Round up my number

Question

^Sandbox

Given two positive integers a and b, return the smallest multiple of b that is greater than or equal to a.

Examples

Definitions: true = is a multiple of b, false = is not a multiple of b.

• f(1,3) = 3 because 1 is false, 1+1 is false, but 1+1+1 is true. Therefore 3 is returned.

• f(5,3) = 6 because 5 is false, but 5+1 is true. Therefore 6 is returned.

• f(9,3) = 9 because 9 is already a multiple of 3.

• f(12,5) = 15 because 12 is false, 12+1 is false, 12+2 is false, but 12+3 is true. Therefore 15 is returned.

Rules

• Input/output can be given by any convenient method.
• You can print it to STDOUT or return it as a function result.
• Either a full program or a function are acceptable.
• Any amount of extraneous whitespace is permitted, provided the characters line up appropriately.
• Standard loopholes are forbidden.
• This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.

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 = 196683; // 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) {
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}

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;
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function getAnswers() {
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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;
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  for (var i = 0; i < langs.length; ++i)
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    var lang = langs[i];
    language = language.replace("{{LANGUAGE}}", lang.lang)
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                       .replace("{{LINK}}", lang.link);
    language = jQuery(language);
    jQuery("#languages").append(language);
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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">

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  </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>
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Answer 1

Python, 17 bytes

lambda a,b:a+-a%b

Try it online!

Since this doesn't use division, it works for both Python 2 and 3. Note that we can't cut out the + sign because a-a%b does a-(a%b), whereas we want to do(-a)%b and add that to a. We could also write it as -a%b+a.

Answer 2

Japt, 2 bytes

cV

Try it

cV     :Implicit input of integers U & V
cV     :Round U up to the nearest multiple of V

With input as an array:

rc

Try it

rc     :Implicit input of array
r      :Reduce by
 c     :Rounding the first element up to the nearest multiple of the second

Less Trivial!, 6 bytes

_vV}aU

Try it

_vV}aU     :Implicit input of integers U & V
_          :Function taking an integer Z as input
 vV        :  Is Z divisible by V?
   }       :End function
    aU     :Starting from U, return the first integer that returns true when passed through that function

ôV ævV

Try it

ôV ævV     :Implicit input of integers U & V
ôV         :Range [U,U+V]
   æ       :Get first element
    vV     :  Divisible by V

Answer 3

Intcode (day 5), 70 66 64 bytes

3,0,3,2,102,-1,2,4,1,0,4,0,1,1,8,1,7,9,0,5,5,5,14,2,1,2,0,4,0,99

Try it online!

Intcode is a programming language invented for Advent of Code 2019. It was first introduced on day 2, then further expanded upon on day 5, which is the version used here.

Since on PPCG, languages are defined by their interpreters, I am using the interpreter that I wrote for my day 5 solution (which was written before this challenge).

Explanation:

3,0,          store first input in cell 0
3,2,          store second input in cell 2
              (cell 1 is a counter, because it's already initialized to zero)
102,-1,2,4,   negate the second input and store in cell 4

              (begin loop)
1,0,4,0,      subtract the second input from the first input
1,1,8,1,      increment the counter (by adding cell 8)
7,9,0,5,      if the first input is greater than zero (cell 9)...
5,5,14,       ... jump to the beginning of the loop (reusing the cell 8 reference)

2,1,2,0,      multiply the counter by the second input
4,0,99        output and quit

Answer 4

MATL, 4 bytes

:ien

Try it online!

Explanation

Consider inputs 12, 5.

     % Implicit input
:    % Range
     % STACK: [1 2 3 4 5 6 7 8 9 10 11 12]
i    % Input
     % STACK: [1 2 3 4 5 6 7 8 9 10 11 12], 5
e    % Reshape as a matrix with that number of rows, padding with zeros
     % STACK: [1  6 11;
               2  7 12;
               3  8  0;
               4  9  0;
               5 10  0]
n    % Number of elements
     % STACK: 15
     % Implicit display

Answer 5

x86 Machine Code, 7 bytes

Since x86 integer division instructions do only floor rounding, this uses the formula: f(a,b) = floor((a+b-1)/b)*b to accomplish ceiling rounding.

02 C3   ADD  AL, BL     ; a = a + b
48      DEC  AX         ; a = a - 1
F6 F3   DIV  BL         ; a = a / b
F6 E3   MUL  BL         ; a = a * b

Input is in AL and BL, output is AL.

Testing

Here is output from a test program for PC DOS that takes interactive input.

Download and test ROUNDMY.COM.

Answer 6

JavaScript (Node.js), 17 bytes

a=>b=>a+(b-a%b)%b

Try it online!

• -1 byte by Neil

Answer 7

Octave, 17 bytes

Same method as this JS answer:

@(a,b)ceil(a/b)*b

Try it online!

Answer 8

Python 2, 23 20 18 bytes

lambda a,b:-a/b*-b

Try it online!

2 bytes thanks to tsh

Answer 9

Haskell, 14 bytes

a%b=a+mod(-a)b

Try it online!

Answer 10

C++ (gcc), 43 41 40 36 25 bytes

#define f(a,b)(b-a%b)%b+a

Try it online!

Saved 4 bytes thanks to @Neil!!!

Saved 11 bytes thanks to @AZTECCO!!!

Answer 11

Jelly, 3 bytes

÷Ċ×

Try it online!

Simply divide, ceiling, multiply.

Answer 12

C (gcc), 47 31 25 bytes

f(a,b){a=a%b?f(++a,b):a;}

Try it online!

-16 bytes thanks to A̲̲ (forgot that return types and parameters would default to int if not defined, whoops!)

-6 thanks to AZTECCO for using a ternary and making the function recursive

Answer 13

Brainfuck, 52 49 bytes

Thanks to Jo King for finding a bug, helping fix it with some extra golfing.

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

Input and output is in character codes.

Try it online!

Explanation

,>>,<<                     [>a, 0,     b,   0]
[>+>->+<[>]>[<+>-]<<[<]>-] [>0, a, b-a%b, a%b]
>>>[<[-<+>]>,]             If a%b is nonzero, add b-a%b to a, and clear a%b
<<.                        Output a+(b-a%b)

Modulus algorithm is straight from the esolang list of brainfuck algorithms.

Answer 14

JavaScript (V8), 22 bytes

This is the function \$(a,b) \mapsto \left\lceil \frac{a}{b} \right\rceil b\$.

a=>b=>Math.ceil(a/b)*b

Try it online!

Answer 15

R, 20 bytes

function(a,b)a+-a%%b

Try it online!

Answer 16

K (oK), 8 bytes

{x+y!-x}

Try it online!

Another port of xnor's Python answer.

Answer 17

Red, 31 bytes

func[a b][round/to/ceiling a b]

Try it online!

Just a built-in round with refinements to, which scales the first argument according to the second, and ceiling, which indicates the direction.

Red, 25 bytes

func[a b][0 - a // b + a]

Try it online!

A port of xnor's Python answer. Don't forget to upvote his answer!

Answer 18

Retina 0.8.2, 28 bytes

\d+
$*
r`1+(\2*) (1+)
$1$2
1

Try it online! Link includes test cases. Explanation:

\d+
$*

Convert the inputs to unary.

r`

Process this stage from right-to-left, so that \2 gets to match \$ b \$ first, allowing us to refer to it when matching within \$ a \$.

1+(\2*) (1+)

Match \$ b \$ in \2, then as many copies of \$ b \$ as we can, without completely exhausting \$ a \$. \1 thus matches \$ b \lfloor \frac {a - 1} b \rfloor = b \lceil \frac {a - b} b \rceil \$

$1$2

Add that to \$ b \$ to give \$ b \lceil \frac a b \rceil \$.

1

Convert to decimal.

Answer 19

Ruby, 13 bytes

->a,b{-a%b+a}

Try it online!

Answer 20

Prolog (SWI) + CLPFD, 30 bytes

f(R,X,Y):-R#=_*Y,R-Y#<X,R#>=X.

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f(R,X,Y)

f is a predicate taking in a variable R, and two integers X and Y.

The CLPFD constraints are expressed using the #-prefixed operators. , means and, so f is true iff all these constraints are satisfied, i.e.,

1) R#=_*Y R is a multiple of Y , and

2) R-Y#<X R-Y < X , and

3) R#>=X R >= X .

Answer 21

Charcoal, 10 bytes

≔±ＮθＩ×θ÷Ｎθ

Try it online! Link is to verbose version of code. Takes \$ b \$ first and \$ a \$ second. Explanation:

≔±Ｎθ

Input \$ b \$, negate it, and save the result.

Ｉ×θ÷Ｎθ

Input \$ a \$ and compute \$ -b \lfloor \frac a {-b} \rfloor \$, then cast to string for implicit print.

Answer 22

GolfScript, 6 bytes

.~)@%+

Try it online!

Function which consumes A and B from the stack, assuming B is below A.

Explanation:

           #Example input: A = 12, B = 5          stack is  5  12
.~)        #Duplicate A and negate the duplicate  stack is  5  12 -12
   @       #Pull 3rd from top to top              stack is 12 -12   5
    %      #Second to top mod top                 stack is 12 3
     +     #Add remaining values                  stack is 15

Answer 23

Javascript (ES6), 19 bytes

a=>b=>a%b?a-a%b+b:a

Naive attempt, pretty sure this can be golfed further.

Answer 24

Java, 99 bytes

static int r(int a,int b){int t=a%b<1||b%a<1?a:r(a+1,b);return t;}

Answer 25

JavaScript, 15 bytes

a=>b=>--a+b-a%b

More expressively, the answer is a => b => a + b-((a-1)%b) - 1, but by decrementing a we can handle both the "subtract one" steps at once, saving a couple of bytes. Then we just get lucky with operator presidencies.

Answer 26

Java (JDK), 15 bytes

b->a->--a+b-a%b

Try it online!

Answer 27

Perl 5, (111 bytes with footer) 52 bytes

sub f{($a,$b)=@_;$c=$b;$c++until!($c%$b)&&$c>=$a;$c}

Try it online!

Takes a and b as arguments to a subroutine named f and increases $c until it is both a multiple of $b and greater than or equal to $a. Then returns $c (the last value evaluated in a subroutine is the return value).

Answer 28

J, 6 bytes

]*>.@%

Try it online!

]         the right argument
 *        multiplied by
  >.@     the ceiling of
     %    the left argument dividided by the right one

If we can swap the order of the aruments:

J, 5 bytes

]+[|-

Try it online!

Answer 29

Pyth, 8 bytes

AQ+G%_GH

Try it online!

Simple port of the Python answer from xnor so please give that answer an upvote.

Debugged:

==================================================
assign('Q',eval_input())
assign('[G,H]',Q)
imp_print(plus(G,mod(neg(G),H)))
==================================================

Answer 30

Keg, -hr, 17 9 6 bytes

¿±¿%¿+

A port of my W answer which is a port of Xnor's python answer

Answer History

9 bytes -rR

&¿{:⑻$%|⑹

Explained

&¿{:⑻$%|⑹
&           #Store the implicit input (the number to round) in the register
 ¿          #Take the number to 'round by'
  {         #While:
   :⑻      #Duplicate the 'round by' and push the register without emptying it
     $%  #Swap the two and perform modulus, checking the result against 0
       |⑹  #    Increment the register

Answer History

17 bytes

&¿{:&:&$%|&1+&}&

Try it online!

Answer 31

APL+WIN, 8 bytes

Prompts for b followed by a:

b×⌈⎕÷b←⎕

Try it online! Courtesy of Dyalog Classic

Answer 32

MathGolf, 3 bytes

/ü*

Similar approach as other answers: \$\lceil\frac{a}{b}\rceil b\$.
I/O as floats with .0, with the inputs in the order a b.

Try it online.

Explanation:

/    # Divide the second (implicit) input by the first (implicit) input
     #  i.e. 3.0 5.0 → 1.666...
 ü   # Ceil it
     #  → 2
  *  # Multiply it by the first (implicit) input
     #  → 6.0
     # (output the entire stack joined together implicitly as result)

Answer 33

Runic Enchantments, 10 bytes

ii:},'cA*@

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ii            Read a, read b
  :}          Dup b, rotate to bottom of stack
    ,         Divide
     'cA      Ceiling
        *     Multiply
         @    Output and terminate

Answer 34

SimpleTemplate, 61 bytes

This was quite fun! Sadly, I had to do the dumb way :(

{@fn F A,B}{@whileA is notmultipleB}{@incA}{@/}{@returnA}{@/}

Creates a function that iterates until it finds a number that's multiple of the 2nd number.

This was actually way smaller than doing it without a function.

---

Ungolfed:

{@fn F numberA, numberB}
    {@while numberA is not multiple of numberB}
        {@inc by 1 numberA}
    {@/}
    {@return numberA}
{@/}

Should be mostly straightforward.

You can see it running on https://ideone.com/MR7t1E (golfed version only).

Answer 35

W, 7 4 bytes

_bm+

-3 bytes thanks to @A

Answer History

0a-bma+

Yet another port of xnor's Python answer.

Answer 36

Wolfram Language (Mathematica), 7 bytes

Ceiling

Just pointing out that Mathematica's builtin, which is most commonly called with one argument, actually does exactly the desired operation when called with two arguments.

Answer 37

Pyth, 5 bytes

+%_QE

Try it online!

Port of @xnor's answer

How it works

+%_QE
+    (Q)- The sum of the implicit first input and...
  _Q    - The negative of the first input...
 %  E   - Modulo the second input

Answer 38

TI BASIC, 8 bytes

-Bint(-A/B

"-" is the negative symbol here, distinct from the minus symbol on the calculator. TI BASIC doesn't have a token for a rounding up operation, we have to pick between iPart( and int( or take another route with fPart( or remainder( on OS 2.53+. int( rounds -2.5 down to -3 where iPart( would return -2, so we can use int( with negation to turn rounding down into rounding up.

Disp is implicit because the result is the final line, and the function can be a full program easily by adding Prompt A,B.