34
\$\begingroup\$

A binary multiple of a positive integer k is a positive integer n such that n is written only with 0s and 1s in base 10 and n is a multiple of k. For example, 111111 is a binary multiple of 3.

It is easy to show that a positive integer has infinitely many binary multiples. See here for a construction proof of one binary multiple for each k. Multiplying by powers of 10 you get infinitely many more.

Your task

Given a positive integer k, return the smallest binary multiple of k.

Input

A positive integer k.

Output

A positive integer n, the smallest binary multiple of k.

Test cases

2 -> 10
3 -> 111
4 -> 100
5 -> 10
6 -> 1110
7 -> 1001
8 -> 1000
9 -> 111111111
10 -> 10
11 -> 11
12 -> 11100
13 -> 1001
14 -> 10010
15 -> 1110
16 -> 10000
17 -> 11101
18 -> 1111111110
19 -> 11001
20 -> 100
100 -> 100

This is so shortest submission in bytes, wins! If you liked this challenge, consider upvoting it... And happy golfing!


This is the first challenge of the RGS Golfing Showdown. If you want to participate in the competition, you have 96 hours to submit your eligible answers. Remember there is 450 reputation in prizes! (See 6 of the rules)

Otherwise, this is still a regular challenge, so enjoy!

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  • 1
    \$\begingroup\$ Related: One-zero dividend. I think this isn't a duplicate though because in that challenge all the 1's had to come before all the 0's and you didn't have to output the smallest instance. \$\endgroup\$ – xnor Feb 24 at 14:54
  • \$\begingroup\$ @xnor thanks! Looking forward for your submission \$\endgroup\$ – RGS Feb 24 at 14:57
  • \$\begingroup\$ This is OEIS A004290. \$\endgroup\$ – Giuseppe Feb 25 at 16:39
  • 1
    \$\begingroup\$ So which answer surprised you the most? \$\endgroup\$ – S.S. Anne Feb 27 at 14:08
  • 1
    \$\begingroup\$ @a'_' ahaha incredible. Are you going to post an answer in Roj? \$\endgroup\$ – RGS Mar 1 at 11:33

45 Answers 45

12
\$\begingroup\$

05AB1E, 6 bytes

∞b.ΔIÖ

Try it online! or verify all test cases (courtesy of @KevinCruijssen)


Explanation

∞b           - Infinite binary list
  .Δ         - Find the first value such that..
    IÖ       - It's divisible by the input
|improve this answer|||||
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  • \$\begingroup\$ Short answer, good job! Do you think you can add the TIO link with all the test cases? \$\endgroup\$ – RGS Feb 24 at 13:49
  • \$\begingroup\$ @RGS Here ya go. ;) \$\endgroup\$ – Kevin Cruijssen Feb 24 at 13:51
  • \$\begingroup\$ Thanks @KevinCruijssen was going to have to go work out how to do that ;) \$\endgroup\$ – Expired Data Feb 24 at 13:51
  • \$\begingroup\$ @ExpiredData Ah, shall I delete it so it'll be a learning curve for you? ;) \$\endgroup\$ – Kevin Cruijssen Feb 24 at 13:52
  • 2
    \$\begingroup\$ @KevinCruijssen nah I would have just gone and stolen it from another one of your answers \$\endgroup\$ – Expired Data Feb 24 at 13:53
11
\$\begingroup\$

Python 2, 42 bytes

f=lambda k,n=0:n*(max(`n`)<'2')or f(k,n+k)

Try it online!


Full program, same length:

a=b=input()
while'1'<max(`b`):b+=a
print b

Try it online!

|improve this answer|||||
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  • \$\begingroup\$ Nice Python submission! If the full program is the same length and doesn't exceed the Recursion Depth, why don't you use it as the "main" submission? :) \$\endgroup\$ – RGS Feb 24 at 13:54
9
\$\begingroup\$

MATL, 10 bytes

`@YBUG\}HM

Try it online! Or verify all test cases.

Explanation

`      % Do...while
  @    %   Push iteration index (1-based)
  YB   %   Convert to binary string (1 gvies '1', 2 gives '10,  etc).
  U    %   Convert string to number ('10' gives 10). This is the current
       %   solution candidate
  G    %   Push input
  \    %   Modulo. Gives 0 if the current candidate is a multiple of the
       %   input, which will cause the loop to exit
}      % Finally: execute on loop exit
  H    %   Push 2
  M    %   Push input to the second-last normal function (`U`); that is,
       %   the candidate that caused the loop to exit, in string form
       % End (implicit). If top of the stack is 0: the loop exits.
       % Otherwise: a new iteration is run
       % Display (implicit)
|improve this answer|||||
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  • 12
    \$\begingroup\$ "Ah, why is there a bug in my program? Hm..." \$\endgroup\$ – Luis Mendo Feb 24 at 9:48
  • \$\begingroup\$ I removed the bug but I don't think I fixed it :( +1! \$\endgroup\$ – RGS Feb 24 at 13:57
8
\$\begingroup\$

JavaScript (ES6), 37 bytes

Looks for the smallest \$n\$ such that the decimal representation of \$p=n\times k\$ is made exclusively of \$0\$'s and \$1\$'s.

f=(k,p=k)=>/[2-9]/.test(p)?f(k,p+k):p

Try it online! (some test cases removed because of recursion overflow)


JavaScript (ES6), 41 bytes

Looks for the smallest \$n\$ such that \$k\$ divides the binary representation of \$n\$ parsed in base \$10\$.

k=>(g=n=>(s=n.toString(2))%k?g(n+1):s)(1)

Try it online! (all test cases)

|improve this answer|||||
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8
\$\begingroup\$

PHP, 38 bytes

while(($n=decbin(++$x))%$argn);echo$n;

Try it online!

Counts up n in binary and divides it's decimal representation by k until there is no remainder; indicating the first, smallest multiple.

|improve this answer|||||
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  • 1
    \$\begingroup\$ Thanks for your PHP submission and thorough test case coverage! +1 \$\endgroup\$ – RGS Feb 24 at 14:39
8
\$\begingroup\$

R, 50 38 bytes

-4 bytes thanks to Giuseppe.

grep("^[01]+$",(k=scan())*1:10^k)[1]*k

Try it online!

It follows from this blog post (linked in the question) that the smallest binary multiple of \$k\$ is smaller than \$2\cdot10^{k-1}\$; this answer uses the larger bound \$k\cdot10^k\$ instead.

Creates a vector of all multiples of \$k\$ between \$k\$ and \$k\cdot10^k\$. The regexp gives the indices of those made only of 0s and 1s; select the first index and multiply by \$k\$ to get the answer.

Will time out on TIO for input greater than 8, but with infinite memory it would work for any input.

|improve this answer|||||
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  • \$\begingroup\$ Use grepl! 40 bytes. I came here to comment a grepl approach based on my SNOBOL answer and found your regex to be better :-) I had !grepl("^[10]+$",F) for the record. \$\endgroup\$ – Giuseppe Feb 24 at 18:38
  • \$\begingroup\$ @Giuseppe Thanks! I knew about grepl but somehow convinced myself that it led to a longer answer... Your version is cleaner! \$\endgroup\$ – Robin Ryder Feb 24 at 22:07
  • 1
    \$\begingroup\$ @Giuseppe Changed again, this time using your original regexp in another grep solution. \$\endgroup\$ – Robin Ryder Feb 24 at 23:45
  • 1
    \$\begingroup\$ Quite clever! Glad I could be of assistance with that! :-) \$\endgroup\$ – Giuseppe Feb 25 at 1:27
7
\$\begingroup\$

Charcoal, 19 bytes

≔1ηW﹪IηIθ≔⍘⊕⍘粦²ηη

Try it online! Link is to verbose version of code. Explanation:

≔1η

Start at 1.

W﹪IηIθ

Repeat until a multiple of n is found, treating the values as base 10.

≔⍘⊕⍘粦²η

Convert from base 2, increment, then convert back to base 2.

η

Output the result.

|improve this answer|||||
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7
\$\begingroup\$

Haskell, 44 38 36 bytes

Saved two bytes by using filter as suggested by @ovs.

f k=filter(all(<'2').show)[0,k..]!!1

Try it online!

This checks all multiples of k and is slow for input 9 and 18.

I much prefer this version which defines the list of all "binary" numbers and searches for the first multiple of k among them. It quickly handles all test cases, but needs 52 bytes:

b=1:[10*x+d|x<-b,d<-[0,1]]
f k=[m|m<-b,mod m k<1]!!0

Try it online!

|improve this answer|||||
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  • \$\begingroup\$ 2 bytes shorter with filter. \$\endgroup\$ – ovs Feb 24 at 10:44
  • \$\begingroup\$ @ovs Thanks, that still applies to my newer version. Somehow I always feel that list comprehension must be shorter... \$\endgroup\$ – Christian Sievers Feb 24 at 10:55
7
\$\begingroup\$

T-SQL, 57 bytes

This script is really slow when using 18 as input

DECLARE @z INT = 18

DECLARE @ int=1WHILE
@z*@ like'%[^10]%'SET @+=1PRINT @z*@

T-SQL, 124 bytes

T-SQL doesn't have binary conversion

This will execute fast:

DECLARE @ int=1,@x char(64)=0,@a int=2WHILE 
@x%@z>0or @x=0SELECT
@x=left(concat(@%2,@x),@),@a-=1/~@,@=@/2-1/~@*-~@a
PRINT @x

Try it online

|improve this answer|||||
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  • \$\begingroup\$ I just can't seem to accept that SQL solves these challenges +1... What is the T- in T-SQL? \$\endgroup\$ – RGS Feb 24 at 14:00
  • \$\begingroup\$ @RGS sql really is horrible for golfing, but it is the only language I know well enough to golf at this level, sql is also it is quite readable(maybe not in this case, but that is not the sql part). T stands for Transact. \$\endgroup\$ – t-clausen.dk Feb 24 at 14:17
  • \$\begingroup\$ maybe it wasn't clear, but I am impressed by your submissions! I wasn't trying to put you down or anything! keep up the really interesting work :D \$\endgroup\$ – RGS Feb 24 at 14:31
  • \$\begingroup\$ @RGS not to worry, I took it as a compliment. It's cool that you post so many interesting questions. \$\endgroup\$ – t-clausen.dk Feb 24 at 14:42
  • \$\begingroup\$ Thanks for your feedback! I try to, at least :) See you next challenge! \$\endgroup\$ – RGS Feb 24 at 14:44
6
\$\begingroup\$

Python 3.8 (pre-release),, 75\$\cdots\$ 50 52 bytes

Saved 2 bytes thanks to Mukundan!!!
Added 2 bytes to fix error kindly pointed out by Giuseppe.

f=lambda k,n=1:(i:=int(f"{n:b}"))%k and f(k,n+1)or i

Try it online!

|improve this answer|||||
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  • \$\begingroup\$ you can save 2 bytes by doing f'{n:b}' instead of bin(n)[2:] Try it online! \$\endgroup\$ – Mukundan Feb 24 at 11:50
  • \$\begingroup\$ @Mukundan Clever use of f-strings - thanks! :-) \$\endgroup\$ – Noodle9 Feb 24 at 11:55
5
\$\begingroup\$

Bash + Unix utilities, 52 bytes

for((n=1;n%$1;));do n=`dc<<<2dio1d$n+p`;done
echo $n

Try it online!

This counts in binary, views the resulting numbers in base 10, and stops when a multiple of the input is reached.

|improve this answer|||||
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5
\$\begingroup\$

Retina 0.8.2, 68 bytes

.+
$*1:1,1;
{`^(1+):\1+,(.+);
$2
T`d`10`.1*;
,0
,10
1+,(.+)
$1$*1,$1

Try it online! Somewhat slow so no test suite. Explanation:

.+
$*1:1,1;

Initialise the work area with n in unary, k in unary, and k in decimal.

{`^(1+):\1+,(.+);
$2

If n divides k then delete everything except the result. This causes the remaining matches to fail and eventually the loop exits because it fails to achieve anything further.

T`d`10`.1*;
,0
,10

Treat k as a binary number and increment it.

1+,(.+)
$1$*1,$1

Regenerate the unary conversion of k.

|improve this answer|||||
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5
\$\begingroup\$

Bash + Core utilities, 40 bytes

seq $1 $1 $[10**$1]|grep ^[01]*$|head -1

Try it online!

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Nice bash submission +1! Also harnessing the power of regexs, aren't you? \$\endgroup\$ – RGS Feb 24 at 14:44
  • \$\begingroup\$ This is really nice. Is it mathematically guaranteed that the solution will be less than or equal to 10^input? \$\endgroup\$ – Jonah Feb 28 at 14:03
  • 1
    \$\begingroup\$ @Jonah Yes, in the question, RGS posted a link to a proof that there is always some value that works, and the proof actually yields that upper bound. Here’s the link: mathspp.com/blog/binary-multiples \$\endgroup\$ – Mitchell Spector Feb 28 at 14:58
  • 1
    \$\begingroup\$ @Jonah In fact, the upper bound proven is 10^input - 1 but it would take more bytes in the bash code to stop at that point. Of course, there's no harm done by producing the one extra value. \$\endgroup\$ – Mitchell Spector Feb 28 at 15:39
  • \$\begingroup\$ @Jonah Sorry, the upper bound proven is the number consisting of n 1s, which is even less than I had said. \$\endgroup\$ – Mitchell Spector Feb 28 at 19:49
5
\$\begingroup\$

Japt, 11 9 bytes

_¤%U}f1 ¤

Try it

|improve this answer|||||
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  • \$\begingroup\$ Really cool solution, and it is crazy fast +1! Is it fast because of the algorithm used or because of the Japt machinery? \$\endgroup\$ – RGS Feb 24 at 14:01
  • \$\begingroup\$ I think this fails for 11. \$\endgroup\$ – my pronoun is monicareinstate Feb 25 at 9:10
  • \$\begingroup\$ It's probably because 11 is crossed out in the program length part. \$\endgroup\$ – petStorm Feb 28 at 1:41
5
\$\begingroup\$

W, 7 6 bytes

-1 because I realized that W has operator overloading over t.

•B⌡≡kü

Uncompressed:

*Tt!iX*

repl.it is quite slow and you need to type in the program in code.w.

Explanation

        % For every number in the range
    i   % from 1 to infinity:
     X  % Find the first number that satisfies
*       %     Multiply the current item by the input
 T      %     The constant for 10
  t     %     Remove all digits of 1 and 0 in the current item
        %     Both operands are converted to a string, just like in 05AB1E.
   !    %     Negate - checks whether it contains only 1 and 0.

      * % Multiply that result with the input (because it's the counter value).
```
|improve this answer|||||
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  • 1
    \$\begingroup\$ Thanks for your W submission! I never understood why you include some code, and then right after it you say "uncompressed" and then include some other code. What does it mean? \$\endgroup\$ – RGS Feb 24 at 14:43
  • 2
    \$\begingroup\$ The first code snippet is a compressed form of the W program, compressed in order to use up all 256 codepoints of CP437. However, there's also another non-compressed form that is human-writable (I compress my source code from on this form). The next code snippet is a human-readable non-compressed form of the W program (hence the "uncompressed"), which is part of the explanation. \$\endgroup\$ – petStorm Feb 24 at 14:46
  • \$\begingroup\$ Ok, I think I understand better now. \$\endgroup\$ – RGS Feb 24 at 14:48
  • \$\begingroup\$ Where is the "get smallest" part? \$\endgroup\$ – Jonathan Allan Feb 24 at 19:43
  • 1
    \$\begingroup\$ @JonathanAllan Whoops, I forgot to add that W's behavior is different with an infinity value. Nice catch. There isn't an online interpreter with a permalink for W yet, I'm waiting for Dennis to recover. \$\endgroup\$ – petStorm Feb 27 at 2:11
4
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Java 10, 62 59 58 bytes

n->{var r=n;for(;!(r+"").matches("[01]+");)r+=n;return r;}

Try it online.

Explanation:

n->{             // Method with long as both parameter and return-type
  var r=n;       //  Result-long, starting at the input
  for(;!(r+"").matches("[01]+");)
                 //  Loop as long as `r` does NOT consists of only 0s and 1s
    r+=n;        //   Increase `r` by the input
  return r;}     //  After the loop is done, return `r` as result

This method above only works for binary outputs \$\leq1111111111111111111\$. For arbitrary large outputs - given enough time and resources - the following can be used instead (99 70 64 bytes):

n->{var r=n;for(;!(r+"").matches("[01]+");)r=r.add(n);return r;}

Try it online.

|improve this answer|||||
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  • \$\begingroup\$ Isn't the reason the first code fails numeric type limitations, and doesn't everybody pretend they do not exist? \$\endgroup\$ – my pronoun is monicareinstate Feb 24 at 13:03
  • \$\begingroup\$ @mypronounismonicareinstate You're correct. I just figured I'd add a lambda function using java.math.BigInteger instead of int/long to prove it is possible to support an arbitrary large output if you'd want to. \$\endgroup\$ – Kevin Cruijssen Feb 24 at 13:05
4
\$\begingroup\$

Perl 5, 21 +2 (-ap) = 23 bytes

$_+=$F[0]while/[^01]/

Run with -a and -p, input to stdin. Just keeps repeatedly adding the input for as long as the result contains anything other than digits 0 and 1.

Try it online!

|improve this answer|||||
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4
\$\begingroup\$

Ruby, 42 40 36 bytes

->k{z=k;z+=k until z.digits.max<2;z}

Try it online!

Very slow for 18, but ultimately gets the job done.

4 bytes golfed down by G B.

|improve this answer|||||
\$\endgroup\$
4
\$\begingroup\$

C (gcc), 80 bytes

r,m,n;b(h){for(r=0,m=1;h;h/=2)r+=h%2*m,m*=10;h=r;}f(k){for(n=1;b(++n)%k;);b(n);}

This can probably be improved some but I have yet to find a way.

Converts the integer into binary and checks if it's a multiple.

Brute-force.

Try it online!

|improve this answer|||||
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  • 1
    \$\begingroup\$ Very clever how f() only needs to call b() (which assigns return value to parameter) rather than assign return value to parameter! \$\endgroup\$ – Noodle9 Feb 24 at 19:52
  • \$\begingroup\$ @Noodle9 I wish I could omit the double-call but div puts the result of modulo in eax (which means eax will always be 0 after the loop). \$\endgroup\$ – S.S. Anne Feb 24 at 19:54
  • \$\begingroup\$ Do you mean the modulo result is in edx? \$\endgroup\$ – 640KB Feb 27 at 3:09
  • \$\begingroup\$ @640KB Yes. I didn't look up the docs beforehand. \$\endgroup\$ – S.S. Anne Feb 27 at 14:03
4
\$\begingroup\$

Jelly,  8  7 bytes

‘¡DṀḊƊ¿

Try it online!

How?

Note that in Jelly empty lists are falsey, while other lists are truthy. Also dequeue, , is a monadic atom which removes the first item from a list, but when presented with only an integer Jelly will first convert that integer to a list by forming the range [1..n] thus yields [2..n].

‘¡DṀḊƊ¿ - Link: integer, k
      ¿ - while...
     Ɗ  - ...condition: last three links as a monad:
  D     -     decimal digits   e.g. 1410 -> [1,4,1,0]  or 1010 -> [1,0,1,0]
   Ṁ    -     maximum                       4                     1
    Ḋ   -     dequeue (implicit range of)   [2,3,4]               []
        -                                   (truthy)              (falsey)
 ¡      - ...do: repeat (k times):
‘       -     increment

For some reason, when the body of a while loop, ¿, is a dyad each iteration sets the left argument to the result and then sets the right argument to the value of the left argument, so the 6 byte +DṀḊƊ¿ does not work. (For example given 3 that would: test 3; perform 3+3; test 6; perform 6+3; test 9; perform 9+6; test 15; perform 15+9; etc...)


Previous 8:

DḂƑȧọð1#

(DḂƑ could be DỊẠ too.)

An alternative 8:

DṀ+%Ịð1#
|improve this answer|||||
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4
\$\begingroup\$

Python 3, 50 48 bytes

f=lambda k,n=0:n*({*str(n)}<={*"01"})or f(k,n+k)

Try it online!

Explanation

  • n represents the current multiple of k.
  • {*str(n)}<={*"01"} checks if n only contains digits 0 or 1. This is done by creating a set of characters of n, then checks if that set is a subset of \$\{0,1\}\$.
  • n*({*str(n)}<={*"01"}) or f(k,n+k) is arranged such that the recursive call f(k,n+k) is only evaluated when n is 0 or n is not a binary multiple of k. Here multiplication acts as a logical and.
|improve this answer|||||
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4
\$\begingroup\$

J, 24 23 22 bytes

+^:(0<10#@-.~&":])^:_~

Try it online!

Add the number to itself + while ^:...^:_ the following is true:

(0<10#@-.~&":]) - Something other than the digits 0 and 1 appear in the stringified number.

|improve this answer|||||
\$\endgroup\$
  • 1
    \$\begingroup\$ I like your submission. Can't even tell why \$\endgroup\$ – RGS Feb 28 at 13:47
3
\$\begingroup\$

Burlesque, 13 bytes

rimo{>]2.<}fe

Try it online!

9 & 18 do work, but take a while because they're such large multiples. So I've taken them out of tests.

ri    # Read to int
mo    # Generate infinite list of multiples
{
 >]   # Largest digit
 2.<  # Less than 2
}fe   # Find the first element s.t.
|improve this answer|||||
\$\endgroup\$
3
\$\begingroup\$

Wolfram Language (Mathematica), 49 bytes

(x=#;While[Or@@(#>1&)/@IntegerDigits@x,x=x+#];x)&

Try it online!

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ This fails on 11 (and, likely, 9) - the code should output the smallest binary multiple, and this one outputs the first one that has both zeroes and ones. 0-byte fix \$\endgroup\$ – my pronoun is monicareinstate Feb 24 at 13:58
  • \$\begingroup\$ This does not work for 3! Ping me as soon as you fix it! \$\endgroup\$ – RGS Feb 24 at 14:14
  • \$\begingroup\$ @RGS Ping...... \$\endgroup\$ – S.S. Anne Feb 24 at 18:10
  • \$\begingroup\$ @S.S.Anne thanks! My pinger is broken \$\endgroup\$ – J42161217 Feb 24 at 19:58
3
\$\begingroup\$

MathGolf, 9 bytes

0ô+_▒╙2<▼

Try it online. (Test cases n=9 and n=18 are excluded, since they time out.)

Explanation:

0          # Start with 0
        ▼  # Do-while false with pop,
 ô         # using the following 6 commands:
  +        #  Add the (implicit) input-integer to the current value
   _       #  Duplicate it
    ▒      #  Convert it to a list of digits
     ╙     #  Pop and push the maximum digit of this list
      2<   #  And check if this max digit is smaller than 2 (thus 0 or 1)
           # (after the do-while, the entire stack joined together is output implicitly)
|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Question: if you answered 10 seconds ago, how come you already golfed it down from 10 to 9 bytes..? +1 for the mathgolf submission though \$\endgroup\$ – RGS Feb 24 at 14:45
  • 1
    \$\begingroup\$ @RGS Yeah, could have just gone with the 9-byter, haha. I golfed it while adding the explanation (thus before posting) and realizing it could be shorter. EDIT: I've removed the strike-through 10. \$\endgroup\$ – Kevin Cruijssen Feb 24 at 14:46
  • \$\begingroup\$ If I golf my answer within the grace period then I never strike through because there's no way to recover the previous version. \$\endgroup\$ – Neil Feb 26 at 10:25
3
\$\begingroup\$

Stax, 9 bytes

ü◘ø⌠Δ>0↔å

Port of my MathGolf answer. This is only my second Stax answer, so there might be a shorter alternative.

Try it online or try it online unpacked (10 bytes).

Explanation (of the unpacked version):

0             # Start at 0
 w            # While true without popping, by using everything else as block:
  x+          #  Add the input-integer
    c         #  Duplicate the top of the stack
     E        #  Convert it to a list of digits
      |M      #  Get the maximum of this list
        1>    #  And check that it's larger than 1
              # (after the while, the top of the stack is output implicitly as result)
|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Cool Stax submission +1! Was stax created by someone from CG.SE? \$\endgroup\$ – RGS Feb 24 at 15:19
  • \$\begingroup\$ @RGS yep, by recursive. :) \$\endgroup\$ – Kevin Cruijssen Feb 24 at 16:44
  • \$\begingroup\$ Nice! Didn't know about that meta question. \$\endgroup\$ – RGS Feb 24 at 17:20
  • \$\begingroup\$ I'm not a really good Stax golfer. Run and debug it \$\endgroup\$ – petStorm Feb 25 at 2:14
  • \$\begingroup\$ @a'_' That unfortunately won't work for inputs that are already correct, like 1, 10, 11,100, etc. \$\endgroup\$ – Kevin Cruijssen Feb 25 at 8:23
3
\$\begingroup\$

Red, 52 bytes

func[n][i: 0 until[""= trim/with to""i: i + n"01"]i]

Try it online!

Red, 54 bytes

func[n][i: 0 until[parse to""i: i + n[any["0"|"1"]]]i]

Try it online!

|improve this answer|||||
\$\endgroup\$
3
\$\begingroup\$

C# (Visual C# Interactive Compiler), 53 49 bytes

turns out Max can cast chars to codes implicitly

I tried a recursive version, but it was longer. Can't think of a way to replace the loop with LINQ...

a=>{int r=a;while($"{r}".Max()>49)r+=a;return r;}

Try it online!

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ I don't think LINQ will be better \$\endgroup\$ – Expired Data Feb 24 at 14:10
  • 1
    \$\begingroup\$ @ExpiredData Well, that "can't think of a way" part was implicitly followed by "without doubling the byte count" :) \$\endgroup\$ – my pronoun is monicareinstate Feb 24 at 14:13
  • \$\begingroup\$ It's not that far off 16 bytes more.. maybe there's a way to golf it down. \$\endgroup\$ – Expired Data Feb 24 at 14:21
  • \$\begingroup\$ Thanks for your C# submission! What is the LINQ thing? \$\endgroup\$ – RGS Feb 24 at 14:37
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    \$\begingroup\$ @RGS: Language INtegrated Query (LINQ) is an series of APIs in .Net that allow you to handle collections with a query language. It has two syntaxes: 1. SQL-like syntax 2. Fluent syntax on the IEnumerable<T> interface. See here for the official documentation: docs.microsoft.com/en-us/dotnet/standard/using-linq \$\endgroup\$ – Malivil Feb 25 at 16:43
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GolfScript, 31 bytes

:i;1{).{2 base}:b~{`+}*~i%}do b

Worked for 2-15 wen I tried them out, didn't bother doing more. May post explanation later, just wanted to get a brute submission down.

Try it online!

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Gaia, 10 9 bytes

⟨:$2…⁻⟩+↺

Try it online!

		| implicit input, n
⟨     ⟩		| (1) monadic link:
 :$		| dup, and get decimal digits
   2…⁻		| remove all 1s and zeros
	↺	| if the result is truthy (non-empty)
       +	| add n and repeat from (1)
		| implicitly print result.

Times out on n=9...

Gaia, 10 bytes

1⟨bdĖ⟩#ebd

Try it online!

Somewhat more interesting and also a lot faster; finds the first integer where: converting it to binary and interpreting as decimal digits are divisible by the input (and ebd converts it to the decimal form).

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