40
\$\begingroup\$

This challenge, while probably trivial in most "standard" languages, is addressed to those languages which are so esoteric, low-level, and/or difficult to use that are very rarely seen on this site. It should provide an interesting problem to solve, so this is your occasion to try that weird language you've read about!

The task

Take two natural numbers a and b as input, and output two other numbers: the result of the integer division a/b, and the remainder of such division (a%b).

This is : shortest answer (in bytes), for each language, wins!

Input/Output

  • 0<=a<=255, 1<=b<=255. Each of your inputs (and outputs too) will fit in a single byte.
  • You may choose any format you like for both input and output, as long as the two numbers are clearly distinguishable (e.g. no printing the two results together without a delimiter)

Examples

a,b->division,remainder
5,7->0,5
5,1->5,0
18,4->4,2
255,25->10,5

Note: Builtins that return both the result of the division and the remainder are forbidden. At least show us how your language deals with applying two functions to the same arguments.

Note 2: As always, an explanation of how your code works is very welcome, even if it looks readable to you it may not be so for someone else!


The Catalogue

The Stack Snippet at the bottom of this post generates the catalogue 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 + 2 (-p flag) = 45 bytes

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

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

/* Configuration */

var QUESTION_ID = 114003; // Obtain this from the url
// It will be like https://XYZ.stackexchange.com/questions/QUESTION_ID/... on any question page
var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";
var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 8478; // This should be the user ID of the challenge author.

/* App */

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, 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.toLowerCase() > b.lang_raw.toLowerCase()) return 1;
    if (a.lang_raw.toLowerCase() < b.lang_raw.toLowerCase()) 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);
  }

}
body {
  text-align: left !important;
  display: block !important;
}

#answer-list {
  padding: 10px;
  width: 290px;
  float: left;
}

#language-list {
  padding: 10px;
  width: 500px;
  float: left;
}

table thead {
  font-weight: bold;
}

table td {
  padding: 5px;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<link rel="stylesheet" type="text/css" href="https://cdn.sstatic.net/Sites/codegolf/all.css?v=ffb5d0584c5f">
<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>

\$\endgroup\$
21
  • \$\begingroup\$ Can I reverse the arguments, i.e. instead of providing a b providing b a instead? \$\endgroup\$ Mar 27 '17 at 13:29
  • \$\begingroup\$ @EriktheOutgolfer: You may choose any format you like for both input and output, as long as the two numbers are clearly distinguishable \$\endgroup\$
    – Emigna
    Mar 27 '17 at 13:30
  • \$\begingroup\$ @Emigna Yeah, I was not sure if reversing didn't make them indistinguishable though. \$\endgroup\$ Mar 27 '17 at 13:32
  • \$\begingroup\$ @EriktheOutgolfer if you know that they need to be reversed you have no problem in distinguishing them :) \$\endgroup\$
    – Leo
    Mar 27 '17 at 13:45
  • \$\begingroup\$ Unfortunately, the BF algorithm doesn't work if the divisor is 1. \$\endgroup\$
    – mbomb007
    Mar 27 '17 at 13:45

79 Answers 79

2
\$\begingroup\$

Gol><>, 11 bytes

I:I:@$S,N%h

Online interpreter

\$\endgroup\$
2
\$\begingroup\$

MATL, 5 bytes

/k&G\

Try it out at MATL Online!

Explanation

        % Implicitly grab the two inputs as numbers
/       % Divide them
k       % Round down the result
&G      % Grab the two inputs again
\       % Compute the remainder
\$\endgroup\$
0
2
\$\begingroup\$

Pyke, 3 bytes

'f%

Try it online!

\$\endgroup\$
2
\$\begingroup\$

Ouroboros, 15 bytes

r.r.@/Inao\%n1(

Takes the numbers in reverse order (e.g. 10 42). Try it here.

Explanation

r.r.             Read a number, duplicate, read a number, duplicate
    @            Rotate a copy of the first number to the top of the stack
     /I          Divide and truncate to integer
       n         Output as number
        ao       Push 10 and output as character (newline)
          \%     Swap the remaining two values and take the mod
            n    Output as number
             1(  Push 1 and swallow that many characters from the end of the program,
                 halting execution
\$\endgroup\$
2
\$\begingroup\$

Ruby, 27 bytes

a,b=$<.map &:to_i
p a/b,a%b

Reads from standard input, stops when it reaches EOF, and only considers the two integers on the first two lines of its input.

\$\endgroup\$
1
  • 2
    \$\begingroup\$ You don't have to use stdin and stdout for I/O. Typically, the golfiest Ruby solutions are anonymous lambdas: try ->a,b{[a/b,a%b]} for 16 bytes! \$\endgroup\$
    – m-chrzan
    Mar 27 '17 at 21:25
2
\$\begingroup\$

dc, 12 bytes

?sadla/rla%f

Explanation:

?            # Read input as space-separated integers, push to stack
 sa          # Store second to register a
   d         # Duplicate first
    la       # Load second
      /      # Push first / second
       r     # Swap (stack now first/second, first)
        la   # Load second
          %  # Push first % second
           f # Print stack (first/second, first%second, but in reverse)

Try it online!

With builtins, it's just ?~f. I also tried reading each integer on its own line, but it's the same length (?d?dsa/rla%f).

\$\endgroup\$
2
\$\begingroup\$

Alice, 12 10 bytes

 \%.\
O:io

Try it online!

Explanation

Alice does have a divmod built-in, which would allow a solution in at most 7 bytes:

</
iMOP

(The P could be any character whatsoever, so I thought I'd put the image of Apple-branded cleaning products in your head. You're welcome.)

But without that built-in, things certainly get a bit more interesting.

Since Alice also has separate built-ins for division and modulo, the program is entirely linear, but the instruction pointer takes a fairly tricky path through the code. Here is the breakdown of the program flow:

\    Reflect to NE. Switch to Ordinal
     Reflect off boundary --> SE.
i    Read all input as a string.
     Reflect off boundary --> NE.
.    Duplicate the input string.
     Reflect off boundary --> SE.
     Reflect off corner --> NW.
.    Duplicate the input string. (Irrelevant)
     Reflect off boundary --> SW.
i    Try to read more input, but that just pushes "".
     Reflect off boundary --> NW.
\    Reflect to S. Switch to Cardinal.
:    Implicitly discard the empty string and convert a copy of the input 
     to the two integers it contains and compute the result of their integer division.
     IP wraps back to the first line.
\    Reflect to NW. Switch to Ordinal.
     Reflect off boundary --> SW.
O    Implicitly convert the division result to a string and print it with a 
     trailing linefeed.
     Reflect off corner --> NE.
\    Reflect to E. Switch to Cardinal.
%    Implicitly convert a copy of the input to the two integers it contains
     and compute the result of their modulo.
.    Duplicate the modulo result. (Irrelevant)
\    Reflect to NE. Switch to Ordinal.
     Reflect off corner --> SW.
o    Implicitly convert the modulo reuslt to a string and print it without a
     trailing linefeed.

At this point we're basically done. The IP does a few more rounds through the code and pushes and prints a few more empty strings, but none of that affects the output. Eventually, the code will attempt a division by zero which terminates the program. The instructions executed by Alice after the last meaningful o are:

%::%o\\i..i\:

The exact movement of the IP through the grid and what exactly these commands do to the stack are left as an exercise to the reader...

\$\endgroup\$
2
  • \$\begingroup\$ I'm glad you wrote this answer :) I've tried to play with the layout and came up with this alternative solution for the same byte count (but maybe a weirder flow) \$\endgroup\$
    – Leo
    Apr 11 '17 at 21:04
  • \$\begingroup\$ @Leo Ah nice. I had some earlier solutions using 1,, but they were longer. Also, nice work figuring out 1, (as opposed to ~) on your own! :D \$\endgroup\$ Apr 11 '17 at 21:22
2
\$\begingroup\$

x86-16 machine code, 5 bytes (non-competing?)

88 1E 0105  MOV  BYTE PTR[AAM1+1], BL   ; modify second byte of AAM opcode for divisor
        AAM1:
D4 ?        AAM                         ; ASCII Adjust AX After Multiply

Input numbers in AL and BL. Output division in AH, remainder in AL.

This "abuses" the x86 "ASCII adjust after multiply" instruction intended to facilitate multiplication of binary coded decimal values (BCD), to convert results to base-10 values. These instructions were deprecated in x64, however as an under-documented feature this behavior can be altered by using self-modifying code to perform arbitrary base conversion of variable input data.

Caveat:

The challenge does state Builtins that return both the result of the division and the remainder are forbidden. These operations are always done at the same time in x86 machine code so not clear if this would disqualify the platform. This answer, however, does not use the x86's DIV instruction. Question posed to OP. :)

\$\endgroup\$
2
\$\begingroup\$

Whitespace, 68 bytes

[S S S N
_Push_0][S N
S _Dupe_0][T    N
T   T   _Read_STDIN_as_integer][T   T   T   _Retrieve_input1][S N
S _Dupe_input1][S N
S _Dupe_input1][T   N
T   T   _Read_STDIN_as_integer][T   T   T   _Retrieve_input2][T S T S _Integer_divide][T    N
S T _Print_as_integer][S S S T  S T S N
_Push_10_newline][T N
S S _Print_as_character][S S S N
_Push_0][T  T   T   _Retrieve_input1][S N
S _Dupe_input1][T   T   T   _Retrieve_input2][T S T T   _Modulo][T  N
S T _Print_as_integer]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs and new-lines only).

Prints with a newline delimiter (could alternatively be a tab for the same byte-count).

Explanation in pseudo-code:

Integer a = STDIN as integer
Integer b = STDIN as integer
Print a integer-divided by b
Print "\n"
Print a modulo-b

Example program flow (inputs 255 and 25):

Command   Explanation              Stack          Heap                STDIN  STDOUT  STDERR

SSSN      Push 0                   [0]
SNS       Duplicate 0              [0,0]
TNTT      Read STDIN as integer    [0]            [{0:255}]           255
TTT       Retrieve from heap #0    [255]          [{0:255}]
SNS       Duplicate 255            [255,255]      [{0:255}]
SNS       Duplicate 255            [255,255,255]  [{0:255}]
TNTT      Read STDIN as integer    [255,255]      [{0:255},{255:25}]  25
TTT       Retrieve from heap #255  [255,25]       [{0:255},{255:25}]
TSTS      Integer divide (255/25)  [10]           [{0:255},{255:25}]
TNST      Print as integer         []             [{0:255},{255:25}]         10
SSSTSTSN  Push 10                  [10]           [{0:255},{255:25}]
TNSS      Print as character       []             [{0:255},{255:25}]         \n
SSSN      Push 0                   [0]            [{0:255},{255:25}]
TTT       Retrieve from heap #0    [255]          [{0:255},{255:25}]
SNS       Duplicate 255            [255,255]      [{0:255},{255:25}]
TTT       Retrieve from heap #255  [255,25]       [{0:255},{255:25}]
TSTT      Modulo (255%25)          [5]            [{0:255},{255:25}]
TNST      Print as integer         []             [{0:255},{255:25}]         5
                                                                                     error

Program stops with an error because no exit is defined.

Since the inputs are guaranteed to be non-negative, I'm using the first input as heap-address to store the second input to save bytes.

\$\endgroup\$
2
\$\begingroup\$

MathGolf, 4 bytes

‼/%]

Inputs in the order b a, output as a list containing two integers.

Try it online.

Explanation:

‼   # Apply the following two commands to the stack separately:
 /  #  Division (which will be integer division if the given arguments are integers)
 %  #  Modulo
    #  (both builtins will use the implicit input-integers for their two required arguments)
  ] # Then wrap all values on the stack into a list
    # (after which the entire stack joined together is output implicitly as result)

For an actual character-separated string, we could use one of these instead for 5 bytes: ‼/% \ (space separator); ‼/%n\ or ‼/%]n (newline separator); ‼/%⌂\ (* separator).

\$\endgroup\$
1
\$\begingroup\$

PowerShell, 38 31 bytes

$a,$b=$args;($a-$a%$b)/$b;$a%$b

Try it online!

Sheesh, this is icky. So, PowerShell (helpfully) returns floating point values when doing division if it doesn't divide evenly. Sometimes, this is a Good Thing, but other times (like here) it's very not. So, you'd figure "Oh, let's just toss an [int] cast and call it good, right?" Nope. Casting from a [double] to an [int] in PowerShell does banker's rounding, so for input 5, 7 we would get 1 back, not 0. As a result, we need to subtract the remainder (from the modulo), then calculate the division, and then calculate the modulo again. Yay!

Both results are left on the pipeline, and output is implicit.

Saved 7 bytes thanks to PeterTaylor being smarter than me.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ ($a-$a%$b)/$b? \$\endgroup\$ Mar 27 '17 at 14:09
  • \$\begingroup\$ @PeterTaylor Oh. Right. facepalm \$\endgroup\$ Mar 27 '17 at 14:41
1
\$\begingroup\$

GolfScript, 6 bytes

.~/p~%

Try it online!

I'm literally on an answer spree!

-3 thanks to Peter Taylor.

\$\endgroup\$
2
  • 2
    \$\begingroup\$ Why evaluate so early if it forces you into such complicated stack manipulations? .~/p~% and similarly for CJam. \$\endgroup\$ Mar 27 '17 at 14:12
  • \$\begingroup\$ @PeterTaylor That is very cool. \$\endgroup\$ Mar 27 '17 at 14:41
1
\$\begingroup\$

Batch, 37 bytes

@set/ad=%1/%2,m=%1%%%2
@echo %d% %m%
\$\endgroup\$
1
  • \$\begingroup\$ Dang, I'm happy Peter gave me a golf, or this would've been shorter than PowerShell and I would've been embarrassed! :D \$\endgroup\$ Mar 27 '17 at 18:35
1
\$\begingroup\$

Microscript, 11 bytes

ivissCl/pl%
\$\endgroup\$
1
\$\begingroup\$

brainfuck, 47 bytes

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

Input/output in byte-values.

Try it online! - Computes 109,10->10,9, which in characters corresponds to m,\n->\n,\t

This is simply the esolang brainfuck divmod algorithm, with a wrapper for doing input/output and dealing with the case where divisor=1. I'm sure that someone could modify the algorithm to make it work with any input without the need for my boilerplate, but at least now there's a brainfuck answer^^

\$\endgroup\$
1
\$\begingroup\$

Cardinal, 34 32 31 bytes

%:~:v
,0.M#
-
-
>8\ < <
^+/'Jx^

Try it online!

Input is in the format b first, a second
Output is in the format a%b, a/b

Explanation:

%:~:v
    #

Sets second input as active value, first input as inactive value then sends this pointer in two directions

,0.M

Gets the active value mod inactive value, outputs then outputs a space

  .
-
-
>8\ < <
^+/'Jx^

Counts the number of times the inactive value can be subtracted from the active value before it falls to <=0 before outputting the count.

\$\endgroup\$
1
\$\begingroup\$

QBIC, 12 bytes

::?a'\`b,a%b

Explanation:

::     Gets two integers from the command line, a and b
?      PRINT
 a'\`b   Integer division in QBasic is '\', however, '\' is the command for ELSE in QBIC
         To tell QBIC not to swap in ELSE at the '\', we need a code literal.
         Everything from the ' to the `is passed on to QBasic without being parsed.
 ,a%b    Also print a MOD b
\$\endgroup\$
1
\$\begingroup\$

ReRegex, 75 Bytes.

#import math
div<(_*),\1_+(,(_*))?>/$3/s(\d+),(\d+)e/$1\/$2,$1%$2/s#input
e

Explination

This is made up of the following regex match and replaces...

  • div<(_*),\1_*(,(_*))?> / $3
  • s(\d+),(\d+)e / $1/$2,$1%$2

And the code line is simply

s(?#input)e.

The s...e takes the input in the form of 123,123 and wraps it, so we know to execute a function on it. In particular, line 2, which changes 123,456 to 123/456,123%456. Then, almost all of it is taken care of by the math library, However, the latest version contains an issue with the fact that integer division is actually ceiling, instead of floor'd division. So the first replacement corrects that.

Currently no TIO, but the GitHub is available.

\$\endgroup\$
1
\$\begingroup\$

J, 9 bytes

(<.@%,|~)

Usage:

   5 (<.@%,|~) 7
0 5
   5 (<.@%,|~) 1
5 0
   18 (<.@%,|~) 4
4 2
   255 (<.@%,|~) 25
10 5

Explanation:

In J a verb is like a function. Verbs are monadic (f x) or dyadic (x f y) and builtin verbs share the same symbol for different monadic and dyadic cases (e.g. x % y is division, % x is reciprocal).

(<.@% , |~) is a train of 3 verbs (fork in J terminology). A fork is: x (f g h) y = (x f y) g (x h y). Our verbs are:

  • |~ - | is modulo in J, but arguments are for some reason reverse to what you'd expect. We need the adverb ~ to reverse the arguments: x f~ y = y f x

  • <. @ % - x % y is division. @ is a conjunction denoting function composition: x (f @ g) y = f (x g y) (J applies right-to-left). <. x is floor.

  • , - Append.

The parenthesis are needed to form the 3-verb train. No parenthesis would parse as x <.@% (, (|~ y)).

\$\endgroup\$
1
\$\begingroup\$

QBasic 4.5, 23 bytes

INPUT a,b
?a\b,a MOD b

This speaks for itself: get two numbers (input needs to be separated by comma) and print their integer division and modulo. QBasic doesn't have a shorthand for MOD, the %symbol is reserved for defining integers.

\$\endgroup\$
1
\$\begingroup\$

I, 5 bytes

/.m,%

/ divide the arguments

.m apply minimum

,% append modulus

Try it online!

\$\endgroup\$
1
\$\begingroup\$

REXX, 20 bytes

arg a b
say a%b a//b
\$\endgroup\$
1
\$\begingroup\$

Fortran 95, 45 bytes

function f(i,j)
write(*,*)i/j,modulo(i,j)
end
\$\endgroup\$
1
\$\begingroup\$

jq, 25 characters

def f(a;b):a/b|floor,a%b;

Sample run:

bash-4.3$ jq -n 'def f(a;b):a/b|floor,a%b;f(255;25)'
10
5

On-line test

\$\endgroup\$
1
\$\begingroup\$

Gema, 25 characters

* *=@div{*;*} @mod{$1;$2}

Sample run:

bash-4.3$ gema '* *=@div{*;*} @mod{$1;$2}' <<< '255 25'
10 5
\$\endgroup\$
1
\$\begingroup\$

Mathematica, 32 31 bytes

Thanks to Martin Ender for calming the code down to the tune of 1 byte!

0@#2//.a_@b_/;b>=#:>(a+1)[b-#]&

Just to mess with the language. Pure function taking the two positive integer arguments in the opposite (counterintuitive) order, and returning the quotient q and the remainder r in the same style, q[r], as in Martin Ender's Mathematica answer. While that answer is shorter, this one is ... more contrary? It implements repeated subtraction on expressions of the form a[b].

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AHK, 35 bytes

a=%1%
b=%2%
Send,% a//b ","Mod(a,b)

AutoHotkey assigns numbers 1-n as variable names for the incoming parameters. It causes some problems when you try to use those in functions because it thinks you mean the literal number 1 instead of the variable named 1. The best workaround I can find is to assign them to different variables.

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Forth, 4 bytes

/MOD

Top of stack will be the quotient, then the remainder.

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TI-Basic (TI-84 Plus CE): 18 bytes

Prompt A
Prompt B
{int(A/B),remainder(A,B

Prompt, A, B, newline, {, int(, /, ,, and ) are all one-byte tokens, but remainder( is a two-byte token.

Prompt prompts you for the numbers.

int(A/B) computes the floored division of A and B

remainder(A,B computes remainder when A is divided by B

{ causes the values of int(A/B) and remainder(A,B) to be stored to a list. This list is implicitly returned, as it is the last evaluated value in the program.

The returned list is printed as {divison remainder}, with a space between the numbers.

Note: TI-Basic does not require closing parentheses or brackets in most cases; one exception being here with int(A/B): the ) is needed so as not to pass the value after the comma as a second argument to int( (which would raise an Error: Syntax).

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  • \$\begingroup\$ You can save 2 bytes with Prompt A,B:int({A/B,remainder(A,B \$\endgroup\$
    – lirtosiast
    Apr 6 '17 at 18:42
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Pip, 8 bytes

Pa//ba%b

Try it online!

Explanation:

           The input is read automatically into the vars a and b
     a%b   calculate a%b. Since this is the last thing we do, this is printed implicitly.
 a//b      We also want to show a // b (double slashes is integer div).
P          So we need an explicit PRINT command.
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