Is it a leap year?

Question

This challenge is quite simple. You will take an input which will be a year from 1801 to 2400, and output if it is a leap year or not.

Your input will have no newlines or trailing spaces:

1954

You will output in any way that you like that clearly tells the user if it is or isn't a leap year (I will accept y or n for yes/no)

You can get a list of leap years here: http://kalender-365.de/leap-years.php I would note that leap years are not ever four years always. 1896 is a leap year, but 1900 is not. The years that follow this "skip" are:

1900
2100
2200
2300

Test cases:

1936 ->  y
1805 ->  n
1900 ->  n
2272 ->  y
2400 ->  y 

EDIT: This is based on a standard Gregorian calendar: http://www.epochconverter.com/date-and-time/daynumbers-by-year.php

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 = 50798; // 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>

Answer 1

APL, 16 14 12 characters

Returns 0 for a leap year, 1 for a non-leap year.

≥/⌽×4 25 4⊤⎕

Try this solution on tryapl.org. Note that I have changed the solution to the dfn {≥/⌽×4 25 4⊤⍵} as tryapl.com does not support ⎕ (take user input). Note that ⎕ is an empty box, not a missing character.

The same solution in J:

4 25 4>:/@|.@:*@#:]

Explanation

Dyadic ⊤ (encode) represents its right argument in the base specified by its left argument. I use base 4 25 4 in this solution. This represents the year y as a polynomial

y mod 400 = 100 a + 4 b + c where b < 100 and c < 4.

Let propositions α, β, and γ represent if a, b, and c are non-zero: Proposition γ is false if y is dividable by 4, β ∧ γ is false if y is dividable by 100 and α ∧ β ∧ γ is false if y is dividable by 400.

A truth table (* representing “don't care”) were proposition Δ represents if y is a leap-year obtains:

α β γ | Δ
0 0 0 | 1
1 0 0 | 0
* 1 0 | 1
* * 1 | 0

The following statement expresses Δ in α, β, and γ:

Δ = ¬((α → β) → γ)).

Due to the structure of this statement, one can express ¬Δ as the reduction ≥/⌽α β γ where ≥ implements ←. This leads to the answer I am explaining right now.

Answer 2

Javascript (ES6), 21 characters

The standard rule is that y is a leap year if 4 divides y and if either 100 doesn't divide y or 400 does divide y. In code,

y%4 == 0 && (y%100 != 0 || y%400 == 0)

There's no need for that 100 and 400. Instead it suffices to check whether 16 or 4 divides y, with 16 chosen if 25 divides y, 4 otherwise. Golfed, this becomes

!(y%(y%25?4:16))

A javascript function that implements this is 21 characters long:

l=y=>!(y%(y%25?4:16))

Perl, 28 26 characters

Same idea, but in perl.

$_=$_%($_%25?4:16)?"n":"y"

Run using the -lp options. For example,

perl -lpe '$_=$_%($_%25?4:16)?"n":"y"'

With the test set as input, this produces

1936
y
1805
n
1900
n
2272
y
2400
y

Answer 3

Pyth, 11 bytes

!%|F_jQ*TT4

This full program read from STDIN and prints True for leap years and False otherwise.

Thanks to @Jakube for suggesting Pyth and basically porting my CJam code.

Verify the test cases yourself in the Pyth Compiler/Executor.

How it works

     jQ*TT   Returns the evaluated input in base 10 × 10.
  |F_        Swaps the digit order and reduces using logical OR.
             So far, we've achieved 1954 -> [19, 54] -> 54 || 19.
!%        4  Returns the logical NOT of the result modulo 4.
             This prints True for multiples of 4 and False otherwise.

Answer 4

CJam, 12 bytes

rS+2m<~e|4%!

This full program read from STDIN and prints 1 for leap years and 0 otherwise.

Verify the test cases yourself in the CJam interpreter.

How it works

r   e# Read from STDIN.
S+  e# Append a space.
2m< e# Rotate two characters to the left.
~   e# Evaluate.
    e# So far, we achieved "1954" -> "54 19" -> 54 19.
e|  e# Logical OR; keep the leftmost non-zero integer.
4%! e# Logical NOT of the kept integer modulo 4.
    e# This pushes 1 for multiples of 4 and 0 otherwise.

Answer 5

TI-BASIC, 20 17 16 13

Because it is tokenized, TI-BASIC is often competitive at simple math challenges, but not this one since there is no "divisible" command. Maybe it is after all, but this is still longer than CJam and Pyth.

This uses David Hammond's method.

not(fPart(Ans/4/4^not(fPart(sub(Ans

Old code at 16 bytes:

not(fPart(Ans/16not(fPart(sub(Ansnot(fPart(Ans/4

Ungolfed:

not(fPart(Ans/16) and not(fPart(Ans/100) and not(fPart(Ans/4))))

fPart( is "fractional part"; exponentiation has higher precedence than division. In TI-BASIC, close-parens are optional.

I use undocumented behavior of the sub( command, usually used to get a substring: when its argument is a number instead of a string, it divides the number by 100. It will work on a TI-83 or 84 series calculator.

20 -> 17 by rearranging code to allow removal of close-parens; 17 -> 16 by replacing 400 with 16; 16 -> 13 by using David Hammond's idea.

Answer 6

Pyth, 19 15 14 bytes

xFm!%Q^d2[2TyT

Way too easy. Try it online: Demonstration or Test harness

edit: Missed, that you can print Truthy/Falsy values, instead of n/y. -4 byte

edit 2: Used the square root idea of Martin. -1 byte

Explanation

                 implicit: Q = input number
         [         generate a list with the numbers:
          2          2
           T         10
            yT       2*10 = 20
  m              map each number d to:
   !%Q^d2          not(Q mod d^2) // True if Q % d^2 == 0 otherwise False
xF               fold by xor

Answer 7

Regex, 83 62 38

Thanks to Toby for tips about combining both halves of the regex.

If we focus on 1801..2400 range only and assume input are integers:

(?!00)([02468][048]|[13579][26])(00)?$

Test in Ruby (^ = \A and $ = \Z because Ruby) for the desired range:

r = /(?!00)([02468][048]|[13579][26])(00)?\Z/
(1801..2401).each do |year|
  leap = year % 4 == 0 && ((year % 100 != 0) || (year % 400 == 0))
  leap_regex = !year.to_s[r].nil?
  if leap != leap_regex
    print 'Assertion broken:', year, " ", leap, " ", leap_regex, "\n"
  end
end

---

(Bonus) for something that should work not only for 1801..2400, but for any non-negative year:

^\d*(0000|(?!00)([13579][26]|(^|[02468])[048])(00)?)$

Test in Ruby (^ = \A and $ = \Z because Ruby) for first 100000 years:

r = /\A\d*(0000|(?!00)([13579][26]|(\A|[02468])[048])(00)?)\Z/
100000.times do |year|
  leap = year % 4 == 0 && ((year % 100 != 0) || (year % 400 == 0))
  leap_regex = !year.to_s[r].nil?
  if leap != leap_regex
    print 'Assertion broken:', year, " ", leap, " ", leap_regex, "\n"
  end
end

Answer 8

Haskell, 19 bytes

f i=gcd 80i>gcd 50i

Try it online!

Answer 9

Stackylogic, 226 bytes

Yes, that is right. I made a program in Stackylogic (non-TC), which was invented by Helka Homba, for the challenge found here.

Stackylogic has only binary input, so 10 (or more, any more digits will be ignored) bit binary must be used (least significant bit inputted first). Any dates outside the specified range might fail, as it simply checks what the inputted number is: it doesn't cover unnecessary dates

Not only is this my first challenge with stackylogic, but the first challenge with stackylogic at all.

Get ready for this mess:

1
0
1?
010
1?0
010
1?10
?1010
001010
?1010
?1010
?010
?10
?0
0
?
110
?10
11010
?010
11010
?1010
001010
?1010
?1010
?1010
?1010
?010
?0
110
?10
11010
?010
1010
01010
01010
?010
?0
110
?0
110
?0
110
1?0
?10
0?10
?10
?0
01
?
?<
0

This took me so long to make, because Stackylogic is the most confusing language I have encountered, and extremely unreadable: you have to know how the rest of the program has executed before you can read the current section being edited. I even had to add spaces for readability while creating it.

Meagre explanation

This is a simple explanation of what it does.

Stackylogic does not have any mathematical functions, so that made this harder. I had to hardcode most of it, to check if it was a specific number.

First, this program will do a NOR of the least significant bits, discarding them in the process. this means that if it is divisible by 4, it will proceed to the main part of the program, otherwise output 0.

Second, the pointer is carried over to the labyrinth of stackylogic, from here, if the next two bits are zero, it will instantly output 1 (as then it is divisible by 16, and so a leap year despite any other conditions), other wise it will check if it is not any of the numbers that is divisible by 4 but not a leap year, between 1801 and 2400.

To explain in detail, would involve making this post many times longer than it already is

Answer 10

Pip, 13 bytes

This one was more interesting than it at first appeared. It took some finagling, but I was finally able to replace those lengthy references to 400 with 4 and the h variable (=100).

!(a%h?aa/h)%4

Outputs 1 for leap year, 0 for non-leap year. Explanation:

               a is command-line argument (implicit)
  a%h?aa/h     If a is divisible by 100, divide it by 100; otherwise, leave it alone
 (        )%4  The result mod 4 is 0 if it's a leap year, nonzero otherwise
!              Negate and (implicitly) print

Answer 11

JavaScript (ES6) 27

The rule: (y%4==0) && (y%100!=0 || y%400==0)

Golfed: !(y%100<1&&y%400||y%4) (mainly using De Morgans's law)

A function implementing the rule:

l=y=>!(y%100<1&&y%400||y%4)

A test (run in Firefox) just to be sure:

l=y=>!(y%100<1&&y%400||y%4)

for(o=[],i=0;i<700;i++)
  y=i+1800,
  x=l(y),
  o[i/100|0]=(o[i/100|0]||'')+y+(x?' <b>Y</b>':' <i>N</i>')+'\n'
    
R.innerHTML='<td>'+o.join('</td><td>')+'</td>'
console.log(o[1])

td { white-space: pre; font-family: monospace; padding: 8px}

b { color: red; }
i { color: blue; }

<table>
  <tr id=R></tr>
</table>

Answer 12

IBM System Z assembler - 56 bytes.

(96 bytes of source. Previously 712 384 202 bytes of source, 168 byte executable).

Smaller version still. No longer saves caller's registers, changes to literal storage, changed addressing mode.

 l        CSECT      
         using l,15 
         l  5,y     
         n 5,f      
         bnz r      
         xr 4,4     
         l 5,y      
         d 4,c      
         ch 4,i     
         bne i      
         n 5,f      
         bnz r      
i        dc h'0'    
r        b  *       
y        dc f'2004' 
f        dc f'3'    
c        dc f'100'  
         end 

New version. This will ABEND with a S0C1 if it's a leap year, and loop if it isn't. Hopefully that fulfils the requirement of indicating the result.

l        CSECT             
         ASMDREG           
         SYSSTATE archlvl=2
         IEABRCX  DEFINE   
         save  (14,12)     
         larl  r9,s        
         using s,r9        
         st 13,w+4         
         la 13,w           
         st 13,w+8         
         la 5,2004         
         st 5,y            
         n 5,=f'3'         
         bnz r             
         xr r4,r4          
         l 5,y             
         d r4,=f'100'      
         ch r4,=h'0'       
         bne i             
         n 5,=f'3'         
         bnz r             
i        dc h'0'           
r        b  0              
s        dc 0d'0'          
y        ds f              
w        ds 18f            
         ltorg             
         end  

OK, so not the shortest (although it might be once we look at the actual executed code plus the interpreter size...)

leapyear CSECT                                                
         ASMDREG                                              
         SYSSTATE archlvl=2                                   
         IEABRCX  DEFINE                                      

         save  (14,12)                                        

         larl  r9,staticArea                                  
         using staticArea,r9                                  
         st r13,w_savea+4       .Save callers savearea        
         la r13,w_savea         .Address my savearea          
         st r13,w_savea+8         . and save it               

         open  (O,OUTPUT)             .open file              

         la r5,1936             .r5 = input year              
         st r5,years            .Save year                    

         cvd r5,double          .Convert year to p-decimal    
         mvc edarea,=xl8'4020202020202120' .Move in edit mask 
         ed edarea,double+4      .Make packed decimal year printable                              
         mvc outrec(4),edarea+4  .Move year string to output area                             
         bas r10,isitleap       .Call leap year routine       

         close (O)              .Close files            
         b return               .Branch to finish

isitleap ds 0h                                                      
         mvi outrec+5,c'N'      .Set default value                                   
         n r5,=f'3'             .Are last 2 bits 0 (Divisible by 4)?
         bnz notleap            .No - not leap                      
         xr r4,r4               .Clear R4                           
         l r5,years             .Reload r5 with year                
         d r4,=f'100'           .divide r4/r5 pair by 100           
         ch r4,=h'0'            .Remainder 0?                       
         bne isleap             .No - leap year                     
         n r5,=f'3'             .Quotient divisible by 4?           
         bnz notleap            .No - not leap                      

isleap   ds    0h                                                   
         mvi outrec+5,c'Y'      .Move in leap year indicator                                    

notleap  ds    0h                                                   
         put O,outrec           .Print output record                                    
         br r10                 .Return to caller                   

* Program termination                                               
return   ds 0h                                                      
         l r13,w_savea+4         .Restore callers savearea          
         return (14,12),,rc=0    .Restore registers and return    
* storage areas                                                     
staticarea  dc 0d'0'                                                
outrec      ds cl10                                                 
years       ds f                                                    
w_savea     ds 18f                save area                         
edarea      ds cl8                    .edit area                    
double      ds d                                                    
* Macros and literals                                               
         print nogen                                                
O        dcb   recfm=F,lrecl=6,dsorg=PS,ddname=O,macrf=PM           
         print gen                                                  
*                                                                   
         ltorg                         literal storage              
         end  

Output:

ABEND S0C1 for a leap year, S222 (when the CPU time has run out) if not.

1936 Y 1805 N 1900 N 2272 Y 2400 Y

(when run multiple times)

Answer 13

CJam, 18 16 bytes

q~[YAK]f{2#%!}:^

Gives 1 (truthy) for leap years and 0 (falsy) otherwise.

Run all test cases here.

Explanation

q~                 e# Read and eval input.
  [YAK]            e# Push an array containing 2, 10, 20 (the square roots of the
                   e# relevant divisors).
       f{    }     e# Map this block onto that array, also passing in the input year.
         2#        e# Square the divisor.
           %!      e# Modulo followed by logical negation. Gives 1 if the year is divisible
                   e# by the given divisor and 0 otherwise.
                   e# At this point we have one of the following arrays:
                   e#   [0 0 0] - not a leap year
                   e#   [1 0 0] - a leap year
                   e#   [1 1 0] - not a leap year
                   e#   [1 1 1] - a leap year
              :^   e# Reduce XOR onto this array, which gives 1 if there is an odd number
                   e# of 1s and 0 if there's an even number.

Answer 14

R, 29

!(Y=scan())%%4&Y%%100|!Y%%400

Test run

> !(Y=scan())%%4&Y%%100|!Y%%400
1: 1936
2: 1805
3: 1900
4: 2272
5: 2400
6: 2200
7: 
Read 6 items
[1]  TRUE FALSE FALSE  TRUE  TRUE FALSE

Answer 15

Mathematica, 40 27 bytes, 17 chars

#∣4∧(#∣100<U+F523>#∣400)

Uses 17 chars, but 27 bytes. Thanks to @alephalpha for the tip. Note that the vertical bars are actually U+2223 for divides. The <U+F523> should be replaced with the corresponding character.

Answer 16

Python2 - 37 bytes

g=lambda x:(x%4or x%400and x%100<1)<1

Note that if a is a nonnegative integer, then a<1 is a short way of writing not bool(a). The last <1 thus effectively converts the expression in the parentheses to a boolean and negates the result.

Applying the function g to an integer n between 1801 and 2400 will return True if n is a leap year, and False otherwise.

Answer 17

sed, 55

s/00$//
y/0123456789/yNnNyNnNyN/
/N.$/y/ny/yn/
s/.\B//g

• First line divides exact centuries by 100.
• Second line gives 'N' to odd digits, 'y' to 4s, and 'n' to non-4s.
• Third line swaps 'y' and 'n' if there's an odd penultimate digit (because 10 is 2 mod 4)
• Final line deletes all but the last character

Note that non-leap years may be printed as n or N depending on whether they are even or odd. I consider this a creative interpretation of the rule which allows alternatives to 'yes' and 'no' without specifying that they have to be consistent.

Answer 18

Julia, 30 28 bytes

y->(y%4<1&&y%100>0)||y%400<1

This creates an unnamed function that accepts an integer argument and returns a boolean value. To call it, give it a name, e.g. f=y->....

Ungolfed:

function f(y)
    (y % 4 == 0 && y % 100 != 0) || y % 400 == 0
end

Example:

julia> for y in [1936, 1805, 1900, 2272, 2400] println(f(y)) end
true
false
false
true
true
true

Answer 19

C, 81

I can do shorter, but this one neatly sticks to 'char' types, without parsing the argument (e.g. with atoi):

main(c,v)char**v;{char*p=*v+9;p-=2*(96==*p+p[1]);putchar("ynnn"[(2**p^p[1])&3]);}

It must be invoked with a name 4 characters long, because it makes the standard assumption that arguments immediately follow the program name, separated by NULs. Furthermore, it assumes that the single argument is encoded in ASCII and has no leading space.

Explanation:

main(c,v)
char**v;
{
    char *p = *v+9;
    if (p[0] + p[1] == '0'+'0')
        p -= 2;
    putchar("ynnn"[((*p << 1) ^ p[1])&3]);
}

*v+9 is the position of the 'tens' digit in v[1]+2.

If the 'tens' and 'units' characters add to 96, we end in 00, so back up two characters, so that 'tens' and 'units' point to the century number.

Now xor 'units' with twice the 'tens', mod 4. This works because 10==±2 mod 4, so the lower bit of the 'tens' can just toggle bit 1 of the 'units'. We use the result as an index into our remainders table, printing y only if the modular result is zero.

Answer 20

Jelly, 7 bytes

ọ4,25>/

Try it online!

How it works

ọ4,25>/  Main link. Argument: n (1801 - 2400)

ọ4,25    Test how many times n is divisible by 4 and 25.
     >/  Verify that the order of 4 is higher than the order of 25.

Answer 21

Befunge-98, (41 bytes)

&:4%#v_:aa*%#v_28*%|
"y",;<;@,"n";>;  ;#[

Simplicity is awesome.

Answer 22

Python, 37 bytes

def c(s):return s%16*(s%25<1)<(s%4<1)

Answer 23

C#, 23 bytes

y=>y%25<1?y%16<1:y%4<1;

Try it online!

Full source, including test cases:

using System;

namespace CountingLeapYears
{
    class Program
    {
        static void Main(string[] args)
        {
            Func<int,bool>s=y=>y%25<1?y%16<1:y%4<1;
            Console.WriteLine(s(1936)); //y
            Console.WriteLine(s(1805)); //n
            Console.WriteLine(s(1900)); //n
            Console.WriteLine(s(2272)); //y
            Console.WriteLine(s(2400)); //y
        }
    }
}

Answer 24

C, 37 34 30 bytes

l(y){y=y%(y%25?4:16)?110:121;}

Wandbox

Answer 25

Java 8, 49 45 22 20 bytes

n->n%(n%25<1?16:4)<1

-2 bytes thanks to @OlivierGrégoire.

Try it online.

Some 22 bytes solutions:

n->n%25<1?n%16<1:n%4<1

Try it online.

n->(n%25<1?n%16:n%4)<1

Try it online.

java.time.Year::isLeap

Try it online.

Explanation:

n->                // Method with integer parameter and boolean return-type
   n%          <1  //  Return whether the integer is divisible by:
     (n%25<1?      //   If the input is divisible by 25:
             16    //    Check if its divisible by 16
            :      //   Else:
             4)    //    Check if its divisible by 4 instead

Answer 26

KDB(Q), 27 bytes

{0=x mod(4 400)0=x mod 100}

Explanation

               0=x mod 100      / boolean of 100 divisibility
        (4 400)                 / 0b -> 4, 1b -> 400
 0=x mod                        / boolean of 4/400 divisibility
{                         }     / lambda

Test

q){0=x mod(4 400)0=x mod 100}1936 1805 1900 2272 2400
10011b

Answer 27

PHP - 45 bytes

$b=$argv[1]%400;echo !$b|!($b%4)&!!($b%100);

Nothing that special really, just abusing type-juggling.

Answer 28

T-SQL 37 22 bytes

Saved 15 bytes thanks to BradC's comment.

The usual hardcoded variable for lack of a stdin.

e.g.

DECLARE @ NVARCHAR(4) = '2016'

Then the solution is:

PRINT ISDATE(@+'0229')

Answer 29

05AB1E, 9 7 bytes

т‰0Kθ4Ö

Try it online or verify all test cases.

Explanation:

т‰         # Divmod the (implicit) input by 100
           #  i.e. 1900 → [19,00]
           #  i.e. 1936 → [19,36]
           #  i.e. 1991 → [19,91]
           #  i.e. 2000 → [20,00]
  0K       # Remove all 0s
           #  i.e. [19,00] → [19]
           #  i.e. [19,36] → [19,36]
           #  i.e. [19,91] → [19,91]
           #  i.e. [20,00] → [20]
    θ      # Pop and get the last item
           #  i.e. [19] → 19
           #  i.e. [19,36] → 36
           #  i.e. [19,91] → 91
           #  i.e. [20] → 20
     4Ö    # Check if it's divisible by 4 (and output the result implicitly)
           #  i.e. 19 → 0 (falsey)
           #  i.e. 36 → 1 (truthy)
           #  i.e. 91 → 0 (falsey)
           #  i.e. 20 → 1 (truthy)

Answer 30

Javascript ES6, 32, 29, 26

Any of following lines works:

f=y=>new Date(y,2,0).getDate()&1
g=y=>!(y&3)^!(y%100)>!(y%400)
h=y=>!(y&3|y%100<1&&y%400)