# 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?) – AShelly Sep 24 '15 at 20:47
• @AShelly Only when running – Beta Decay Sep 24 '15 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. – Timwi Sep 24 '15 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. – Geobits Sep 25 '15 at 0:50
• A "vanilla fizzbuzz" sounds delicious. – Reinstate Monica -- notmaynard Sep 25 '15 at 15:12

# Python 91 Bytes

for n in range(100):
c=""
if n%3==0:c+="fizz"
if n%5==0:c+="buzz"
if c=="":c=n
print c


# Ruby, 72 bytes

(1..100).each{|n|puts "#{n%3==0?'Fizz':''}#{n%5==0?'Buzz':n%3!=0?n:''}"}


Explanation

For each number from 1 to 100, print 'Fizz' if the number mod 3 is 0, then if the number mod 5 is 0, print 'Buzz' else if the number mod 3 is 0, print the number.

• Hello, and welcome to PPCG! Great first post! – NoOneIsHere Jun 3 '16 at 14:58

# Groovy, 61 bytes

100.times{a=++it%3?"":"fizz"
a+=it%5?"":"buzz"
println a?:it}


## Mathematica, 61 Bytes

Heavily inspired by this post on the mathematica stack exchange

fizzbuzz[#,,fizz,,buzz][[#~GCD~15~Mod~15]]&~Array~100//Colum‌​n

• This doesn't work properly for me on 10.2, but I assume that's a version variation as {#,,"fizz","fizzbuzz","buzz"}[[#~GCD~15~Mod~11]]&~Array~100//Column does – Mark S. Aug 4 '17 at 0:42

# Plain Javascript (no console.log() & no alert()), 64 bytes

for(f=b='';f++<100;b+=(f%5?z||f:z+'Buzz')+'\n')z=f%3?'':'Fizz';b


Just copy/paste into any javascript console and hit enter.

# Lua, 126 bytes

for i=1,100 do if i%15==0 then print"FizzBuzz"elseif i%3==0 then print"Fizz"elseif i%5==0 then print"Buzz"else print(i)end;end

• I don't know Lua too well, but can you remove a few spaces (100do and 0then twice) – Zacharý Aug 3 '17 at 20:36

# Java, 153 145 bytes

enum f{;static{for(int i=1;i<101;i++)System.out.println(i%3>0&i%5>0?i:(i%3<1&i%5<1?"FizzBuzz":(i%3<1?"Fizz":(i%5<1?"Buzz":0))));System.exit(0);}}


Not as long as I expected it to be, maybe because I ommited main function and used an enum. Quite overcomplicated print statement.

enum f {;                                       // Required because it is an enum
static {                                    // static block
for (int i = 1; i < 101; i++) {         // loop from 1 to 100 inclusive
boolean by3 = i % 3 == 0;
boolean by5 = i % 5 == 0;

if (!by3 && !by5)                   // Not divisible by 3 and 5
System.out.println(i);          // Just show the number
else if (by3 && by5)                // Divisible by both
System.out.println("FizzBuzz"); // Show FizzBuzz
else if (by3)                       // Divisible only by 3
System.out.println("Fizz");     // Show Fizz
else                                // Divisible only by 5
System.out.println("Buzz");     // Show Buzz
}
System.exit(0);                         // Required because there can't be output to stderr and it would crash trying to run main function
}
}


Ungolfed not modified:

enum f {;
static {
for (int i = 1; i < 101; i++)
System.out.println(i % 3 > 0 & i % 5 > 0 ? i : (i % 3 < 1 & i % 5 < 1 ? "FizzBuzz" : (i % 3 < 1 ? "Fizz" : (i % 5 < 1 ? "Buzz" : 0))));
System.exit(0);
}
}


EDIT: Saved 8 bytes by replacing ==0 with <1, !=0 with >0 and && with &

## K, 52 Bytes

-1@{,/$$[#i:&~.q.mod[x;3 5];FizzBuzz i;x]}'1+!100;  Thanks # Turtlèd, 221 bytes 99;,:[*,l]u[*d{*u,dr}ul' l[ l]ru]u3;[ [ r]l[ ' l]ll"Fizz";]<97:>5[ (*[ r]l[ ' l]ll)(zr)"Buzz";]<104:>[ lll'0rrr[ r]l{*' l{*l}{ l}(9'0l( r"1!")(9"10!"))(8'9)(7'8)(6'7)(5'6)(4'5)(3'4)(2'3)(1'2)(0'1)(!'0):{ l}}[ l]drrrr{zd}]  due to a bug (having two close curly brackets next to each other causes the second one to match to the first ones open bracket), the Try it online version is different for now (I put a space between the two close curly brackets) 99;,:[*,l]u[*d{*u,dr}ul' l[ l]ru]u3;[ [ r]l[ ' l]ll"Fizz";]<97:>5[ (*[ r]l[ ' l]ll)(zr)"Buzz";]<104:>[ lll'0rrr[ r]l{*' l{*l}{ l}(9'0l( r"1!")(9"10!"))(8'9)(7'8)(6'7)(5'6)(4'5)(3'4)(2'3)(1'2)(0'1)(!'0):{ l} }[ l]drrrr{zd}]  Try it online! I might write an explanation if I get time motivated # REXX 81 Bytes f.="" f.0="buzz" t.="" t.0="fizz" do i=1 to 100 f=i//5 t=i//3 say overlay(t.t||f.f,i) end  • You can golf away four characters by removing the empty strings: f.= instead of f.="". – idrougge May 16 '17 at 11:54 • I tested f.= on a couple of online interpreters and they didn't like it. – theblitz May 16 '17 at 12:06 • All right, I only tested it with Regina. – idrougge May 16 '17 at 12:11 # Ruby, 68 65 bytes q='Buzz';(1..100).map &->e{p e%3<1?'Fizz'+(e%5<1?q:''):e%5<1?q:e}  • If you've made your program one byte shorter, you'll want to change the header to 64 bytes :) – numbermaniac May 16 '17 at 21:05 • Why map &->e{ and not map{|e| ? – daniero Jun 5 '17 at 21:19 # braingasm, 40 bytes The language is coming along nicely. A few recent features allows for a quite decent FizzBuzz, if I may say so myself: 100[>#3p["Fizz".+]#5p["Buzz".+]z[#:]10.]  Here's how it works: 100[ One hundred times: > Go to the next cell. #3p[ If current cell number is divisble by 3: "Fizz". Print "Fizz". + Increment current cell ] #5p["Buzz".+] Same thing for 5 and "Buzz". z[ If the current cell is 0 (hasn't been incremented): #: Print current cell number ] 10. Print a newline ]  • Hmm nice language... Is there an online interpreter or is it all offline? – Beta Decay May 27 '17 at 16:29 • @BetaDecay Thanks :) No, nothing online yet. I'm focusing mostly on implementing all my ideas, and to document things once they're a little bit more stable. – daniero May 27 '17 at 16:41 # tcl, 72 time {puts [expr [incr i]%3?i%5?i:"":"Fizz"][expr i%5?"":"Buzz"]} 100  I think it is more golfable, to avoid the repetition of i%5 demo # MATLAB, 172 bytes s=char(arrayfun(@(n){num2str(n)},[1:100 1e7])');s(3:3:end,1:4)=repmat('Fizz',33,1);s(5:5:100,1:4)=repmat('Buzz',20,1);s(15:15:100,:)=repmat('FizzBuzz',6,1);disp(s(1:100,:))  • @TaylorScott Thanks for the edit – Luis Mendo Aug 20 '17 at 18:31 # Python 2.7, 59 Bytes for i in range(100):print i%3//2*'Fizz'+i%5//4*'Buzz'or i+1  Explanation: for i in range(100)  This will generate a list of numbers from 1 to 100 and assign it to the i variable. print i%3//2*'Fizz'+i%5//4*'Buzz'or i+1  Note that this loop will repeat 100 times, and in each time, the loop will do what this line is commanding. • "This will generate a list of numbers from 1 to 100" I think you meant 0 to 99, otherwise you wouldn't need to use i+1 :) – Aaron Aug 25 '17 at 15:46 • You don't need //; with integer arguments, / performs integer division in Python 2. -~i saves an additional byte. tio.run/##K6gsycjPM/r/Py2/… – Dennis Aug 25 '17 at 15:49 # AsciiDots, 152 bytes  # /"" \~\ @\-\ [=]@*-\ /@----*-~& |0/---*[>] |@\<>_"Buzz"@1) |*-~/ /-/ ||[%]-\>#100) v\-*#5*/ |.// | \<>_"Fizz"@1) |/-~/ @|[%]-\ 0\-*#3/ *-[+] \#1/  Try it online! Python 2, 57 bytes for i in range(100):print i%3/2*"Fizz"+i%5/4*"Buzz"or i+1  Not the most beautiful code but it works at least with python -c  Extremely late answer and for a language that has already a bit too many answers, however, i've not seen a single function answer that actually RETURNS the expected string, instead of printing it to console, so here it is: # Javascript 73 bytes (i=0,r="")=>i>99?r:f(++i,r+=(i%5?i%3?i:'Fizz':i%3?'Buzz':'FizzBuzz')+' ')  # Funky, 59 bytes for(i=1i<100i++)print((("Fizz"*1&!i%3)+("Buzz"*1&!i%5))ori)  Try it online! # Acc!!, 260 bytes Count i while 100-i { i%3/2*2 Count f while _/2 { Write 70 Write 105 Write 122 Write 122 1 } _+i%5/4*2 Count b while _/2 { Write 66 Write 117 Write 122 Write 122 1 } Count d while 1-_ { (i+1)/10 Count z while _ { Write 48+_ 0 } Write 48+(i+1)%10 1 } Write 10 }  Try it online! ### Explanation Once you know how Acc!! works, this is pretty straightforward. Count while loops increment their variable from 0 as long as their condition is true. Write takes an ASCII code and outputs the corresponding character. Bare expressions like i%3/2*2 are assigned to the accumulator, _. Division is integer division. Our code iterates i from 0 through 99 (thus, the number in question on each iteration is actually i+1). • If i%3 is 2 (i.e. i+1 is divisible by 3), we want to print Fizz. • Set the accumulator to i%3/2*2, which is 0 if i%3 is less than 2, and 2 otherwise. • Loop with condition _/2; we enter the loop iff the accumulator was set to 2 last step. • Inside the loop, write Fizz. • Set the accumulator to 1, breaking out of the f loop while keeping a record that we Fizzed. • If i%5 is 4 (i.e. i+1 is divisible by 5), we want to print Buzz. Same approach as above, except we now add i%5/4*2 to the existing accumulator value to preserve a possible 1 from the Fizz step. • If we printed Fizz and/or Buzz, the accumulator is now 1; otherwise, it is 0. If it is 0, we need to output the number. • Loop with condition 1-_: we enter the loop iff the accumulator is not 1. • Set accumulator to (i+1)/10, the 10's digit. • If that result is not 0, write the 10's digit, and set accumulator to 0 to break out of the z loop. • Write the 1's digit, (i+1)%10. • Set accumulator to 1 to break out of the d loop. • Finally, write a newline. # nodejs repl, 60 bytes for(i=0;++i<101;util.puts(i%5?s||i:s+'Buzz'))s=i%3?'':'Fizz'  With latest node, this prints a deprecation warning to the console. This presumably violates the rule about printing to STDERR, but using an older version of node will fix this. • Answers are not required to be in the latest version of their language, so this is still valid. – pppery Apr 9 at 2:22 # APL Nars 93 chars h;i;j i←0 L:j←0⋄i+←1⋄→A×⍳0≠3∣i⋄⍞←'Fiz'⋄j←1 A:→B×⍳0≠5∣i⋄⍞←'Buz'⋄j←1 B:→C×⍳j=0⋄→D C:⍞←i D:⎕←''⋄→L×⍳i≤99  Indented:  ∇ h;i;j i←0 L:j←0⋄i+←1⋄→A×⍳0≠3∣i⋄⍞←'Fiz'⋄j←1 A: →B×⍳0≠5∣i⋄⍞←'Buz'⋄j←1 B: →C×⍳j=0⋄→D C: ⍞←i D: ⎕←''⋄→L×⍳i≤99 ∇  # C, 152 bytes #include<stdio.h> #define p printf int i,f;main(){while(i++<100){if(i%3==0&&(f=1))p("Fizz");if(i%5==0&&(f=1))p("Buzz");if(f==0||(f=0))p("%d",i);p("\n");}}  Ungolfed one: #include<stdio.h> #define p printf int i,f; main() { while(i++<100) { if(i%3==0&&(f=1))p("Fizz"); if(i%5==0&&(f=1))p("Buzz"); if(f==0||(f=0))p("%d",i); p("\n"); } }  • This is not ok: main(){while(i++<100){(i%3<1&&p("fizz"))+(i%5<1&&p("buzz"))<1&&p("%d",i);p("\n");} ? – RosLuP Dec 7 '17 at 19:01 # C (gcc), 101 bytes #define P printf main(i){for(i=0;i++<100;P("\n"))(i%3<1&&P("fizz"))+(i%5<1&&P("Buzz"))<1&&P("%u",i);}  It could have one indefinite behavior in the use + operation, the order of evaluation could be swapped. Try it online! • You can change #define P printf to #define P printf( to cut off three bytes. – Steadybox Dec 8 '17 at 7:25 • 94 bytes – Steadybox Dec 8 '17 at 7:32 # ><>, 59 bytes 0v!oo:oo"Fiz"\ 1<oan?*/!?%5\!?%3;?)"c"::::::+ o"Buzz"/0oo  Try It Online Prints a trailing newline ### How it works 0v... .<... Initialises the stack with 0 ..... ..... 1<................;?)"c"::::::+ Increment the counter and duplicate it a looooot ..... End the program if the counter is larger than 99 (ascii value of c) 0..oo:oo"Fiz"\ .............\!?%3... Check if the counter is divisible by 3 and print Fizz if so ..... Also push a 0 to the stack .......o Swap the pushed 0 if it exists (otherwise it swaps two copies of the counter) ......./!?%5... Check if the counter is divisible by 5 and print Buzz if so o"Buzz"/0oo Also push a 0 to the stack .......... Multiply the top two values of the stack. ..oan?*... If the counter was divisible by 5 or 3 print the number .......... Print a newline and loop around again  As time goes, the stack fills up, with 2 copies of the counter for every Fizz and Buzz and 3 for each FizzBuzz. This is because of the extra copy of the counter that isn't printed and the extra 0s being pushed to the stack. # Kotlin, 115 bytes fun main(a:Array<String>){(1..100).map{println(when(it%15){0->"FizzBuzz";3,6,9,12->"Fizz";5,10->"Buzz";else->it})}}  Try it online! ## face, 152 bytes (FizzBuzz%d@ )\F*,c'Fo>m+*1+m=*m?*m1+m3+m4+m5+11334455BF"B4%B"%4N%"N3:~0?%=+3?=31?w=F4>:3%=+5?=51?w=B4>:5??+p="%c+w="=>:+w=N1>+++1*=55*==4+==1-=+=?=~  Annotated: (FizzBuzz%d@ ) \F*,c'Fo> ( obtain data pointer, a number, and stdout ) m+*1+ ( counter variable, goes from 1 to 100 ) m=*m?* ( result and temp variables ) m1+m3+m4+m5+11334455 ( relevant constants ) BF"B4%B"%4N%"N3 ( get pointers to Fizz, Buzz, %d, and \n ) :~ 0? ( has either Fizz or Buzz been printed? ) %=+3?=3 ( skip to label 3 if not divisible by 3 ) 1?w=F4>:3 ( otherwise, print "Fizz" ) %=+5?=51?w=B4>:5 ( same for "Buzz" ) ??+p="%c+w="=>:+ ( if we didn't print anything, print the number ) w=N1> ( print a newline ) +++1 ( increment the counter ) *=55*==4+==1-=+= ( set the result variable to 100-n ) ?=~  Try it online! (The trailing newline is required on TIO due to a bug that's been fixed in a newer version of face.) # J, 73 Bytes }.":('Fizz'"_)('Buzz'"_)('FizzBuzz'"_)@.((0:=3&|)++:@(0:=5&|))"0 i.101  Definitely not the greatest, looking for ways to improve this. ### Explanation:  "0 i.101 | To each number from 0 to 100 ((0:=3&|)++:@(0:=5&|)) | 1 if divisble by three, 2 if by 5, 3 if by both @. | Using the result of the previous verb, select a verb from the following gerund ":('Fizz'"_)('Buzz'"_)('FizzBuzz'"_) | Apply the appropriate Fizz-Buzz string }. | Remove the first entry (since we counted 0)  # 17, 229 bytes 17 was made after the challenge was made, but not designed to be any good at it. This code could probably be golfed a bit smaller, but it is a reasonably hard language to do anything with. Golfed version(removed unneeded space and duplicated numbers where useful): 0{2 # 1 + : 2 @ 5g == 1 + 0 @}1{2 # 3 % ! 3 * 2 # 5 % ! 5 * + 10 + 0 @}13{42  63  73 :   a  0 0 @}15{3f  6f  73 :   a  0 0 @}18{42  63  73 : : :   3f  6f    a  0 0 @}10{2 #$$ a $0 0 @}}777{0 2 @ 0 1 @ 0 0 @}  Ungolfed version: 0 { 2 # 1 + : 2 @ 5g == 1 + 0 @ } 1 { 2 # 3 % ! 3 * 2 # 5 % ! 5 * + 10 + 0 @ } 13 { 42$ 63 $73$ 73 $a$
0 0 @
}

15 {
3f $6f$ 73 $73$ a $0 0 @ } 18 { 42$ 63 $73$ 73 $3f$ 6f $73$ 73 $a$
0 0 @
}

10 {
2 #  a \$
0 0 @
}

777 {
0 2 @
0 1 @
0 0 @
}


Explanation:

Block 0: Load value at 2, add 1, duplicate, store at 2, if value == 100 push 1, else 0, add 1, store at 0.

Block 1: Will only be ran after block zero if value != 100. Load from 2, mod 3, not, times 3. If value at 2 is a multiple of 3 it will be 3, else 0. Same again for 5, add them and 17(looks like 10, but remember it uses base 17) and stores value at 0.

Block 13(or 20): Will only be run after 1 if value is a multiple of 3, not 5. Push and print ascii values for Fizz\n

Block 15(or 22): Will only be run after 1 if value is a multiple of 5, not 3. Push and print ascii values for Buzz\n

Block 18(or 25): Will only be run after 1 if value is a multiple of 3 and 5. Push and print ascii values for FizzBuzz\n

Block 10(or 17): Will only run after 1 if value is not multiple of 3 or 5. Prints numeric form of value.

Block 777(or 2149): Like main for C++, first block to be run

# kavod, 688 bytes

535 9}122 2}122 2}105 2}70 2}4 2}122 3}122 3}117 3}66 3}4 3}1 0><-+0>~0~><3 66 9}165.86?<1+>><-+2 0 86 9}478.><-+><5 100 9}165.120?<1+>><-+3 0 120 9}478.><-+<140?><-+><139 9}392.0><-+1+><10#100 158 9}258.162?47.9{.171 9}178.><-+9{.0 8}0 8}8}9}8{9{1+8{>8}9}8{9{>214 9}276.218?241.><-+<7}><<+8{1+8}8}7{196.><-+<1-+<-8{1-9{.-265 9}318.1 273 9}296.9{.-283 9}318.0 1-293 9}296.9{.-303 9}318.307?313.><-9{.1+9{.323?9{.><1+331?359.8}9}8{9{1-345?355.8}9}8{9{325.1+363.1-363.8}9}8{9{><-+9{.0 8}9}8{9{-9{.396?451.><0 406 9}276.457?><-+418 9}421.9{.425?445.10 433 9}178.439 9}421.48+#9{.~0~9{.48+#9{.><-+45#469 9}378.475 9}421.9{.8}~><8}489?499.1-8}>#8{485.><-+8{><8}513?526.1-<8}9}8{9{509.8{+8{~9{.


Try it online!

Formatted better:

535 9}122 2}122 2}105 2}70 2}4 2}122 3}122 3}117 3}66 3}4 3}1 0><-+0>~0~><3 66 9}165.86?<1+>>
<-+2 0 86 9}478.><-+><5 100 9}165.120?<1+>><-+3 0 120 9}478.><-+<140?><-+><139 9}392.0><-+1+>
<10#100 158 9}258.162?47.9{.171 9}178.><-+9{.0 8}0 8}8}9}8{9{1+8{>8}9}8{9{>214 9}276.218?241.
><-+<7}><<+8{1+8}8}7{196.><-+<1-+<-8{1-9{.-265 9}318.1 273 9}296.9{.-283 9}318.0 1-293 9}296.
9{.-303 9}318.307?313.><-9{.1+9{.323?9{.><1+331?359.8}9}8{9{1-345?355.8}9}8{9{325.1+363.1-363
.8}9}8{9{><-+9{.0 8}9}8{9{-9{.396?451.><0 406 9}276.457?><-+418 9}421.9{.425?445.10 433 9}178
.439 9}421.48+#9{.~0~9{.48+#9{.><-+45#469 9}378.475 9}421.9{.8}~><8}489?499.1-8}>#8{485.><-+8
{><8}513?526.1-<8}9}8{9{509.8{+8{~9{.


Ok so I can only really explain this vaguely, but stacks 2 and 3 hold the strings Fizz and Buzz. The program loops is roughly equivalent to the following code:

push 0 ;dummy

:fb
drop
copy 0
dup
call divides 3
jif not3
copy
drop
call puts 2 0

:not3
drop

dup
call divides 5
jif not5
copy
drop
call puts 3 0

:not5
drop

jif cont
drop
dup
call putn
push 0

:cont
drop
dup

putc 10

call gt 100
jnt fb

ret


...just translated into pseudo-ops codes. Yeah. Then the rest of the program is defining sanity functions.

J 9}N. usually means "call function at N, and return to J afterwards"; 9{. signifies this return.