36
\$\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\$
  • \$\begingroup\$ Can I reverse the arguments, i.e. instead of providing a b providing b a instead? \$\endgroup\$ – Erik the Outgolfer 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\$ – Erik the Outgolfer 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

75 Answers 75

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.

| improve this answer | |
\$\endgroup\$
  • 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).

| improve this answer | |
\$\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...

| improve this answer | |
\$\endgroup\$
  • \$\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\$ – Martin Ender Apr 11 '17 at 21:22
1
\$\begingroup\$

Batch, 37 bytes

@set/ad=%1/%2,m=%1%%%2
@echo %d% %m%
| improve this answer | |
\$\endgroup\$
  • \$\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\$ – AdmBorkBork Mar 27 '17 at 18:35
1
\$\begingroup\$

Microscript, 11 bytes

ivissCl/pl%
| improve this answer | |
\$\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^^

| improve this 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.

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

Pyke, 3 bytes

'f%

Try it online!

| improve this answer | |
\$\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
| improve this answer | |
\$\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.

| improve this answer | |
\$\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)).

| improve this answer | |
\$\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.

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

I, 5 bytes

/.m,%

/ divide the arguments

.m apply minimum

,% append modulus

Try it online!

| improve this answer | |
\$\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].

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

Forth, 4 bytes

/MOD

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

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

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).

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ You can save 2 bytes with Prompt A,B:int({A/B,remainder(A,B \$\endgroup\$ – lirtosiast Apr 6 '17 at 18:42
1
\$\begingroup\$

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.
| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

NO!, 118 bytes

It is too insane to do this challenge with NO! which is why it's non-competing (not really, the language is too young to compete in the bloodthirsty arena known as PPCG)

NOOOOOOOOOOOO?NOOOOOOOOOOO
NOOOOOOOOOOOO?NOOOOOOOOOOO
NOOOOO?yes!yess
NOOOOOOOOOOOOOOOOO?yes!yess
NOOOOOOOO?nooo!noooo

Please say you don't want an explanation (one added anyway)

NOOOOOOOOOOOO?                   Create an integer from 
              NOOOOOOOOOOO                              STDIN

NOOOOOOOOOOOO?                   Create an integer from 
              NOOOOOOOOOOO                              STDIN

NOOOOO?                          Divide               by 
       yes!                             line 1 result
           yess                                          line 2 result

NOOOOOOOOOOOOOOOOO?              Modulo               by
                   yes!                 line 1 result
                       yess                              line 2 result

NOOOOOOOO?                       Output               and
          nooo!                         line 3 result
               noooo                                      line 4 result

A yes is used to denote a line number (number determined by number of ss) but the output command takes either numbers or line numbers as arguments and noo!nooo is shorter than yess!yesss

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

Triangular, 21 19 bÿtes

$\:A@$U.vS/p%_<%mp<

Prints the result of division, a newline, then the remainder. Formats into this triangle:

     $
    \ :
   A @ $
  U . v S
 / p % _ <
% m p < ÿ ÿ

ÿ is the no-op that is automaticallÿ inserted when there is no code to fill the smallest possible triangle.

Trÿ it online!

Explanation:

I don't know whÿ \ is the first character... Removed it and managed to save two bÿtes. Still don't know whÿ it was there in the first place. ಠ_ಠ

Directionals:

     .
    \ .
   . . .
  . . v .
 / . . . <
. . . <

This is how the interpreter sees the code, without directionals:

$:$S_%pUA@pm%
  • $: reads integer input to the stack and duplicates it. Stack contains input1 input1.
  • $S reads another input to the stack, then stashes it in memorÿ. Stack contains input1 input1 input2, and memorÿ contains input2.
  • _ divides the top two stack values. Now the stack contains input1 input1/input2.
  • % prints the top of stack as an integer.
  • p pops the top of stack. Now the stack contains input1.
  • U pulls memorÿ onto the stack. Now the stack contains input1 input2.
  • A pushes 10 to the stack. Stack contains input1 input2 10.
  • @ prints it as ASCII (newline).
  • p pops it - stack contains input1 input2.
  • m performs modulus on the top two stack values. Stack now contains input1%input2.
  • % prints the top of stack as an integer.
  • the IP then runs off the plaÿing field, and the program terminates.
| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

Implicit, 9 bytes

$$|/%;_@9

Try it online!

$$|/%;_@9
$$         read two integer inputs
  |        duplicate stack
   /       division
    %      print
     ;     pop
      _    modulus
       @9  print HTAB (delimiter)
| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

Whispers v2, 77 bytes

> Input
> Input
>> 1÷2
>> ⌊3⌋
>> 1%2
>> Output 4
>> Output 5
>> Then 6 7

Try it online!

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

Keg, 12 10 bytes

Other than the creator, I think only I am actively using it. Therefore it is rarely seen on this site; in addition it is esoteric.

Takes 2 inputs separated with newlines.

¿¿:^:"$/"%

Explanation

¿¿#          Takes 2 integer inputs: e.g.     stack [1, 2]
  :#         Duplicate the top item.          stack [1, 2, 2]
   ^#        Reverse the stack.               stack [2, 2, 1]
    :#       Duplicate the stack.             stack [2, 2, 1, 1]
     "#      Right shifts (put top to bottom) stack [1, 2, 2, 1]
      $#     Swap top two items               stack [1, 2, 1, 2]
       /#    Divide top two                   stack [1, 2, 0.5]
         "#  Right shifts                     stack [0.5, 1, 2]
          %# Modulos top two items            stack [0.5, 1]

# Stack is implicitly outputted

TIO

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

33, 11 bytes

OcOcsdoilro

Try it online!

Explanation

OcOc        | Harvest input from user
    s       | Store "a" for later use
     doi    | Print the integer division followed by a newline
        l   | Get "a" back
         ro | Print the remainder of the division
| improve this answer | |
\$\endgroup\$
1
\$\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.

| improve this answer | |
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1
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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).

| improve this answer | |
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0
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Wonder, 13 bytes

@@[/#1#0%#1#0

Usage:

((@@[/#1#0%#1#0)10)5

Simply a curried lambda that returns a list with both the division and modulus. I'm looking for a golfier way to do this, something that doesn't involve lambdas.

Be sure to do tK1000 beforehand to view list items in the output.

| improve this answer | |
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0
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PHP, 35 Bytes

[,$a,$b]=$argv;echo$a/$b^0,_,$a%$b;
| improve this answer | |
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0
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Forth, 25 bytes

: f 2dup / -rot mod ;

Try it online

Input like a b, output like remainder quotient.

| improve this answer | |
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0
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Elixir, 24 bytes

&{div(&1,&2),rem(&1,&2)}

Anonymous function which uses the capture operator and returns a tuple containing the results.

Full program with test cases (and yes, the . in the function call is required!):

f =
&{div(&1,&2),rem(&1,&2)}

# test cases:
IO.inspect f.(5,7)      # 0,5
IO.inspect f.(5,1)      # 5,0
IO.inspect f.(18,4)     # 4,2
IO.inspect f.(255,25)   # 10,5

Try it online on Elixir Playground.

| improve this answer | |
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0
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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.

| improve this answer | |
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  • \$\begingroup\$ ($a-$a%$b)/$b? \$\endgroup\$ – Peter Taylor Mar 27 '17 at 14:09
  • \$\begingroup\$ @PeterTaylor Oh. Right. facepalm \$\endgroup\$ – AdmBorkBork Mar 27 '17 at 14:41

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