# 1, 2, Fizz, 4, Buzz

## Introduction

In our recent effort to collect catalogues of shortest solutions for standard programming exercises, here is PPCG's first ever vanilla FizzBuzz challenge. If you wish to see other catalogue challenges, there is "Hello World!" and "Is this number a prime?".

## Challenge

Write a program that prints the decimal numbers from 1 to 100 inclusive. But for multiples of three print “Fizz” instead of the number and for the multiples of five print “Buzz”. For numbers which are multiples of both three and five print “FizzBuzz”.

## Output

The output will be a list of numbers (and Fizzes, Buzzes and FizzBuzzes) separated by a newline (either \n or \r\n). A trailing newline is acceptable, but a leading newline is not. Apart from your choice of newline, the output should look exactly like this:

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
16
17
Fizz
19
Buzz
Fizz
22
23
Fizz
Buzz
26
Fizz
28
29
FizzBuzz
31
32
Fizz
34
Buzz
Fizz
37
38
Fizz
Buzz
41
Fizz
43
44
FizzBuzz
46
47
Fizz
49
Buzz
Fizz
52
53
Fizz
Buzz
56
Fizz
58
59
FizzBuzz
61
62
Fizz
64
Buzz
Fizz
67
68
Fizz
Buzz
71
Fizz
73
74
FizzBuzz
76
77
Fizz
79
Buzz
Fizz
82
83
Fizz
Buzz
86
Fizz
88
89
FizzBuzz
91
92
Fizz
94
Buzz
Fizz
97
98
Fizz
Buzz


The only exception to this rule is constant output of your language's interpreter that cannot be suppressed, such as a greeting, ANSI color codes or indentation.

## Further Rules

• This is not about finding the language with the shortest approach for playing FizzBuzz, this is about finding the shortest approach in every language. Therefore, no answer will be marked as accepted.

• Submissions are scored in bytes in an appropriate preexisting encoding, usually (but not necessarily) UTF-8. Some languages, like Folders, are a bit tricky to score--if in doubt, please ask on Meta.

• Nothing can be printed to STDERR.

• Feel free to use a language (or language version) even if it's newer than this challenge. If anyone wants to abuse this by creating a language where the empty program generates FizzBuzz output, then congrats for paving the way for a very boring answer.

Note that there must be an interpreter so the submission can be tested. It is allowed (and even encouraged) to write this interpreter yourself for a previously unimplemented language.

• If your language of choice is a trivial variant of another (potentially more popular) language which already has an answer (think BASIC or SQL dialects, Unix shells or trivial Brainfuck derivatives like Alphuck and ???), consider adding a note to the existing answer that the same or a very similar solution is also the shortest in the other language.

• Because the output is fixed, you may hardcode the output (but this may not be the shortest option).

• You may use preexisting solutions, as long as you credit the original author of the program.

• Standard loopholes are otherwise disallowed.

As a side note, please don't downvote boring (but valid) answers in languages where there is not much to golf; these are still useful to this question as it tries to compile a catalogue as complete as possible. However, do primarily upvote answers in languages where the authors actually had to put effort into golfing the code.

## Catalogue

var QUESTION_ID=58615;var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";var COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk";var OVERRIDE_USER=30525;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)}}
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:700}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="//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>

• Nothing can be printed to STDERR. Is this true only when running, or also when compiling (assuming that is a separate step?) Sep 24, 2015 at 20:47
• @AShelly Only when running Sep 24, 2015 at 20:48
• I’m not sure I like the fact that you hardcoded the 100 into the challenge. That way, a program that just generates the expected output is a valid entry, but is not interesting for this challenge. I think the challenge should expect the program to input the number of items to output. Sep 24, 2015 at 23:28
• @Timwi While I agree that it would make it (only slightly) more interesting, I've very often seen FizzBuzz as strictly 1 to 100 (on Wikipedia and Rosetta Code, for example). If the goal is to have a "canonical" FB challenge, it makes sense. Sep 25, 2015 at 0:50
• A "vanilla fizzbuzz" sounds delicious. Sep 25, 2015 at 15:12

1
9 10 11 12
13

# APL, 56 50 bytes

⊣{1+⍵⊣⎕←∊2↑((0=3 5|⍵)/'Fizz' 'Buzz'),''(⍕⍵)}⍣100⊢1


Note: The very first ⊣ should suppress output in GNU APL. Replace it with → to get correct results in Dyalog, etc, or with portable assignment to some variable X← adding one byte.

# A+, 51 bytes

(x←100)do↓⊃2↑((0=3 5|1+x)/4⊂'FizzBuzz'),⌽2↑<1↓⍕1+x;

• These seem quite different, so why not put them in different answers? Nov 9, 2015 at 7:47
• While I wouldn't call A+ a “trivial variant,” it's a direct derivative of APL and I think any solution in one would translate to similar byte count in another. Also, I just updated it making the answers work exactly the same way. Nov 9, 2015 at 9:22

# ShapeScript, 57 bytes

0'1+0?3%1<"Fizz"*1?5%1<"Buzz"*+"#0?"1?_1<*!@"
"@'554***!#


Try it online!

### How it works

0          Push 0 (accumulator).
'          Push a string that, when evaluated, does the following:
1+         Increment the accumulator.
0?         Push a copy.
3%1<       Check if the remainder of its division by 3 is zero.
"Fizz"*    Push "Fizz" for True, "" for False.
1?         Push another copy of the accumulator.
5%1<       Check if the remainder of its division by 5 is zero.
"Buzz"*    Push "Buzz" for True, "" for False.
+          Concatenate the potential fizzes and buzzes.
"          Push a string that, when evaluated, does the following:
#          Discard the topmost stack item.
0?         Push a copy of the item below it (accumulator).
"
1?         Push a copy of the concatenation.
_1<        Check if its length is zero.
*!         Execute "#0?" once for True, zero times for False.
@          Swap the generated output with the accumulator.
"          Push "\n".
"
@          Swap it with the accumulator.
'
554**      Push 5 * 5 * 4 = 100.
*!         Execute the '...' string 100 times.


# Javascript ES6, 77 chars

"\n".repeat(100).replace(/(?=\n)/g,(m,i)=>(++i%3?m:"Fizz")+(i%5?m:"Buzz")||i)


Test:

"\n".repeat(100).replace(/(?=\n)/g,(m,i)=>(++i%3?m:"Fizz")+(i%5?m:"Buzz")||i) == document.querySelector("pre").textContent

• Here's my ES6 attempt, with parts stolen from yours (89 chars): "0".repeat(99).split(0).map((m,i)=>{return(++i%3?m:"Fizz")+(i%5?m:"Buzz")||i}).join("\n") Jan 7, 2016 at 6:48
• @starbeamrainbowlabs, why do you use {return ...} instead of expression? Jan 7, 2016 at 10:21
• Because I couldn't get the expression to work :( Jan 7, 2016 at 16:20
• Save a byte by using template strings + literal newline at front. Jan 29, 2016 at 0:34

# Elixir, 182 bytes

import Stream
f=fn(n)->(zip(cycle(["","","fizz"]),cycle(["","","","","buzz"]))|>zip(iterate(1,&(&1 + 1)))|>map(fn{{"",""},n}->n
{{p,o},_}->p<>o end))|>take(n)|>Enum.each(&IO.puts/1)end


LiveDemo

Calling: f.(100)

# ///, 229 bytes

/x/Buzz//b/
x//f/
Fizz/1
2f
4bf
7
8fb
11f
13
14fx
16
17f
19bf
22
23fb
26f
28
29fx
31
32f
34bf
37
38fb
41f
43
44fx
46
47f
49bf
52
53fb
56f
58
59fi
61
62f
64bf
67
68fb
71f
73
74fx
76
77f
79bf
82
83fb
86f
88
89fx
91
92f
94bf
97
98fb

• Out-golfed May 15, 2017 at 2:58
• @Challenger5 Well done. May 15, 2017 at 2:59
• @LeakyNum I have A CJam script that finds common substrings in the input. It's useful for /// golfing. May 15, 2017 at 5:55

## JavaScript (using external library) (90 bytes)

x=>_.Range(1,100).Select(x=>x%3==0?(x%5==0?"FizzBuzz":"Fizz"):x%5==0?"Buzz":x).WriteLine()


Code explanation: Creates a range starting at 1 and extending 100 elements. Select maps the item according to the predicate. First check if divisible by 3, if so check if divisible by 5. If so, select "FizzBuzz", else select "Fizz". Otherwise,check if divisible by 5. If so, select "Buzz", else just select the number. WriteLine is built into the library, and it writes each element on its own line

# DUP, 76 bytes

[$$3/%[]['F,'i,'z,'z,]?\5/%[]['B,'u,'z,'z,]?*[.][]?10,]c:0[100<][1+c;!]# Explanation of program flow, code blocks: DUP has two default truth values: -1 (true, all bits set) and 0 (false, no bits set). For conditional evaluations every nonzero value is accepted as true, comparisons themselves lead to either -1 or 0. DUP has two conditional expressions, if-then-else and a while loop: [execute if not false][execute if false]?. E.g. 1['t,]['f,]? prints t to STDOUT, because 1 is not false. [condition][execute while not false]#. E.g. 5[0>][.1-]# prints the numbers 5,4,3,2,1 to STDOUT. The condition block tests the top stack value if it’s greater than 0, the execute block prints the number and decrements it. The condition that’s checked is always the topmost stack element, which gets popped off the stack during the check. [$$3/%${dup,dup,push 3, moddiv, pop, dup} [] {if condition true do nothing} ['F,'i,'z,'z,]? {if condition false print Fizz} \5/%$                               {swap, push 5, moddiv, pop, dup}
[]                                  {if condition true do nothing}
['B,'u,'z,'z,]?                   {if condition false print Buzz}
*                                   {mul}
[$.] {if condition true dup, print to STDOUT} []? {if condition false do nothing} 10, {push 10, print char to STDOUT (newline)} ]c: {define function c} 0 {push 0} [$100<]                                 {while dup < push 100 true}
[1+c;!]#                         {do push 1, add, execute function c}


Step by step:

instruction             stack
0                       0
[
$0,0 100 0,0,100 < 0,-1 ] 0 [ 1 0,1 + 1 increment counter c; 1,0 0=start address of function c ! execute c [$$3 1,1,1,3 / 1,1,1,0 mod,div operator (n mod 3, n div 3) % 1,1,1 /% → mod$                       1,1,1,1         result of n%3 used as flag for later check
[]                      1,1,1           condition true→do nothing
['F,'i,'z,'z,]?                         skipped
\                       1,1,1           swap top and 2nd stack elements
5                       1,1,1,5
/                       1,1,1,0         mod,div operator
%                       1,1,1           /% → mod
$1,1,1,1 n%5 flag [] 1,1,1 condition true → do nothing ['B,'u,'z,'z,]? skipped * 1,1 multiply %3 and %5 flags, result is only >0 if neither %3 nor %5 are 0. [ 1 if condition (flag)>0$                      1,1             dup
.]                    1               print number to STDOUT: '1'
[]?                                     skipped

back to while loop

[
$1,1 100 1,1,100 < 1,-1 ] 1 [ 1 1,1 + 2 increment counter c; 2,0 0=start address of function c ! execute c ...  A full description of all operators and program flow can be found at my GitHub repository, together wtih my Julia interpreter of DUP. An online Javascript implementation showing useful debug information can be found here. A rather terse introduction can be found on the esolangs.org DUP page. # S.I.L.O.S, 141 bytes lbla a+1 c=1-(a%3)*(a%5) if c b printIntNoLine a lblb c=(a%3) if c d print Fizz lbld c=(a%5) if c c print Buzz lblc printLine b=100-a if b a  Try it online! Unfortunately SILOS is incredibly verbose, and an interpreter bug cost me a few extra bytes, but I think SILOS is "superior" only to scratch in this challenge. # Fourier, 81 bytes |Fizz|F|Buzz|B1~i100(i~X%15{0}{1~XFB}X%5{0}{1~XB}X%3{0}{1~XF}X{i}{Xo}10ai^~i)  Try it on FourIDE Who knew such a simple task would yield such a mammoth of a program! Pseudocode of the program: Function F { Print "Fizz" } Function B { Print "Buzz" } i = 1 While i != 100 X = i If X % 15 == 0 X = 1 Function F Function B End if If X % 5 == 0 X = 1 Function B End If If X % 3 == 0 X = 1 Function F End If If X == i Print X End if Print "\n" i += 1 End while  • you say function a in the ungolfed, do you mean function F? May 15, 2017 at 23:39 • also I don't think you get to call it mammoth if it fits on screen in one line, and the challenge actually isn't that simple May 15, 2017 at 23:40 ## ///, 209 208 bytes /I/ EFizz //H/FizzE//G/ Fizz //E/Buzz /1 2G4I7 8GE11G13 14 H16 17G19I22 23GE26G28 29 H31 32G34I37 38GE41G43 44 H46 47G49I52 53GE56G58 59 H61 62G64I67 68GE71G73 74 H76 77G79I82 83GE86G88 89 H91 92G94I97 98GBuA  • is it just me or is some of this redundant? there isn't much point in shortening Buzz to BuC if the bytes saved by that in general are less than the bytes added May 15, 2017 at 4:34 # Micro, 72 bytes :i{i1+:i"":d i3/i3%=if("Fizz":d,) i5/i5%=if(d"Buzz"+:d,) d:\i100=if(,a)}  # Jq 1.5, 90 bytes def f:if.%15<1then"FizzBuzz"elif.%5<1then"Buzz"elif.%3<1then"Fizz"else. end;range(100)+1|f  Try it online! # Modula-2, 222 bytes MODULE F;IMPORT InOut;VAR i:INTEGER;BEGIN FOR i:=1 TO 100 DO CASE i*i*i*i MOD 15 OF 0:InOut.WriteString("FizzBuzz")|1:InOut.WriteInt(i,0)|6:InOut.WriteString("Fizz")|10:InOut.WriteString("Buzz")END;InOut.WriteLn END END F.  Tested with Amsterdam Compiler Kit. The program needs 4-byte INTEGER. If using 2-byte INTEGER, i*i*i*i overflows and the program fails with a case error. With indentation: MODULE F; IMPORT InOut; VAR i: INTEGER; BEGIN FOR i := 1 TO 100 DO CASE i * i * i * i MOD 15 OF 0: InOut.WriteString("FizzBuzz") | 1: InOut.WriteInt(i, 0) | 6: InOut.WriteString("Fizz") | 10: InOut.WriteString("Buzz") END; InOut.WriteLn END END F.  ACK compiles the PIM3 dialect of Modula-2 and comes with an InOut module. Other compilers for Modula-2 or Oberon might have shorter syntax or come with shorter modules. • I don't like having 3 calls to InOut.WriteString, but adding a PROCEDURE p(s:ARRAY OF CHAR);BEGIN InOut.WriteString(s)END p; would cost 61 bytes and save only 48 bytes in the calls. • I would write FOR i:=1TO 100DO but ACK rejects it as a syntax error. • I had used i*i*i*i MOD 15 in my AppleScript answer. # MY-BASIC, 116 bytes A response. For i=1 To 100 S="" If i Mod 3=0 Then S="Fizz" If i Mod 5=0 Then S=S+"Buzz" If S="" Then Print i; Else Print S; Next  Try it online! # Ada (GNAT), 196 bytes procedure GNAT.IO.F is begin for I in Integer range 1..100 loop Put((if I mod 3=0 then"Fizz"else"")&(if I mod 5=0 then"Buzz"else""));if I mod 3*I mod 5/=0 then Put(I);end if;New_Line;end loop;end;  Try it online! 189 bytes if extraneous whitespace is allowed: procedure GNAT.IO.F is begin for I in Integer range 1..100 loop Put(if I mod 3=0 then(if I mod 5=0 then"FizzBuzz"else"Fizz")else(if I mod 5=0 then"Buzz"else I'Image));New_Line;end loop;end;  Try it online! # 05AB1E, 25 bytes тLv”FizzÒÖ”#y3 5‚ÖÏJy‚éθ,  Try it online! I figured I'd add a fourth 05AB1E answer to this challenge. This uses my favourite "vectorise is-divisible and then repeat the fizbuzz list accordingly" approach. ## Explained тLv”FizzÒÖ”#y3 5‚ÖÏJy‚éθ, тL # The range 1 .. 100 v # For each N in that range: ”FizzÒÖ”# # Push the list ["Fizz", "Buzz"] y3 5‚Ö # Push N % [3, 5] == 0 (vectorised) Ï # "Multiply" the two lists as you would in Vyxal J # And concatenate that into a single string y‚ # And pair that with N é # Sorting that by length, so that the value we want is always last θ, # And print it with a newline  # D, 115 bytes import std;void main(){string x;for(int i;i++<100;x=((i%3?"":"Fizz")~(i%5?"":"Buzz")),writeln(x?x:i.to!string)){};}  Try it online! ### Adapted from this code: import std;void main(){auto fizzbuzz(in uint i){string r;if(i%3==0)r~="Fizz";if(i%5==0)r~="Buzz";if(r.length==0)r~=i.to!string;return r;}enum r=1.iota(101).map!fizzbuzz;r.each!writeln;}  # Fortran (GFortran), 187 bytes DO I=1,100 IF(MOD(I,3)==0.AND.MOD(I,5)==0)THEN;PRINT'(A)','FizzBuzz' ELSEIF(MOD(I,3)==0)THEN;PRINT'(A)','Fizz' ELSEIF(MOD(I,5)==0)THEN;PRINT'(A)','Buzz' ELSE;PRINT'(I0)',I ENDIF;ENDDO;END  Try it online! # JavaScript (code.golf), 62 56 bytes for(i=0;++i<101;print(i%5?f||i:f+'Buzz'))f=i%3?'':'Fizz'  This is a code.golf version, which allows usage of print instead console.log. ### Here is the un-code.golfed version: for(i=0;++i<101;console.log(i%5?f||i:f+'Buzz'))f=i%3?'':'Fizz'  # Nim, 72 bytes for i in 1..100:echo max(["Fizz",""][i*i%%3]&["Buzz",""][ord i%%5>0],$i)


Try it online!

# Pascal (FPC), 180 bytes

program f(o);var i:integer;begin for i:=1to 100do if i mod 15=0then writeln('FizzBuzz')else if i mod 3=0then writeln('Fizz')else if i mod 5=0then writeln('Buzz')else writeln(i)end.


Try it online!

# Morse Code, 1777 bytes

.---- / ..--- / ..-. .. --.. --.. / ....- / -... ..- --.. --.. / ..-. .. --.. --.. / --... / ---.. / ..-. .. --.. --.. / -... ..- --.. --.. / .---- .---- / ..-. .. --.. --.. / .---- ...-- / .---- ....- / ..-. .. --.. --.. -... ..- --.. --.. / .---- -.... / .---- --... / ..-. .. --.. --.. / .---- ----. / -... ..- --.. --.. / ..-. .. --.. --.. / ..--- ..--- / ..--- ...-- / ..-. .. --.. --.. / -... ..- --.. --.. / ..--- -.... / ..-. .. --.. --.. / ..--- ---.. / ..--- ----. / ..-. .. --.. --.. -... ..- --.. --.. / ...-- .---- / ...-- ..--- / ..-. .. --.. --.. / ...-- ....- / -... ..- --.. --.. / ..-. .. --.. --.. / ...-- --... / ...-- ---.. / ..-. .. --.. --.. / -... ..- --.. --.. / ....- .---- / ..-. .. --.. --.. / ....- ...-- / ....- ....- / ..-. .. --.. --.. -... ..- --.. --.. / ....- -.... / ....- --... / ..-. .. --.. --.. / ....- ----. / -... ..- --.. --.. / ..-. .. --.. --.. / ..... ..--- / ..... ...-- / ..-. .. --.. --.. / -... ..- --.. --.. / ..... -.... / ..-. .. --.. --.. / ..... ---.. / ..... ----. / ..-. .. --.. --.. -... ..- --.. --.. / -.... .---- / -.... ..--- / ..-. .. --.. --.. / -.... ....- / -... ..- --.. --.. / ..-. .. --.. --.. / -.... --... / -.... ---.. / ..-. .. --.. --.. / -... ..- --.. --.. / --... .---- / ..-. .. --.. --.. / --... ...-- / --... ....- / ..-. .. --.. --.. -... ..- --.. --.. / --... -.... / --... --... / ..-. .. --.. --.. / --... ----. / -... ..- --.. --.. / ..-. .. --.. --.. / ---.. ..--- / ---.. ...-- / ..-. .. --.. --.. / -... ..- --.. --.. / ---.. -.... / ..-. .. --.. --.. / ---.. ---.. / ---.. ----. / ..-. .. --.. --.. -... ..- --.. --.. / ----. .---- / ----. ..--- / ..-. .. --.. --.. / ----. ....- / -... ..- --.. --.. / ..-. .. --.. --.. / ----. --... / ----. ---.. / ..-. .. --.. --.. / -... ..- --.. --..


# Prolog (SWI), 225229 165 bytes

f(N,'FizzBuzz'):-N mod 15<1. f(N,'Fizz'):-N mod 3<1,N mod 5>0. f(N,'Buzz'):-N mod 3>0,N mod 5<1. f(N,N):-N mod 3>0,N mod 5>0. :-between(1,100,X),f(X,Y),\+writeln(Y).


Try it online!

###### Thanks to @JoKing (redundant init and h), I lost 64 bytes!

:-initialization m. f(N,'FizzBuzz'):-N mod 3=:=0,N mod 5=:=0. f(N,'Fizz'):-N mod 3=:=0,N mod 5=\=0. f(N,'Buzz'):-N mod 3=\=0,N mod 5=:=0. f(N,N):-N mod 3=\=0,N mod 5=\=0. m:-between(1,100,X),f(X,Y),write(Y),nl,fail,h(0). m:-h(1).

• What is the point of the h calls? also the initialization part? Some simple golfing for 165 bytes (though it can be half the size)
– Jo King
May 3 at 6:40
• @JoKing Sorry, I'm not that good at Prolog. I got really into it one day but then lost interest in it. Thanks for the tip! May 3 at 16:16

# Raku, 44 bytes

say 'Fizz'x$_%%3~'Buzz'x$_%%5||\$_ for 1..100


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# V (vlang.io), 137 bytes

fn main(){for i in 1..101{if i%15==0{println('FizzBuzz')}else if i%3==0{println('Fizz')}else if i%5==0{println('Buzz')}else{println(i)}}}


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#### Ungolfed code:

fn main () {
mut i := 1
for i < 101 {
match true {
i % 15 == 0 {
println('FizzBuzz')
} i % 3 == 0 {
println('Fizz')
} i % 5 == 0 {
println('Buzz')
} else {
println(i)
}
}
i++
}
}


# Fig, $$\31\log_{256}(96)\approx\$$ 25.517 bytes

jcnnna100 5'"Buzz"3'FOa9+"Fizz"


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Hmm yes irrational bytes. "It'll be shorter" they said. "It's the future of golf".

8.5 bytes longer than legitimate vybuzz, 5.5 more than Jelly, 15.5 more than the optimal vybuzz/Fact fizzbuzz.

## Explained

jcnnna100 5'"Buzz"3'FOa9+"Fizz"
na100 5'"Buzz"               # Every 5th item of the range [1, 100] replaced with "Buzz"
n              3'             # To every 3rd item of that:
+"Fizz"  #   Prepend the word "Fizz"
FOa9         #   And remove all digits from it. This should just be FcD (because cD is allegedly all digits), but turns out that cD is [A-z0-9], which is allegedly what cN is supposed to be, but cN errors out. Thanks Seggan very cool.
jcn                              # Join all that on newlines

• Not amused. After fixing about 8 bugs I got the n solution one char shorter smh Sep 8 at 2:29
• this prints BuzzFizz Sep 8 at 3:30
• @Seggan well the jokes on you because I got it 3 characters shorter Sep 8 at 3:53
• Golfing in somebody else's language as a means of bragging about your own language is in very poor taste, IMO. Sep 8 at 4:15
• @DLosc maybe, but in this case, it's more along the lines of friendly banter Sep 8 at 4:17
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