623
\$\begingroup\$

Note to challenge writers as per meta consensus: This question was well-received when it was posted, but challenges like this, asking answerers to Do X without using Y are likely to be poorly received. Try using the sandbox to get feedback on if you want to post a similar challenge.


It's 2017 2018 2019 2020 2021 2022 2023 2024 already, folks, go home.

Woo, 10 years of this challenge!

So, now that it's 2014, it's time for a code question involving the number 2014.

Your task is to make a program that prints the number 2014, without using any of the characters 0123456789 in your code, and independently of any external variables such as the date or time or a random seed.

The shortest code (counting in bytes) to do so in any language in which numbers are valid tokens wins.


Leaderboard:

var QUESTION_ID=17005,OVERRIDE_USER=7110;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="<h1>"+e.body.replace(OVERRIDE_REG,"")+"</h1>")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/<a/.test(o)&&(o=jQuery(o).text()),s[o]=s[o]||{lang:e.language,user:e.user,size:e.size,link:e.link}});var t=[];for(var o in s)s.hasOwnProperty(o)&&t.push(s[o]);t.sort(function(e,s){return e.lang>s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;
body{text-align:left!important}#answer-list,#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="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><div id="language-list"> <h2>Winners 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><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table>

\$\endgroup\$
16
  • 24
    \$\begingroup\$ Even though numbers are ignored in brainfuck, I thought I'd post one anyway. 32 Chars: ++++++[>++++++++<-]>++.--.+.+++. \$\endgroup\$ Apr 1, 2015 at 21:37
  • 7
    \$\begingroup\$ Brainfuck isn't a valid language for this challenge. \$\endgroup\$
    – Joe Z.
    Apr 1, 2015 at 22:49
  • 14
    \$\begingroup\$ I know. That's why I posted it as a comment \$\endgroup\$ Apr 1, 2015 at 22:51
  • 13
    \$\begingroup\$ I wonder if this question gets a small spike in popularity around New Year's. \$\endgroup\$
    – Joe Z.
    Dec 26, 2015 at 23:28
  • 5
    \$\begingroup\$ Waiting for "Come on folks, don't you realize it's 2016?" :) \$\endgroup\$
    – padawan
    Jan 4, 2016 at 23:35

333 Answers 333

1
\$\begingroup\$

Jelly, 10 bytes

⁹⁴×H_⁴Ḥ¤’’

Explanation:

⁹            Set the current value to 256.
 ⁴×          Multiply by 16. The current value is now 4096.
   H         Divide by 2. The current value is now 2048.
    _⁴Ḥ¤     Subtract by 16/2. The current value is now 2016.
        ’’   Decrement twice. The current value is now 2014.
\$\endgroup\$
0
1
\$\begingroup\$

MAWP, 30 28 bytes

!!+!!!!++!*+/!+!!+!+!!++++*:

Try it!

This will be fun to golf.

This is longer than @Lyxal 's answer, but outputs only one time as one number.

\$\endgroup\$
1
  • \$\begingroup\$ 15 bytes \$\endgroup\$
    – lyxal
    Sep 28, 2020 at 9:10
1
\$\begingroup\$

Poetic, 112 bytes

two`s a bad thing
using a two?o,hardly!i am using a poem
a numeric poem?oh,clearly,but a limited one for certain

Try it online!

\$\endgroup\$
1
\$\begingroup\$

DROL, 13 bytes

ziill<ukl<dfo

I wanted to add a language that wasn't already included...

DROL is a limited instruction assembler-like language with only two registers as storage. The language does include numbers, so it think it qualifies for this question. But numbers are only used for loop length indicators, as well as being used as the name of some instructions. It's described on the Esolang Wiki DROL page.

Here's a rundown of what the code does:

z              set R1=0
 ii            increment R1 by 1 twice   (2)
   ll          square R1 three two      (16)
     <         shift left R1            (32)
      u        set R2=R1                (32)
       k       increment R2             (33)
        l      square R1              (1024)
         <     shift left R1          (2048)
          d    decrement R1           (2047)
           f   subtract R1 = R1 - R2  (2014)
            o  print R1 as an integer
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1
\$\begingroup\$

Kotlin, 42 bytes

fun main()=print("ް.".map{it.code}.sum())

I used U+1968 which is ް and a . which is U+46.

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1
\$\begingroup\$

Phooey, 16 bytes

=@+$i>$i=$i<@+$i

Try it online!

                 # Stk  Tape
=@+$i>$i=$i<@+$i # (0) >0  0   (initial state)
=                # (0) >1  0   tape = stack == tape to get 1
 @               #  1  >1  0   push to stack
  +              # (0) >2  0   pop; add stack to tape
   $i            # (0) >2  0   print tape as integer (2)
     >$i         # (0)  2 >0   move tape ptr right, print (0)
        =$i      # (0)  2 >1   same as above to get 1 again, print (1)
           <@    #  2  >2  1   move back, push
             +$i # (0) >4  1   add stack to tape, print (4)

Thank goodness for Phooey's = operator. This would be impossible in Foo.


Phooey, 23 19 bytes

This version actually generates the number 2014 instead of printing 2,0,1, and 4.

=@+@@@+@**@@*@+--$i

Try it online!

                    # stack       | tape
=@+@@@+@**@@*@+--$i #         (0) |    0  initial state
=                   #         (0) |    1  tape = tape == pop() (to get 1)
 @+                 #         (0) |    2  double by adding to self
   @@               #    2     2  |    2  push two copies to the stack
     @+             #    2     2  |    4  double
       @*           #    2     2  |   16  square by multiplying by self
         *          #          2  |   32  multiply by 2
          @         #    2    32  |   32  push 32 for later
           @*       #    2    32  | 1024  square
             @+     #    2    32  | 2048  double
               -    #          2  | 2016  pop and subtract
                -   #         (0) | 2014  pop and subtract
                 $i #         (0) | 2014  print 2014
\$\endgroup\$
1
\$\begingroup\$

ARM assembly, 94 85 bytes (28 bytes compiled)

Textual assembly for ARM mode.

s:subs sb,sb
adc sb,sb
add sb,sb
lsl sl,sb,sb
add sb,sl,lsl sb
rsb sb,sl,lsl sl
bx lr

A function which returns 2014 in sb (r9)

Clobbers sl (r10) and sb (r9).

Expanded version:

        // It feels so empty here...
        .globl s
s:
        // use the quirky way subs affects the flags
        // to set r9 to 1
        subs    r9, r9, r9
        adc     r9, r9, r9
        // double r9 by adding it to itself
        // lsl works just as well
        // r9 = 2
        add     r9, r9, r9
        // r10 = r9 << r9
        // r10 = 2 << 2
        // r10 = 8
        lsl     r10, r9, r9
        // r9 = r9 + (r10 << r9)
        // r9 = 2 + (8 << 2)
        // r9 = 2 + 32
        // r9 = 34
        add     r9, r9, r10, lsl r9
        // r9 = (r10 << r10) - r9
        // r9 = (8 << 8) - 34
        // r9 = 2048 - 34
        // r9 = 2014
        rsb     r9, r9, r10, lsl r10
        // Return
        bx      lr

This uses the same idea as my Phooey answer, of generating 2048, then subtracting 34. While I do have access to push and pop, ARM isn't a stack machine. It is a register machine. Additionally, we have lsl for shifting left which makes a few cases easier.

It is yet another rare case where the inverted carry flag on ARM is useful for something other than 64-bit subtraction, as it allows us to set a register to 1 when combined with adc.

Additionally, this only works in ARM mode: Thumb-2 did not show the return of shifting by register (which kinda was a dumb waste of encoding bits)

It uses the classic register names instead of the format which is r[0-15].

\$\endgroup\$
1
\$\begingroup\$

Python 3, 16 bytes

print(ord('ߞ'))

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Vyxal, 10 3 bytes

»÷∩

Try it Online!

Simply the compressed number 2014

\$\endgroup\$
1
  • \$\begingroup\$ Can't be bothered to download the interpreter, but does !!!!!''.$... work for 11 bytes? \$\endgroup\$
    – Jo King
    Oct 9, 2020 at 4:29
1
\$\begingroup\$

Grok, 13 bytes

iNH`I:P-zP-zq

Alternate 13 Byte solution:

i:H:N`-Yx-zZq
\$\endgroup\$
1
\$\begingroup\$

Branch, 11 bytes

}}#/#}#^}}#

Try it on the online Branch interpreter!

Branch does have numbers in it. } is increment and { is decrement. # outputs as number. / goes to the left child, which is automatically initialized to 0, which is shorter than {{. Finally, ^ goes to the parent, which is 2 when that command is called. Actually, since the current node is 1, and the parent is 2, we could do ^}} or }}} to get 4.

An alternative solution that produces 2014 on the tree itself instead of outputting it character by character:

Branch, 50 bytes

}}^\}}}}}^*^\}}^*{^\}}^*^\/}}}}}}^\}}}}}}}}}^*{^*#

Try it on the online Branch interpreter!

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1
\$\begingroup\$

Deadfish~, 12 bytes

iioddoioiiio

Try it online!

This, this is surprisingly short!

\$\endgroup\$
3
  • \$\begingroup\$ save bytes by squaring instead of incrementing twice at the end \$\endgroup\$
    – user100690
    Apr 24, 2021 at 15:21
  • \$\begingroup\$ This can just be vanilla deadfish, since you don't use ch{}(). Also, iiio can be replaced with iso for 11 bytes. \$\endgroup\$
    – emanresu A
    Apr 25, 2021 at 22:32
  • \$\begingroup\$ And you can get the number 2014: iiisddsddddsdddddddddddo but that's a bit long. \$\endgroup\$
    – emanresu A
    Apr 25, 2021 at 22:34
1
\$\begingroup\$

PICO-8, 24 18 bytes

?ord("⁙")*ord("j")

ord() gets the numerical index of a character; in P8SCII, is at index 19 and j is at index 106, creating the equation \$19*106=2014\$.

\$\endgroup\$
1
\$\begingroup\$

Python 3 (15 Bytes)

print(ord('ߞ'))

Python 3 (24 Bytes)

print(ord('޴')+ord('*'))

JavaScript (28 Bytes)

-~!+[]+[+[]]+-~[]+-~!''-~!''
\$\endgroup\$
1
1
\$\begingroup\$

C++ (gcc), 33 bytes

main(){std::cout<<('j'-'W')*'j';}

Try it online!

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Should #include <iostream> be included as byte count? \$\endgroup\$
    – tsh
    Feb 11, 2022 at 9:26
  • \$\begingroup\$ Might as well go for putting \x13 in your source! -6 bytes \$\endgroup\$
    – M Virts
    Jun 11, 2022 at 21:59
  • \$\begingroup\$ @tsh I agree the include should be part of it. In the same vein you could just switch to C \$\endgroup\$
    – M Virts
    Jun 11, 2022 at 22:01
1
\$\begingroup\$

Pascal, 76 B

See also Delphi.
This program requires a processor supporting ISO standard 10206 “Extended Pascal”, specifically constant definitions may be expressions (ord(true) is an expression).

program p(output);const I=ord(true);begin write(I+I,I-I:i,I:i,I+I+I+I:i)end.

If ord(maxChar) ≥ 2014 we can use just one character in the 43 character program

program q(output);begin write(ord('X'))end.

where X needs to be substituted by the implementation-defined char value having the ordinal value 2014.


For reference, an Extended Pascal program printing the current year looks like this:

program t(output); var t: timeStamp; begin getTimeStamp(t); writeLn(t.year) end.
\$\endgroup\$
1
\$\begingroup\$

Go, 56 bytes

func f(){for _,c:=range`-+,/`{print(string(c+'F'-'A'))}}

Attempt This Online!

Alternative to that other Go answer. Prints to STDERR.

\$\endgroup\$
1
\$\begingroup\$

tinylisp 2, 5 bytes

(h"ߞ

Try it at Replit!

Explanation

The same character-code solution as everybody else. h is short for head, which, when given a string, returns the character code of its first character. The string quote and the function-call parenthesis both auto-close at the end of the line/program.

Non-string solution, 43 bytes

(d T(+(*)(*
(d X(* T T T T T
(-(* X X T)X T

Here, we take advantage of the fact that tinylisp 2's multiplication builtin * is variadic: it can take any number of arguments, including zero. Without arguments, its return value is the product of an empty list, or \$1\$.

(d T(+(*)(*)))

Define T as the sum of \$1\$ and \$1\$, or \$2\$.

(d X(* T T T T T

Define X as the product of five \$2\$s, or \$32\$.

(-(* X X T)X T

Output the product of \$32\$ and \$32\$ and \$2\$, minus \$32\$, minus \$2\$ (the - builtin is also variadic): \$(32 \cdot 32 \cdot 2) - 32 - 2 = 2014\$.

\$\endgroup\$
1
\$\begingroup\$

Logically, 68 bytes.

@w:a,b,c;WRITE(h,a,b,c,l,h,h,l,l)()
@M w(l,h)()w()()w(h)()w(l,l,h)()

Consists of two gate definitions. w writes the 3 passed bits in the form of 00110cba, and then M has 4 w gates for each of the digits.

Whilst logically can use numbers for both argument unrolling and high-low states, it can be avoided pretty easily.

\$\endgroup\$
1
\$\begingroup\$

Thunno 2, 3 bytes

«¬ẓ

Attempt This Online! Simple compressed integer.

Thunno 2, 4 bytes (UTF-8)

'ߞC

Attempt This Online! Charcode.

Thunno, 4 bytes (UTF-8)

'ߞO

Attempt This Online! Basically the same so I won't create another answer for it.

\$\endgroup\$
5
  • \$\begingroup\$ Thunno 2 HHHHHHHHHHHHHHHHHHHHTs, 1 byte: 4 \$\endgroup\$
    – noodle man
    May 27, 2023 at 20:21
  • \$\begingroup\$ @Jacob you goofy - 4 is a digit and no digits allowed in the answer. You too noodles for golf. \$\endgroup\$
    – lyxal
    May 27, 2023 at 23:55
  • \$\begingroup\$ @lyxal oops fixed it: Thunno 2, {"Z"*2014}l, 0 bytes. 🍜 \$\endgroup\$
    – noodle man
    May 28, 2023 at 1:07
  • \$\begingroup\$ (and, in case it’s not clear, those are joke submissions and obviously unfair) \$\endgroup\$
    – noodle man
    May 28, 2023 at 2:58
  • \$\begingroup\$ @Jacob lol, nice ones \$\endgroup\$
    – The Thonnu
    May 28, 2023 at 6:59
1
\$\begingroup\$

!+~%, 8 bytes

++!+~%+!

Stumbled over this challenge and sorry, could not resist. I promise this will not happen again. (-;

An explanation, to make this look serious:

  • ++ twice adds 7 to the initial 6, so we end up at 20, that gets printe by !
  • to the accumulator resetted to 6, + adds 7, and the ~ reverses the result from 13 to 31
  • % does a modulo from the number with a leading 1, so 131 mod 31 happens to be 7.
  • the + adds another 7, and ! prints the 14 behind the 20.
\$\endgroup\$
1
\$\begingroup\$

Vyxal 3, 3 bytes

~¥ᴴ

Try it Online!

because of how the encoding works, vyxal 3 takes 4 bytes to use a base 252 encoding, but you can shave a byte because 2014 is only 2 bytes in base 255 and thus you can use the ~ element which converts 2 elements on the codepage to base 255

\$\endgroup\$
1
\$\begingroup\$

Commodore BASIC V2, tested on the Commodore C64, 37 PETSCII characters. Non-competing

Note that the {CRSR LEFT} in the following "source code" is the cursor left control character.

A=.↑.:A%=π+A:?A♥("╮")"{CRSR LEFT}"R╮(ST▂(A%),A)

Non-obfuscated, this would look like this:

A=.↑.:A%=π+A:PRINTASC("╮")"{CRSR LEFT}"RIGHT$(STR$(A%),A)

or on a Commodore C64 screen, it would look like this:

Commodore C64 answer to "Produce the number 2014 without any numbers in your source code"

\$\endgroup\$
5
  • 1
    \$\begingroup\$ What’s non-competing about this? Looks fine to me \$\endgroup\$
    – noodle man
    Jan 12 at 19:02
  • \$\begingroup\$ So upvote it then? idk... \$\endgroup\$ Mar 14 at 15:01
  • 1
    \$\begingroup\$ I didn’t upvote because I was (and still am) confused what you meant by non-competing, if that means the answer is invalid then why should I upvote it \$\endgroup\$
    – noodle man
    Mar 14 at 15:53
  • \$\begingroup\$ It is not invalid, it is non-competing or just for fun/just for a laugh. \$\endgroup\$ Mar 14 at 16:15
  • 1
    \$\begingroup\$ I guess I don’t get the joke \$\endgroup\$
    – noodle man
    Mar 14 at 20:25
1
\$\begingroup\$

Rattle, 1211 bytes

d|n*+b=+**b

Try it Online!

Explanation

d|            take "d" as input
  n           get the ASCII int value of 'd' (100)
   *          multiply (by 2 without an argument) to get 200
    +         increment by 1 to get 201
     b        add "201" to print buffer
      =+**    reset, increment, multiply by 2, multiply by 2 to get 4
          b   add "4" to print buffer, output implicitly resulting in "2014"
      
\$\endgroup\$
1
\$\begingroup\$

Python 3, 42 24 bytes

print('%x'%(ord('—')))

In UTF-8.

is U+2014.

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Actionscript 3.0, 65 26 bytes

trace(""+"ߞ".charCodeAt())
Edit:

Outputs 2014

\$\endgroup\$
2
  • 2
    \$\begingroup\$ Even though the current year is 2024, this challenge still requires programs to output 2014. :) \$\endgroup\$
    – DLosc
    Apr 22 at 1:14
  • 2
    \$\begingroup\$ @DLosc oops, i'll fix that. Adding a char 78 instead of two 44 should work. \$\endgroup\$ Apr 22 at 5:27
1
\$\begingroup\$

YASEPL, 53 27 bytes

=f$+****++++++=t$+***-+f*f<

explanation:

=f$+****++++++=t$+***-+f*f<      packed
=f$+****++++++                   F = 2 * 2 * 2 * 2 * 2 + 6 = 38
              =t$+***-+f*f       T = 2 * 2 * 2 * 2 - 1 + F * F = 2014
                          <      print out T (which is 2014)
\$\endgroup\$
0
\$\begingroup\$

Objective C

NSDateFormatter *formatter = [[NSDateFormatter alloc] init];
[formatter setDateFormat:@"yyyy"]; 
NSLog(@"%@",[formatter stringFromDate:[NSDate date]]);
\$\endgroup\$
7
  • 1
    \$\begingroup\$ From the comments: Gelatin: “Is it acceptable to use the current year?” Joe Z.: “No, it has to be 2014 exactly.” \$\endgroup\$
    – manatwork
    Jan 1, 2014 at 13:17
  • \$\begingroup\$ Because the question is a code-golf question, please add the character count. \$\endgroup\$
    – ProgramFOX
    Jan 1, 2014 at 13:31
  • 1
    \$\begingroup\$ What happened to you, Smalltalk ? You look...different. \$\endgroup\$
    – bug
    Jan 7, 2014 at 1:17
  • \$\begingroup\$ NSLog(@"%i",'&'*('F'-'A')); \$\endgroup\$ Feb 16, 2017 at 18:57
  • 1
    \$\begingroup\$ @Cœur sorry meant this NSLog(@"%i",'&'*('V'-'!')); \$\endgroup\$ Mar 17, 2018 at 20:08
0
\$\begingroup\$

Perl, 24 bytes

print-ord(A)+ord for U,O

Pure ascii, no nonprinting characters or utf8-only characters used. Uses the 21st and 15th letters of the alphabet to print 20, 14.

\$\endgroup\$
0
\$\begingroup\$

C# (56 characters)

Class P{static void Main(){Console.Write(','*','+'N');}} 
\$\endgroup\$
5
  • \$\begingroup\$ This doesn't print anything. \$\endgroup\$
    – shamp00
    Jan 5, 2014 at 15:05
  • \$\begingroup\$ 26 characters = 4 bytes?? I don't think so... \$\endgroup\$
    – jub0bs
    Jan 6, 2014 at 12:24
  • \$\begingroup\$ Sorry. I just included the logic only. Now I included the whole program. \$\endgroup\$ Jan 7, 2014 at 3:34
  • \$\begingroup\$ You need System.Console to use Console. \$\endgroup\$
    – shamp00
    Jan 8, 2014 at 14:24
  • \$\begingroup\$ using System; will be on the header. So.... \$\endgroup\$ Jan 8, 2014 at 17:50

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