Divisibility test

Question

Task

Given two strictly positive integers n and d as input, determine whether n is evenly divisible by d, i.e., if there exists an integer q such that n = qd.

You may write a program or a function and use any of the our standard methods of receiving input and providing output.

The output should be a truthy or a falsy value; truthy if n is divisible by d, and falsy otherwise.

Your code only has to handle integers it can represent natively, as long as it works for all signed 8-bit integers. However, your algorithm has to work for arbitrarily large integers.

You may use any programming language, but note that these loopholes are forbidden by default.

This is code-golf, so the shortest valid answer – measured in bytes – wins.

Test cases

 n,  d    output

 1,  1    truthy
 2,  1    truthy
 6,  3    truthy
17, 17    truthy
22,  2    truthy
 1,  2    falsy
 2,  3    falsy
 2,  4    falsy
 3,  9    falsy
15, 16    falsy

Leaderboard

The Stack Snippet at the bottom of this post generates the catalog from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

## Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

## Ruby, <s>104</s> <s>101</s> 96 bytes

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

## Perl, 43 + 3 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the snippet:

## [><>](http://esolangs.org/wiki/Fish), 121 bytes

<style>body { text-align: left !important} #answer-list { padding: 10px; width: 290px; float: left; } #language-list { padding: 10px; width: 290px; float: left; } table thead { font-weight: bold; } table td { padding: 5px; }</style><script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="language-list"> <h2>Shortest Solution by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr> </thead> <tbody id="languages"> </tbody> </table> </div> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr> </thead> <tbody id="answers"> </tbody> </table> </div> <table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table><script>var QUESTION_ID = 86149; var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe"; var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk"; var OVERRIDE_USER = 48934; var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, comment_page; function answersUrl(index) { return "https://api.stackexchange.com/2.2/questions/" + QUESTION_ID + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER; } function commentUrl(index, answers) { return "https://api.stackexchange.com/2.2/answers/" + answers.join(';') + "/comments?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + COMMENT_FILTER; } function getAnswers() { jQuery.ajax({ url: answersUrl(answer_page++), method: "get", dataType: "jsonp", crossDomain: true, success: function (data) { answers.push.apply(answers, data.items); answers_hash = []; answer_ids = []; data.items.forEach(function(a) { a.comments = []; var id = +a.share_link.match(/\d+/); answer_ids.push(id); answers_hash[id] = a; }); if (!data.has_more) more_answers = false; comment_page = 1; getComments(); } }); } function getComments() { jQuery.ajax({ url: commentUrl(comment_page++, answer_ids), method: "get", dataType: "jsonp", crossDomain: true, success: function (data) { data.items.forEach(function(c) { if (c.owner.user_id === OVERRIDE_USER) answers_hash[c.post_id].comments.push(c); }); if (data.has_more) getComments(); else if (more_answers) getAnswers(); else process(); } }); } getAnswers(); var SCORE_REG = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/; var OVERRIDE_REG = /^Override\s*header:\s*/i; function getAuthorName(a) { return a.owner.display_name; } function process() { var valid = []; answers.forEach(function(a) { var body = a.body; a.comments.forEach(function(c) { if(OVERRIDE_REG.test(c.body)) body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>'; }); var match = body.match(SCORE_REG); if (match) valid.push({ user: getAuthorName(a), size: +match[2], language: match[1], link: a.share_link, }); else console.log(body); }); valid.sort(function (a, b) { var aB = a.size, bB = b.size; return aB - bB }); var languages = {}; var place = 1; var lastSize = null; var lastPlace = 1; valid.forEach(function (a) { if (a.size != lastSize) lastPlace = place; lastSize = a.size; ++place; var answer = jQuery("#answer-template").html(); answer = answer.replace("{{PLACE}}", lastPlace + ".") .replace("{{NAME}}", a.user) .replace("{{LANGUAGE}}", a.language) .replace("{{SIZE}}", a.size) .replace("{{LINK}}", a.link); answer = jQuery(answer); jQuery("#answers").append(answer); var lang = a.language; lang = jQuery('<a>'+lang+'</a>').text(); languages[lang] = languages[lang] || {lang: a.language, lang_raw: lang.toLowerCase(), user: a.user, size: a.size, link: a.link}; }); var langs = []; for (var lang in languages) if (languages.hasOwnProperty(lang)) langs.push(languages[lang]); langs.sort(function (a, b) { if (a.lang_raw > b.lang_raw) return 1; if (a.lang_raw < b.lang_raw) return -1; return 0; }); for (var i = 0; i < langs.length; ++i) { var language = jQuery("#language-template").html(); var lang = langs[i]; language = language.replace("{{LANGUAGE}}", lang.lang) .replace("{{NAME}}", lang.user) .replace("{{SIZE}}", lang.size) .replace("{{LINK}}", lang.link); language = jQuery(language); jQuery("#languages").append(language); } }</script>

Answer 1

Jelly, 1 byte

ḍ

This took me hours to golf.

Try it online!

Answer 2

Brain-Flak, 72 70 64 62 58 46 bytes

{({}[()]{(<()>)}{}<({}[()]<({}())>)>)}{}{{}}{}

Takes dividend and divisor (in that order) as input and prints the divisor (truthy) or nothing. Since each stack has an implicit, infinite amount of zeroes, empty output should be considered falsy.

While not stack-clean, this solution uses only a single stack.

Try it online!

Thanks to @WheatWizard for golfing off 2 bytes!

How it works

                INPUT: a (dividend), b (divisor)
                INITIAL STACK: n = a, d = b, r = 0
                               An infinite amount of zeroes follows.

{               While n is non-zero:
  (
    {}              Pop n from the stack.
    [()]            Yield -1.
    {               While the top of the stack (initially, d) is non-zero:
      (<()>)          Push 0.
    }
    {}              Pop 0. This will remove d from the stack if d = 0, leaving r
                    on top. We can think of this as performing the assignment
                    (d, r) = (r, d) if d = 0.
    <
      (
        {}              Pop d.
        [()]            Yield -1.
        <
          (
            {}              Pop r.
            ()              Yield 1.
          )               Push r + 1.
        >               Yield 0.
      )               Push d + (-1) + 0 = d - 1.
    >               Yield 0.
  )               Push n + (-1) + 0 + 0 + 0 = n - 1.
}               Each iteration decrements n, swaps d and r if d = 0, decrements d,
                and increments r.
                FINAL VALUES: n = 0
                              d = b - r
                              r = a % b if a % b > 0 else b
{}              Pop n.
{               While the top of the stack is non-zero:
  {}              Pop it.
}               This pops d and r if d > 0 (and, thus, a % b > 0) or noting at all.
{}              Pop d or a 0, leaving r if r = b and, thus, a % b = 0.

Modulus calculation, 42 bytes

The above full program can be modified in a trivial manner to calculate the modulus instead.

{({}[()]<({}[()]<({}())>)>{(<()>)}{})}{}{}

As before, this method is not stack-clean, but it uses only a single stack. A modulus of 0 will leave the stack empty, which is roughly equivalent to leaving 0; each stack contains infinite zeroes.

Try it online!

How it works

Compare the two loops of the divisibility tester and the modulus calculator.

{({}[()]{(<()>)}{}<({}[()]<({}())>)>)}
{({}[()]<({}[()]<({}())>)>{(<()>)}{})}

The only difference is the location of {(<()>)}{}, which swaps d and r if d = 0. To calculate the modulus, we perform this swap after decrementing d and incrementing r.

This change does not affect the outcome if a %b > 0, but if a % b = 0, it leaves (n, d, r) = (0, b, 0) – rather than (n, d, r) = (0, 0, b) – on the stack.

Thus, to obtain the modulus, we only have to pop n and d with {}{}.

Stack-clean modulus calculation, 64 bytes

The 42-byte modulus algorithm is not stack-clean, so it cannot be used as is in all programs. The following version pops dividend and divisor (in that order) from the active stack and pushes the modulus in return. It has no other side effects.

({}(<()>)){({}[()]<(({}()[({})])){{}(<({}({}))>)}{}>)}({}{}<{}>)

This solution is largely based on @WheatWizard's previous 72-byte record, but it saves 6 bytes by never switching stacks.

Try it online!

How it works

             INPUT: a (dividend), b (divisor)
             INITIAL STACK: n = a, b

(
  {}         Pop and yield n = a.
  (<()>)       Push d = 0.
)              Push n + 0 = n.
             STACK: n, d = 0, b
{(           While n in non-zero:
  {}           Pop and yield n.
  [()]         Yield -1.
  <
   ((
     {}         Pop and yield d.
     ()         Yield 1.
     [({})]     Pop b, push it back on the stack, and yield -b.
   ))         Push d + 1 + -b = d + 1 - b twice.
   {          While/if d + 1 - b is non-zero, i.e., if d < b - 1
     {}         Pop d + 1 - b (second copy).
     (<(
       {}         Pop d + 1 - b (first copy).
       ({})       Pop b and push it back on the stack.
     )>)        Push d + 1 - b + b = d + 1, then 0.
   }          If the loop wasn't skipped entirely, pushing 0 breaks out.
              If d < b - 1, it essentially performs the assignment d = d + 1.
              However, if d = b - 1, we get d = d + 1 - b = b - 1 + 1 - b = 0.
              In all cases, we wind up with d = (d + 1) % b.
   {}         Pop 0.
  >         Yield 0.
)}        Push n + -1 + 0 = n - 1. Break if n - 1 = 0.
          STACK: n = 0, d = a % b, b
(
  {}        Pop and yield n = 0.
  {}        Pop and d = a % b.
  <{}>      Pop b, but yield 0.
)         Push 0 + a % b + 0 = a % b.

Answer 3

x86_32 machine code, 8 bytes

08048550 <div7>:
 8048550:   99                      cdq   
 8048551:   f7 f9                   idiv   %ecx
 8048553:   85 d2                   test   %edx,%edx
 8048555:   0f 94 c0                sete   %al

This is my first code golf answer, so hopefully I'm following all the rules.

This first calls cdq to clear out the edx register, then performs signed division on the ecx register, which stores the remainder in edx. The test edx, edx line will set the zero flag if edx is zero, and sete puts a 0 for false if edx was not zero, and puts a 1 for true if edx was 0.

This is just the code snippet that contributes to the byte count, but for testing, here is the C code I wrote with the inline assembly because it's easier this way to handle I/O.

Answer 4

Hexagony, 15, 13, 12 10 bytes

Everybody's favorite hexagon-based language! :D

TL;DR works using magic, unformatted solutions in decreasing byte count:

?{?..>1'%<.@!'/
?{?!1\.'%<@.>
?{?\!1@'%\!(
?{?!1\@'%<

Saved 2 bytes thanks to @MartinEnder's layout wizardry.

@FryAmTheEggman saved 1 byte by using the corners more creatively

Both @MartinEnder and @FryAmTheEggman came up with a 10 byte solution that doesn't print anything for falsely values.

My solution (15):

Unformatted:

?{?..>1'%<.@!'/

Formatted:

  ? { ?
 . . > 1
' % < . @
 ! ' / .
  . . .

@Martin Ender's Solution (13):

Unformatted:

?{?!1\.'%<@.>

Formatted:

  ? { ?
 ! 1 \ .
' % < @ .
 > . . .
  . . .

Explanation:

First, we get the input and take the modulus.

  ? { ?
 . . . .
' % . . .
 . . . .
  . . .

Then, it checks if the modulus is 0 or not. If it is, the IP turns 60 degrees left, bounces off the mirror, sets the cell to 1 and prints.

Then, the IP continues onto the fourth row. When it reaches the >, it turns to the right instead (because the value of the cell is now 1). It goes oob, and comes back in the bottom right corner heading NW. The IP hits the <, goes along the top row, and comes back in the right corner to hit the @, stopping the program.

  . . .
 ! 1 \ .
. . < @ .
 > . . .
  . . .

If the modulus turns out to be positive, the IP turns 60 degrees to the right. Once it goes out the bottom right corner, it continues on the bottom left edge because of Hexagony's wrapping rules. The ' is reused to make the IP go to a cell with 0 in it. The IP then travels along the fourth row, wraps around to the second, hits print, and gets reflected into the <. The rest of the path to the @ is the same.

  . . .
 ! . \ .
' . < @ .
 > . . .
  . . .

That's some serious wizardry.

@FryAmTheEggman's Solution (12):

Unformatted:

?{?\!1@'%\!(

Formatted:

  ? { ?
 \ ! 1 @
' % \ ! (
 . . . .
  . . .

Explanation:

Like the other solutions, it gets the input and takes the modulus.

  ? { ?
 . . . .
' % . . .
 . . . .
  . . .

Then, the IP gets deflected into the bottom corner. If the modulus is positive, it goes on the top left edge. The ? has no more input, so it sets the cell to 0. The ! then prints the 0, and the @ terminates the program.

  ? . .
 \ ! . @
. . \ . .
 . . . .
  . . .

Things are much trickier for when the modulus is 0. First of all, it gets decremented, then reset to 0, then set to 1, then printed. Then, the 1 gets decremented to 0. After that, the program runs like it does at the beginning until it tries to do 0%0. That makes it throw a silent error and quit.

  ? { ?
 . . 1 .
' % \ ! (
 . . . .
  . . .

I really like the silent error trick, but a simpler way would be to replace the ( with / so that the IP passes through the first time, but gets reflected into @ the second.

Collaborative solution (10):

Unformatted:

?{?!1\@'%<

Formatted:

  ? { ?
 ! 1 \ @
' % < . .
 . . . .
  . . .

This program starts out the same as all the other programs, getting the input and modding it.

If the input is 0, the IP turns left when it hits <. It gets deflected into 1!@, which prints 1 and quits.

  . . .
 ! 1 \ @
. . < . .
 . . . .
  . . .

If the input is positive, the IP turns right when it hits <. It exits through the corner, and goes along the top right edge hitting the @ without printing.

  . . ?
 . . . @
. . < . .
 . . . .
  . . .

Answer 5

Brain-flak 102, 98, 96 bytes

(({}<>))<>{({}[()])<>(({}[()])){{}(<({}[({})])>)}{}({}({}))<>}{}<>([{}]{}){<>(([()])())}({}{}())

Eww. Gross. I might post an explanation, but I barely understand it myself. This language hurts my brain.

Try it online!

Thanks to github user @Wheatwizard for coming up with a modulus example. I probably could not have figured that out myself!

Also, the shorter answer is here.

Possibly incorrect explanation:

(({}<>))                    #Push this element onto the other stack
<>                          #Move back to stack one.
{                           #While the top element is non-zero:
 ({}[()])                   #  Decrement the number on top
 <>                         #  Move to the other stack
 (({}[()]))                 #  Push the top element minus one twice
 {                          #  While the top element is non-zero:
  {}                        #    Pop the top element
  (<          >)            #    Push a zero
        ({})                #    Push the second from top element
       [    ]               #    Evalue this second from top element as negative
    ({}      )              #    And push that negative plus the top element
 }
 {}                         #  Pop the top element
 ({}({}))                   #  Push the top element plus the second from the top, AND push the second from top
 <>                         #  Switch stacks
}

{}                          #Pop the stack
<>                          #Switch to the other stack
([{}]{})                    #And push the top element minus the second element.

The rest is pretty straightforward.

{              }            #While the top element is non-zero:
 <>                         #Move to the other stack
   (([()])  )               #Push a negative one
          ()                #AND push the previously pushed value + 1 (e.g. 0)

                 (      )   #Push:
                  {}{}      #The top two elements added together
                      ()    #Plus one

Answer 6

Javascript (ES6) 17 12 11 bytes

a=>b=>a%b<1

• EDIT: Removed 5 bytes because 'a>0' is expected.
• EDIT2: Removed 1 byte thanks to Downgoat.

Answer 7

Mathematica - 17 13 3 bytes

∣

Thanks to @MartinEnder for saving a ton of bytes!

Answer 8

Retina, 12 bytes

^(1+)\1* \1$

Takes space-separated input in unary, like 111111111111 1111 to check if 12 if divisible by 4. Prints 1 (true) or 0 (false).

Try it online!

FryAmTheEggman saved two bytes. Oops, rewrote my answer to take the arguments in the right order. (Then Fry beat me to it in the comments. I’m slow at regex!)

Answer 9

Batch, 20 bytes

@cmd/cset/a!(%1%%%2)

Outputs 1 on success, 0 on failure.

Answer 10

Vim, 11 keystrokes

C<C-r>=<C-r>"<C-Left>%<C-Right><1<cr>

Not bad for a language that only handles strings. :D

Answer 11

C#, 27 13 12 Bytes

a=>b=>a%b<1;

Thanks to TuukkaX for pointing out anonymous lambdas are acceptable. Thanks to David Conrad for pointing me on to currying which I wasn't even aware was a thing.

Short and sweet, since we're only dealing with integers we can use <1 rather than ==0 and save a whole byte.

Answer 12

brainfuck, 53 bytes

Takes input as bytes, output is a byte value of 0x00 or 0x01. It's the DivMod algorithm followed by Boolean negation.

,>,<[->-[>+>>]>[+[-<+>]>+>>]<<<<<]>,>[<+>,]+<[>-<-]>.

Try it online - Has a bunch of extra + near the end so you can see the output in ASCII.

Answer 13

Brain-Flak, 88 86 bytes

(<({}<>)>)<>{({}[()])<>(({}()[({})])){{}(<({}({}))>)}{}<>}<>(({}<{}>)){{}{}(<(())>)}{}

This is a denser version of the original Brain-Flak divisibility test algorithm written by Dr Green Eggs and Iron Man DJMcMayhem and myself.

Here is a brief(ish) explanation of how it works:

  ({}<>)        #Move the top of the stack to the other stack #Start Mod
(<      >)      #Push zero
<>              #Switch stacks
{               #While the top of the stack is not zero
 ({}[()])       #Subtract one from the top of the stack
 <>             #Switch stacks
   {}()         #Pop the top, add one and ...
       [({})]   #Subtract the second element on the stack
 ((          )) #Push twice
 {              #If the top is not zero
  {}            #Pop the duplicate
    ({}({}))    #Add the second element to the first
  (<        >)  #Push zero
 }              #End if
 {}             #Pop the zero
 <>             #Switch back
}               #End While
<>              #Switch to the other stack
 ({}<{}>)       #Remove the second value on the stack         #End Mod
(        )      #Duplicate the result of modulation
{               #If the top is not zero
 {}{}           #Pop the top two elements
 (<(())>)       #Push a one and a zero
}               #End if
{}              #Pop the zero

Try it Online!

Answer 14

LOLCODE, 74 64 bytes

HOW IZ I f YR a AN YR b
BOTH SAEM MOD OF a AN b AN 0
IF U SAY SO

Answer 15

Java 8, 11 bytes

a->b->a%b<1

What the heck, there are JS and C# versions of this, why not a Java version, too?

Usage:

import java.util.function.Function;

public class Program {
    public static void main(String[] args) {
        System.out.printf("%d, %d %b%n", 9, 3, divides(9, 3, a->b->a%b<1));
        System.out.printf("%d, %d %b%n", 3, 9, divides(3, 9, a->b->a%b<1));
    }

    public static boolean divides(int a, int b,
            Function<Integer, Function<Integer, Boolean>> f) {
        return f.apply(a).apply(b);
    }
}

Answer 16

C, 60 Bytes

#include <stdio.h>
main(){int a,b;scanf("%d %d",&a,&b);a%b==0;}

Answer 17

R, 22 20 bytes

a=scan();!a[1]%%a[2]

As usually, reads two numbers from the input that is terminated by an empty line.

Update: thanks to Jarko Dubbeldam for shaving off 2 bytes (despite the fact that his edit was rejected, it was very helpful!).

Answer 18

APL (Dyalog Unicode), 3 bytes

0=|

Try it online!

Is zero equal to the division remainder?

Answer 19

Python, 16 bytes

lambda D,d:D%d<1

Answer 20

GolfScript, 3 bytes

~%!

Explanation:

~    # Evaluate the input
 %   # Take the first modulus the second
  !  # Boolean not

Try it online!

Answer 21

CJam, 6 4 bytes

Saved 2 bytes thanks to Dennis

q~%!

Try it online

q    e# Take in the input
 ~   e# Dump the individual values to the stack
  %  e# Modulus
   ! e# Boolean NOT

Answer 22

Brachylog, 2 bytes

%0

Try it online!

Answer 23

Julia, 9 bytes

D\d=D%d<1

Try it online!

Answer 24

Fortran 95, 78 bytes

function f(i,j)result(k)
integer::i,j,k
k=merge(1,0,MOD(i,j)<1)
end function f

Answer 25

MarioLANG, 121 109 107 bytes

Saved 14 bytes thanks to Martin Ender

;>(-)-)+(([!)
)"=========#[
; +(![-)< )<!+
  ==#==="  "#:
>!< >(+ !![(<
=#"="===##=:"
  !      <
  #======"

Try it online!

Explanation

The algorithm is simply to keep subtracting d from n to see if you can do it an integer number of times and have no remainder.

;
)
;

>
=
 
 

First, the input is collected. n is in the first cell, d in the second.

 >(-)-)+(([!
 "=========#
          )<
           "
 !
 #"="===##=
  
  

This is essentially the main loop. It decrements the first and second cells, and increments the third.

           [!)
           =#[
             !+
             #:
            (<
            :"
 
 

This is the final output. If after the incrementing/decrementing, the first cell is 0, then we've eliminated n. If after this, the second cell (d) is 0, then d went into n evenly. We increment and print (1). Otherwise, move back to the first cell (which is 0) and print it.

 
 
  +(![-)<  
  ==#==="  
 !< >(+ !![
 #"="===##=
  !      <
  #======"

This loop happens if the second cell is 0 after incrementing and decrementing. It copies the third cell to the second cell. The part at the bottom is to bypass the loop if the cell is not 0.

Answer 26

Tcl , 34 bytes

ge stdin a
ge stdin b
exp $a%$b<1

My first /*successful*/ attempt in codegolf ! This code must be executed in Tcl shell , otherwise it will not work.

One byte thanks to @Lynn .

Four bytes thanks to @Lynn and @LeakyNun (now I understand what he meant)!

Answer 27

PHP, 23 22 bytes

<?=$argv[1]%$argv[2]<1

prints 1 for true, empty string (=nothing) for false

call from cli with n and d as arguments

---

10 bytes for ancient PHP: <?=$n%$d<1

Answer 28

J, 3 bytes

0=|

Usage:

2 (0=|) 10 

Will return 1. And is equivalent to pseudocode 10 MOD 2 EQ 0

Note this is very similar to the APL answer, because J is heaviliy inspired by APL

Answer 29

PowerShell v2+, 20 bytes

!($args-join'%'|iex)

Takes input as two command-line arguments $args, -joins them together into a string with % as the separator, pipes that to iex (short for Invoke-Expression and similar to eval). The result is either 0 or non-zero, so we take the Boolean not ! of that result, which means either $TRUE or $FALSE (non-zero integers in PowerShell are truthy). That Boolean is left on the pipeline and output is implicit.

Alternative versions, also 20 bytes each

param($a,$b)!($a%$b)
!($args[0]%$args[1])

Same concept, just slightly different ways of structuring the input. Thanks to @DarthTwon for providing these.

Examples

PS C:\Tools\Scripts\golfing> .\divisibility-test.ps1 24 12
True

PS C:\Tools\Scripts\golfing> .\divisibility-test.ps1 24 13
False

PS C:\Tools\Scripts\golfing> .\divisibility-test.ps1 12 24
False

Answer 30

Haskell, 13 11 bytes

((1>).).mod

This defines a new function (!) :: Integral n => n -> n -> Bool. Since mod n m returns only positive numbers if n and m are positive, we can save a byte by using 1> instead of 0==.

Usage:

ghci> let n!d=1>mod n d
ghci> 100 ! 2
True
ghci> 100 ! 3
False