest version saves 2 bytes thanks to using `As` instead of `"01"` to build a string containing `0` and `1`, as suggested by @aditsu. The full proposed solution in the comment saves another byte, but it looks fairly different from mine, so I didn't want to post it under my name.
And 1 more byte saved by @Dennis.
Explanation:
```
li      Get input and convert to int.
:V      Save it in variable V.
!       Negate the value. Since we saved it in V, we don't need it on the stack anymore.
        But we need 0 on the stack as the start value for y. This conveniently
        accomplishes both with a single operator, since the input is guaranteed to be
        larger than 0.
{       Loop over y.
  )       Increment y.
  _       Copy it.
  V*      Multiply with input in variable V.
  s       Convert to string.
  As      Push the string "10", as the number 10 converted to a string .
  -       Remove 0 and 1 digits. This will result in an empty list if there were only
          0 and 1 digits. The empty list is falsy, and will terminate the loop.
}g      End loop.
```
[Answer]
# Java, 213 bytes
```
import java.math.*;class P{public static void main(String[]a){BigInteger b=new java.util.Scanner(System.in).nextBigInteger(),c,d=c=b.ONE;while(!(b.multiply(c)+"").matches("[01]+"))c=c.add(d);System.out.print(c);}}
```
Uses `BigInteger`s and as such has (for all reasonable intents and purposes) unbounded input size. Not sure about the complexity though, that depends on the growth rate of our function here.
Thanks to geobits and ypnypn for saving a handful of bytes.
[Answer]
# C++11, many bytes, very fast, wow (1.5 s on 1999999998, 0.2 s on 1…10000)
(Golfed Python version below.)
We start with a concept somewhat similar to aditsu’s solution, where we inductively build up a collection of modular remainders reachable in n steps. But instead of waiting until we find remainder 0, we check for two found remainders a and b such that a·10^n + b = 0. This meet-in-the-middle approach halves the depth of the search tree, so it’s much faster on large inputs and uses much less memory.
Some benchmarks:
```
$ echo 99999999 | \time ./decbin
1111111122222222333333334444444455555555666666667777777788888889
0.18user 0.01system 0:00.20elapsed 99%CPU (0avgtext+0avgdata 69360maxresident)k
0inputs+0outputs (0major+16276minor)pagefaults 0swaps
$ echo 999999999 | \time ./decbin
111111111222222222333333333444444444555555555666666666777777777888888889
1.22user 0.04system 0:01.27elapsed 100%CPU (0avgtext+0avgdata 434776maxresident)k
0inputs+0outputs (0major+37308minor)pagefaults 0swaps
$ echo 2147483647 | \time ./decbin
4661316525084584315813
0.00user 0.00system 0:00.01elapsed 72%CPU (0avgtext+0avgdata 5960maxresident)k
0inputs+0outputs (0major+1084minor)pagefaults 0swaps
$ echo 1999999998 | \time ./decbin
555555556111111111666666667222222222777777778333333333888888889444444445
1.42user 0.08system 0:01.50elapsed 100%CPU (0avgtext+0avgdata 544140maxresident)k
0inputs+0outputs (0major+38379minor)pagefaults 0swaps
$ \time ./decbin 10000.out
0.19user 0.00system 0:00.20elapsed 100%CPU (0avgtext+0avgdata 3324maxresident)k
0inputs+264outputs (0major+160minor)pagefaults 0swaps
```
Code:
```
#include <algorithm>
#include <boost/iterator/transform_iterator.hpp>
#include <fstream>
#include <list>
#include <iostream>
#include <string>
#include <utility>
#include <vector>
using namespace boost;
using namespace std;
static inline bool cmp_first_partnered(pair<int, pair<int, int>> a,
                                       pair<int, pair<int, int>> b) {
  return a.first < b.first;
}
static inline bool eq_first_partnered(pair<int, pair<int, int>> a,
                                      pair<int, pair<int, int>> b) {
  return a.first == b.first;
}
static pair<int, int> retrace(int modulus, int place, pair<int, int> state,
                              list<vector<int>>::iterator i,
                              list<vector<int>>::iterator j, string &ret) {
  if (i == j)
    return state;
  state = retrace(modulus, (place * 10LL) % modulus, state, next(i), j, ret);
  int remainder = state.first;
  long long k = state.second * 10LL;
  if (!binary_search(i->cbegin(), i->cend(), remainder)) {
    remainder = ((long long)remainder + modulus - place) % modulus;
    k += 1;
  }
  int digit = k / modulus;
  if (digit != 0 || ret.size())
    ret += '0' + digit;
  return make_pair(remainder, k % modulus);
}
static void mult(int modulus, int x, int y,
                 vector<pair<int, pair<int, int>>>::iterator i,
                 vector<pair<int, pair<int, int>>>::iterator j) {
  if (y - x == 1) {
    for (auto k = i; k != j; k++)
      k->first = (k->first * 10LL) % modulus;
    return;
  }
  int z = (x + y) / 2;
  vector<pair<int, pair<int, int>>>::iterator k = lower_bound(
      i, j, make_pair(int(((long long)modulus * z + 9) / 10), make_pair(0, 0)));
  mult(modulus, x, z, i, k);
  mult(modulus, z, y, k, j);
  inplace_merge(i, k, j,
                [](pair<int, pair<int, int>> a, pair<int, pair<int, int>> b) {
                  return make_pair(a.first, a.second.second) <
                         make_pair(b.first, b.second.second);
                });
}
static string go(int modulus) {
  if (modulus == 1)
    return "1";
  int sequence = 1;
  list<vector<int>> v = {{0}};
  vector<pair<int, pair<int, int>>> partnered;
  int place = 1;
  while (true) {
    v.emplace_back(v.rbegin()->size() * 2);
    vector<int> &previous = *next(v.rbegin()), &current = *v.rbegin();
    auto offset = [modulus, place, sequence](int a) {
      return (a + (long long)place) % modulus;
    };
    auto old_mid =
        lower_bound(previous.cbegin(), previous.cend(), modulus - place),
         new_mid = lower_bound(previous.cbegin(), previous.cend(), place);
    current.resize(
        set_union(new_mid, previous.cend(),
                  make_transform_iterator(previous.cbegin(), offset),
                  make_transform_iterator(old_mid, offset),
                  set_union(previous.cbegin(), new_mid,
                            make_transform_iterator(old_mid, offset),
                            make_transform_iterator(previous.cend(), offset),
                            current.begin())) -
        current.begin());
    int place2 = modulus - (long long)place * place % modulus;
    auto offset_partnered = [modulus, place, place2,
                             sequence](pair<int, pair<int, int>> a) {
      return make_pair((a.first + (long long)place2) % modulus,
                       make_pair((a.second.first + (long long)place) % modulus,
                                 sequence + a.second.second));
    };
    auto old_mid_partnered =
        lower_bound(partnered.cbegin(), partnered.cend(),
                    make_pair(modulus - place2, make_pair(0, 0))),
         new_mid_partnered = lower_bound(partnered.cbegin(), partnered.cend(),
                                         make_pair(place2, make_pair(0, 0)));
    vector<pair<int, pair<int, int>>> next_partnered(partnered.size() * 2 + 1);
    auto i =
        set_union(partnered.cbegin(), new_mid_partnered,
                  make_transform_iterator(old_mid_partnered, offset_partnered),
                  make_transform_iterator(partnered.cend(), offset_partnered),
                  next_partnered.begin(), cmp_first_partnered);
    if (new_mid_partnered == partnered.cend() ||
        new_mid_partnered->first != place2)
      *i++ = make_pair(place2, make_pair(place, sequence));
    next_partnered.resize(
        set_union(new_mid_partnered, partnered.cend(),
                  make_transform_iterator(partnered.cbegin(), offset_partnered),
                  make_transform_iterator(old_mid_partnered, offset_partnered),
                  i, cmp_first_partnered) -
        next_partnered.begin());
    partnered.swap(next_partnered);
    sequence += previous.size();
    place = (place * 10LL) % modulus;
    mult(modulus, 0, 10, partnered.begin(), partnered.end());
    partnered.resize(
        unique(partnered.begin(), partnered.end(), eq_first_partnered) -
        partnered.begin());
    auto with_first = [](int a) { return make_pair(a, make_pair(a, 0)); };
    vector<pair<int, pair<int, int>>> hits;
    set_intersection(partnered.cbegin(), partnered.cend(),
                     make_transform_iterator(current.cbegin(), with_first),
                     make_transform_iterator(current.cend(), with_first),
                     back_inserter(hits), cmp_first_partnered);
    if (hits.size()) {
      pair<int, pair<int, int>> best = *min_element(
          hits.begin(), hits.end(),
          [](pair<int, pair<int, int>> a, pair<int, pair<int, int>> b) {
            return a.second.second < b.second.second;
          });
      string ret = "";
      pair<int, int> state =
          retrace(modulus, 1, make_pair(best.second.first, 0), v.begin(),
                  prev(v.end()), ret);
      retrace(modulus, 1, make_pair(best.first, state.second), v.begin(),
              prev(v.end()), ret);
      return ret;
    }
  }
}
int main(int argc, const char *argv[]) {
  ios_base::sync_with_stdio(false);
  if (argc >= 2) {
    ofstream ofs(argv[1]);
    for (int modulus = 1; modulus <= 10000; modulus++)
      ofs << go(modulus) << '\n';
  } else {
    int modulus;
    cin >> modulus;
    cout << go(modulus) << '\n';
  }
  return 0;
}
```
# Python, 280 bytes (8.6 seconds on 1999999998 with PyPy)
```
n=input()
if n<2:print 1;exit()
d={0:0}
l=[]
k=1
b=x=y=0
while 1:
 for a in[0]+l:
  m=(a+k)%n
  if m not in d:l.append(m);d[m]=b
 k=(k*10)%n;b+=1
 for a in l:
  if(-k*a)%n in d:
   while(a-x)%n:x+=10**d[(a-x)%n]
   while(-y-k*a)%n:y+=10**d[(-y-k*a)%n]
   print(10**b*x+y)/n;exit()
```
[Answer]
# Mathematica 115 bytes
```
p=Drop[Union[FromDigits/@Flatten[Table[Tuples[{0,1},{k}],{k,2,12}],1]],2];
i=Input[];FirstCase[p,x_/;Divisible[x,i]]
```
[Answer]
# R, 45 bytes
```
x=scan();y=2;while(grepl("[2-9]",x*y))y=y+1;y
```
Usage:
```
> x=scan();y=2;while(grepl("[2-9]",x*y))y=y+1;y
1: 2
2: 
Read 1 item
[1] 5
> x=scan();y=2;while(grepl("[2-9]",x*y))y=y+1;y
1: 21
2: 
Read 1 item
[1] 481
> x=scan();y=2;while(grepl("[2-9]",x*y))y=y+1;y
1: 42
2: 
Read 1 item
[1] 2405
```
[Answer]
**Java 156 bytes**
```
public class P{public static void main(String[]a){long x=Long.valueOf(a[0]),y;for(y=2;!(""+x*y).replaceAll("1|0","").isEmpty();y++);System.out.println(y);}}
```
Massive thanks to aditsu :)
[Answer]
# Pyth - ~~12~~ 11 bytes
Uses filter with numeric arg to get first natural number that fulfills predicate, default is 1 which is what we want. Setwise diff to check if only zeros and ones.
```
f!-j*QT10U2
```
[Test Suite](http://pyth.herokuapp.com/?code=f%21-j%2aQT10U2&test_suite=1&test_suite_input=2%0A21&debug=0).
[Answer]
# Java, ~~198~~ ~~193~~ 181 bytes
Thanks to @aditsu for shaving off 5 bytes AND increasing the range of testable numbers!
Note that some values loop negatively due to how Java parses integers. This could be circumvented by BigInteger, but the bonus was simply less valuable.
I know that I'm not going to win, but I hope this inspires other, shorter, answers.
```
class A{public static void main(String[] a){for(long i=1;;i++){try{long b=Long.parseLong(a[0]);if(b*i<0)break;Long.parseLong(b*i+"",2);System.out.println(i);}catch(Exception e){}}}}
```
Ungofled:
```
class A {
   public static void main(String[] a){
      for(long i=1;;i++){ // infinite loop starting at 1
         try{ // if an error is thrown by attempting to parse as binary, restart while adding 1 to i
            long b=Long.parseLong(a[0]); // For later - it was shorter to declare than use twice
            if(b*i<0)break; // Break out of the program if we have looped.
            Long.parseLong(b*i+"",2); // Multiply out and see if it's passable as a binary number, otherwise, throw error and go back to the top of the loop
            System.out.println(b); // print it out
         } catch (Exception e) {} // do nothing on catch
      }
   }
}
```
[Answer]
# C, 107 101 bytes (105 99 bytes for 32-bits)
There is a distinct lack of answers in C on code golf. Indeed, C is not the best choice for writing the smallest possible program, but it's not that bad:
```
main(d,b){char s[9];gets(s);for(b=atoi(s);sprintf(s,"%d",b*d),strspn(s,"01")[s];d++);printf("%d",d);}
```
You can do without the #includes, but then all the function definitions will be implicit. The main drawback is that this causes the assumption that all functions return ints. This is a problem on 64-bit machines for functions that actually return a pointer. If you are on a 32-bit machine, 2 bytes can be shaved off the above solution:
```
main(d,b){char s[9];for(b=atoi(gets(s));sprintf(s,"%d",b*d),strspn(s,"01")[s];d++);printf("%d",d);}
```
Somewhat more readable version:
```
int main()
{
  char s[9];
  gets(s);
  int d = 1;
  int b = atoi(s);
  for (; sprintf(s, "%d", b * d), strspn(s, "01")[s]; d++);
  printf("%d", d);
}
```
[Answer]
# C# time near 5 seconds (1 to 10000)
As requested, here is a golfed C# program answering the original challenge.
Input as command line argument, output to console.
```
using System;using System.Collections.Generic;using System.Numerics;using System.Linq;
class P{static void Main(string[] a){int m,n=int.Parse(a[0]);var d=new Dictionary<int,long>();long b;int h;
for(d[n]=0,b=h=1;;b*=2,h=(h*10)%n)foreach(int k in d.Keys.Reverse())if(!d.ContainsKey(m=(h+k)%n)){
var w=d[k]|b;if(m==0){Console.Write(BigInteger.Parse(Convert.ToString(w,2))/n);return;}d.Add(m,w);}}}
```
Then, as for the bounty: the bounty should go to aditsu, as I think his algorithm cannot be beaten in terms of perfomance. But anatolyg self-answer is amazing too.
Here is my fast implementation in C#. I suppose that in C++ it could be faster (maybe 2x). Compiled and tested with Visual Studio 2010, .NET framework 4, 64 bits, redirecting output to nul. Time : 00:00:05.2604315
```
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Numerics;
using System.Diagnostics;
class Program
{
   static BigInteger Find(int n)
   {
      var d = new Dictionary<int, long>();
      long kb;
      int km;
      d[n] = 0;
      for (kb = km = 1; ; kb *= 2, km = (km * 10) % n)
      {
         foreach (int key in d.Keys.Reverse())
         {
            int m = (km + key) % n;
            if (!d.ContainsKey(m))
            {
               long w = d[key] | kb;
               if (m == 0)
               {
                  return BigInteger.Parse(Convert.ToString(w, 2));
               }
               d.Add(m, w);
            }
         }
      }
   }
   static void Exec(int n, out string sq, out string sa)
   {
      var v = Find(n);
      sq = (v/n).ToString();
      sa = v.ToString();
   }  
   static void Main(string[] args)
   {
      // string n = Console.ReadLine();
      int limit = int.Parse(args[0]);
      string q ="", a = "";
      Stopwatch x = new Stopwatch();
      x.Start();
      for (int n = 1; n <= limit; n++)
      {
         Exec(n, out q, out a);
         Console.WriteLine("{0} {1} {2}", n, q, a);
      }
      x.Stop();
      Console.Error.WriteLine("{0}", x.Elapsed);
   }
}
```
[Answer]
## C with GMP (621 bytes, fast)
I've tried to be fast and short, but favoured fast. This implementation uses a slightly improved version of the number-theoretic speedup I mentioned in a comment on [aditsu's answer](https://codegolf.stackexchange.com/a/61003/194).
Save as `pseudobinary.c` and compile with `gcc pseudobinary.c -lgmp -o pseudobinary`. Note that this allocates so much memory for large inputs that you will need to compile it for a 64-bit platform.
```
#include <gmp.h>
int main(int y,char*z[]){int i,n,b,c,e,f,m,*j,*k,*l,*r,*h;char *d,*s;mpz_t
B,I,Q;i=atoi(z[1]);n=i;for(b=0;n%10<1;++b)n/=10;for(;n%2<1;++b)n/=2;for(;n%5<1;++b)n/=5;if(n<2)--b;d=calloc(n,1);j=calloc(n,sizeof(int));r=calloc(99,sizeof(int));c=2;d[1]=1;*j=r[1]=e=1;l=j+1;for(s=0;!s;++c){r[c]=e=e*10%n;k=l;for(h=j;h<k;h++){f=*h;m=(e+f)%n;if(d[m]<1){*l++=m;if(m<1){s=malloc(99);memset(s,48,99);for(f=c;f;f=d[m=(m+n-r[f])%n])s[c-f]++;s[c]=0;h=k;}d[m]=c;}}}f=strlen(s);s[f]=48;s[f+b]=0;mpz_init_set_str(B,s,10);mpz_init_set_si(I,i);mpz_init(Q);mpz_divexact(Q,B,I);d=mpz_get_str(0,10,Q);printf("%s\n",d);return 0;}
```
### Loop version for timing (751 bytes)
```
#include <gmp.h>
char **v;int main(){int i,n,b,c,e,f,m,*j,*k,*l,*r,*h;char *d,*s;mpz_t
B,I,Q;v=calloc(10001,sizeof(char*));v[1]=s=malloc(99);memset(s,48,99);*s=49;s[1]=0;for(i=0;++i<10001;){n=i;for(b=0;n%10<1;++b)n/=10;for(;n%2<1;++b)n/=2;for(;n%5<1;++b)n/=5;d=calloc(n,1);j=calloc(n,sizeof(int));r=calloc(99,sizeof(int));c=2;d[1]=1;*j=r[1]=e=1;l=j+1;for(;!v[n];++c){r[c]=e=e*10%n;k=l;for(h=j;h<k;h++){f=*h;m=(e+f)%n;if(d[m]<1){*l++=m;if(m<1){v[n]=s=malloc(99);memset(s,48,99);for(f=c;f;f=d[m=(m+n-r[f])%n])s[c-f]++;s[c]=0;h=k;}d[m]=c;}}}free(d);free(j);free(r);s=v[n];f=strlen(s);s[f]=48;s[f+b]=0;mpz_init_set_str(B,s,10);mpz_init_set_si(I,i);mpz_init(Q);mpz_divexact(Q,B,I);d=mpz_get_str(0,10,Q);printf("%s\n",d);free(d);s[f+b]=48;s[f]=0;}return 0;}
```
### Ungolfed loop version
```
#include <gmp.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char **cache;
int main() {
    int i,n,shift,_kb,km,key,m,*ks,*ksi,*nksi,*res,*ii;
    char *d,*s;
    mpz_t B,I,Q;
    cache = calloc(10001,sizeof(char*));
    if (!cache) { printf("Failed to malloc cache\n"); return 1; }
    cache[1]=s = malloc(99);
    memset(s,48,99);
    *s=49;
    s[1]=0;
    for (i=0;++i<10001;) {
        n=i;
        for(shift=0;n%10<1;++shift)n/=10;
        for(;n%2<1;++shift)n/=2;
        for(;n%5<1;++shift)n/=5;
        d = calloc(n,1);
        if (!d) { printf("Failed to malloc d\n"); return 1; }
        ks = calloc(n,sizeof(int));
        if (!ks) { printf("Failed to malloc ks\n"); return 1; }
        res = calloc(99,sizeof(int));
        if (!res) { printf("Failed to malloc res\n"); return 1; }
        _kb = 2;
        d[1] = 1;
        *ks = res[1] = km = 1;
        nksi = ks + 1;
        for(;!cache[n];++_kb) {
            res[_kb] = km = km*10%n;
            ksi = nksi;
            for (ii = ks; ii < ksi; ii++) {
                key = *ii;
                m = (km + key) % n;
                if (d[m] < 1) {
                    *nksi++ = m;
                    if (m < 1) {
                        cache[n] = s = malloc(99);
                        if (!s) { printf("Failed to malloc s\n"); return 1; }
                        memset(s,48,99);
                        for(key=_kb;key;key = d[m = (m + n - res[key]) % n])s[_kb-key]++;
                        s[_kb]=0;
                        ii = ksi; // break
                    }
                    d[m] = _kb;
                }
            }
        }
        free(d);
        free(ks);
        free(res);
        // Add shift * '0'
        s=cache[n];
        key=strlen(s);
        s[key]=48;
        s[key+shift]=0;
        // convert to big integer, divide, print
        mpz_init_set_str(B,s,10);
        mpz_init_set_si(I,i);
        mpz_init(Q);
        mpz_divexact(Q,B,I);
        d = mpz_get_str(0,10,Q);
        if (!s) { printf("Failed to malloc quotient\n"); return 1; }
        printf("%s\n", d);
        free(d);
        // Remove shift * '0'
        s[key+shift]=48;
        s[key]=0;
    }
    return 0;
}
```
[Answer]
# C + GMP, 669
This is really fast for smallish numbers; it starts to choke when the result has more than 64 digits.
```
#include<gmp.h>
#define B(x)(int)((x*(long)k)%n);
int*M,*H,P[99],n,x,p,q=2,e=1,k=10,y,f,z;char*E,C[99];int b(int k,int t){int
j=E[k],a=1<<(j-2);if(j<2){C[t]=49;return 1;}x=(int)((k+n-P[j]*(long)H[k]%n)%n);if(x)b(x,t);return a+b(H[k],t-a);}int
main(){scanf("%d",&n);E=calloc(n+1,1);M=calloc(n+1,4);H=malloc(n*4);M[1]=E[1%n]=P[1]=1;while(!E[0]){P[++e]=k;p=q;for(x=0;++x<p;){y=B(M[x])if(E[n-y]){E[0]=e;H[0]=M[x];break;}}if(!E[x=0])while(++x<p){y=B(M[x])for(z=0;z<p;++z){f=y+M[z];if(f>=n)f-=n;if(!E[f]){E[f]=e;H[f]=M[x];M[q++]=f;}}}k=B(k)}memset(C,48,98);C[99]=0;x=b(0,97);mpz_t
m,r;mpz_init(r);mpz_init_set_str(m,C+98-x,10);mpz_fdiv_q_ui(r,m,n);puts(mpz_get_str(C,10,r));}
```
Version that loops to 10000 (671 bytes):
```
#include<gmp.h>
#define B(x)(int)((x*(long)k)%n);
#define N 10001
int M[N],H[N],P[99],n=0,x,p,q,e,k,y,f,z;char E[N],C[99];int b(int k,int t){int
j=E[k],a=1<<(j-2);if(j<2){C[t]=49;return 1;}x=(int)((k+n-P[j]*(long)H[k]%n)%n);if(x)b(x,t);return a+b(H[k],t-a);}int
main(){while(++n<N){memset(E,M[0]=0,n);M[1]=E[1%n]=P[1]=e=1;q=2;k=10;while(!E[0]){P[++e]=k;p=q;for(x=0;++x<p;){y=B(M[x])if(E[n-y]){E[0]=e;H[0]=M[x];break;}}if(!E[x=0])while(++x<p){y=B(M[x])for(z=0;z<p;++z){f=y+M[z];if(f>=n)f-=n;if(!E[f]){E[f]=e;H[f]=M[x];M[q++]=f;}}}k=B(k)}memset(C,48,98);C[99]=0;x=b(0,97);mpz_t
m,r;mpz_init(r);mpz_init_set_str(m,C+98-x,10);mpz_fdiv_q_ui(r,m,n);puts(mpz_get_str(C,10,r));}}
```
Here are some commands for testing my code as well as my competitors', and the results on my laptop:
```
ls -l *.c*       
-rw-r--r-- 1 aditsu aditsu  669 Oct 27 15:01 mult-aditsu-single.c
-rw-r--r-- 1 aditsu aditsu  671 Oct 27 15:01 mult-aditsu.c
-rw-r--r-- 1 aditsu aditsu 3546 Oct 27 15:01 mult-anatoly.c
-rw-r--r-- 1 aditsu aditsu 6175 Oct 27 15:01 mult-anders.cpp
-rw-r--r-- 1 aditsu aditsu  621 Oct 27 15:01 mult-peter-single.c
-rw-r--r-- 1 aditsu aditsu  751 Oct 27 15:01 mult-peter.c
gcc -w -march=native -O3 mult-aditsu-single.c -lgmp -o mult-aditsu-single
gcc -w -march=native -O3 mult-aditsu.c -lgmp -o mult-aditsu
gcc -w -march=native -O3 mult-peter-single.c -lgmp -o mult-peter-single
gcc -w -march=native -O3 mult-peter.c -lgmp -o mult-peter
gcc -w -march=native -O3 --std=c99 mult-anatoly.c -o mult-anatoly
g++ --std=c++11 -march=native -O3 mult-anders.cpp -o mult-anders
for i in {1..5}; do time ./mult-anders mult-anders.txt; done
./mult-anders mult-anders.txt  0.34s user 0.00s system 99% cpu 0.344 total
./mult-anders mult-anders.txt  0.36s user 0.00s system 99% cpu 0.358 total
./mult-anders mult-anders.txt  0.34s user 0.00s system 99% cpu 0.346 total
./mult-anders mult-anders.txt  0.35s user 0.00s system 99% cpu 0.347 total
./mult-anders mult-anders.txt  0.34s user 0.00s system 99% cpu 0.344 total
for i in {1..5}; do ./mult-anatoly mult-anatoly.txt; done
Time: 0.254416
Time: 0.253555
Time: 0.245734
Time: 0.243129
Time: 0.243345
for i in {1..5}; do time ./mult-peter > mult-peter.txt; done
./mult-peter > mult-peter.txt  0.14s user 0.00s system 99% cpu 0.137 total
./mult-peter > mult-peter.txt  0.15s user 0.00s system 97% cpu 0.153 total
./mult-peter > mult-peter.txt  0.15s user 0.00s system 99% cpu 0.149 total
./mult-peter > mult-peter.txt  0.15s user 0.00s system 99% cpu 0.150 total
./mult-peter > mult-peter.txt  0.14s user 0.00s system 99% cpu 0.138 total
for i in {1..5}; do time ./mult-aditsu > mult-aditsu.txt; done
./mult-aditsu > mult-aditsu.txt  0.06s user 0.00s system 95% cpu 0.058 total
./mult-aditsu > mult-aditsu.txt  0.05s user 0.00s system 97% cpu 0.055 total
./mult-aditsu > mult-aditsu.txt  0.06s user 0.00s system 99% cpu 0.056 total
./mult-aditsu > mult-aditsu.txt  0.05s user 0.00s system 99% cpu 0.054 total
./mult-aditsu > mult-aditsu.txt  0.05s user 0.00s system 98% cpu 0.055 total
md5sum *.txt
6eef8511d3bc5769b5d9218be2e00028  mult-aditsu.txt
6eef8511d3bc5769b5d9218be2e00028  mult-anatoly.txt
6eef8511d3bc5769b5d9218be2e00028  mult-anders.txt
6eef8511d3bc5769b5d9218be2e00028  mult-peter.txt
```
[Answer]
# T-SQL, 164 156 155 154 159 bytes
(-1 byte. Thanks Jonathan!) 
(-1 more because why do I have trailing spaces on lines? SMH) 
(+5 realized my golfing broke things) 
```
create function b(@ int)
returns int
as begin
declare @b varchar(max)='',@i int=@
while @>0SELECT @b=cast(@%2 as varchar)+@b,@/=2
return cast(@b as int)/@i
end
```
I don't know why I keep coming back to these questions where I'm supposed to convert to Binary... T-SQL doesn't know how to do that right. 
In any case, here's a [SQLFiddle](http://sqlfiddle.com/#!6/9064e/1/0). 
Un-golfed: 
```
create function binarySquare(@id int)
returns int 
as BEGIN
```
Most of this stuff is required to write a function in T-SQL, as far as I'm aware. 
```
    declare @bin nvarchar(max) = ''
```
Create a blank string that we're going to store as our binary number.
```
    declare @id2 int = @id
```
Save the input value for use at the end. It seems like there should be a way to use the original input even if we change the value, but I can't find one. 
```
    while @id>0
      BEGIN
        SET @bin = cast(@id%2 as varchar(1)) + @bin
```
So we take our original input, MOD it with 2 to find the remainder, and that's going to be our next smallest digit. For example, 5%2 = 1
```
        SET @id = @id/2
```
Then we take our number, and divide it in half. Because it's an `int` type, it rounds it down to the nearest whole number, so 5/2 = 2. 
 END
We then loop through this until the value is 0. So we end up with 
 5%2 = 1 5/2 = 2
 2%2 = 0 2/2 = 1
 1%2 = 1 1/2 = 0 
which gives us our binary string value of 101. 
```
    declare @binNum int = (SELECT cast(@bin as int))
```
We take our binary string and convert it back to an `int` again.
```
    return @binNum/@id2
```
We return our binary string `int` divided by our original value, per the origin of the question. 
```
END
```
[Answer]
# [Julia](https://julialang.org), 38 bytes (brute force approach)
```
~n=(i=0;while"$((i+=1)n)"⊈"01"end;i)
```
[Attempt This Online!](https://ato.pxeger.com/run?1=m700qzQnM3HBgqWlJWm6FjfV6vJsNTJtDazLMzJzUpVUNDQytW0NNfM0lR51dSgZGCql5qVYZ2pCVQeVFmfmpSsEFGXmlaRxRStoVCrYKtRVWCs4FICFNJRUjTIVDk9XUNU1NMgEyqlmxuQp6ShU6ChUAkmtSk1NhbT8IoUKhcw8BUMrI1OFWIjJMPcAAA)
* -7 bytes (!!) thanks to MarcMush: rewrite with `⊈` instead of `maximum`
* -1 byte thanks to MarcMush: increment `i` inside the loop evaluation
[Answer]
# [Pyth](https://pyth.readthedocs.io), 8 bytes
```
fq`;{`*Q
```
**[Try it here.](https://pyth.herokuapp.com/?code=fq%60%3B%7B%60%2aQ&input=21&debug=0)**
---
# How?
```
fq`;{`*Q   - Full program.
f          - First input where the condition is truthy over [1, ∞)
      *Q   - The product of the input and the current number
     `     - Converted to String.
    {      - Deduplicated.
 q         - Is equal to?
  `;       - The string representation of the innermost lambda variable, T, which is "10".
           - Output implicitly.
```
This pretty much relies on the fact that numbers do not have trailing zeros, and `1` always occurs before `0` in a positive integer only consisting of `0`s and `1`s.
[Answer]
# Ruby, 46 bytes
I should really eliminate the while loop with an alternative loop.
```
n,k=gets,0;$_="#{n.to_i*k+=1}"while/[^01]/;p k
```
Edit: Thanks @manatwork for shaving off 1 byte!
Edit2: Thanks @histocraft for the insane 9 bytes!
Edit: Thanks @manatwork again for shaving off 7 bytes!
[Answer]
# Scala, 114 Bytes
```
val i=scala.io.StdIn.readInt;Stream.from(1).foreach{x=>if((i*x+"").map{_.asDigit}.max<2){print(x);System.exit(0)}}
```
Readable version
```
val i=scala.io.StdIn.readInt
Stream.from(1).foreach{x => 
    if((i*x+"").map{_.asDigit}.max<2) {
        print(x)
        System.exit(0)
    }
}
```
[Answer]
# gawk4 brute force, 28+2 = 30 bytes
```
{while(++n*$0~/[2-9]/);}$0=n
```
Needs to be called with the `-M` option for using big numbers. Of course this is ridiculously slow, using big numbers slows it down even more, but theoretically the input is not limited, and RAM usage is negligible.
### Usage example ( if you got time to waste `;)`)
```
echo 27 | awk -M '{while(++n*$0~/[2-9]/);}$0=n'
```
# gawk4 optimized, 69+2 = 71 bytes
```
{for(;!a[0];NR*=10)for(i in a)a[j=(s=a[i]+NR)%$0]?0:a[j]=s}$0=a[0]/$0
```
Well, this ended up being a clone of aditsu's answer. After looking at [this question](https://math.stackexchange.com/questions/388165/how-to-find-the-smallest-number-with-just-0-and-1-which-is-divided-by-a-give) I was still figuring out how to code the subset-sum part, when I couldn't resist looking at the other answers here.
In awk array elements have the (strange ?) behaviour that if you compare a non-existing element to something it is somehow initialized as empty before being compared (I'll admit that I'm not quite sure about what is happening there). So after checking `!a[0]` the `for(i in a)` loop starts even without initializing `a[$0]` to `0` as aditsu did.
Of course the `-M` option has to be used for this too.
Though it is rather fast it is still remarkably slower than Python. For `79992` this takes around 14 seconds on my 2GHz Core2Duo. And I wouldn't say it works for inputs up to 2^31, because in the worst case it has to build an array of big numbers (gawk4 uses GMP for this), which has the size of the input number. As a 'bonus' large arrays are very very slow in awk...
]