Most creative way to display 42

Question

Douglas Adams was born on March 11, 1952, and died when he was just 49. In honor of this wonderful writer, I challenge you to display 42 in the most creative way possible.

You could print it in the log, via some convoluted method, or display it as ASCII art, or anything! Just come up with a creative method of displaying 42.

Because this a popularity-contest, whichever answer has the most upvotes by March 11, will be declared the winner.

Note: this is not a duplicate. The question it was marked as duplicating was a code-trolling question whose goal was to write code to output 42, not find the most creative way to display it.

Winner: grovesNL! With an astounding 813 votes! Congrats!

Honorable Mentions:

Mr Lister C 228 For the clever use of #define

David Carraher Mathematica 45 For the complicated and convoluted math function to achieve 42

Aschratt Windows Calculator 20 Because, well, it's windows calculator And definitely 1337.

f.rodrigues Python 17 Because of the clever use of using external programs. And MSPaint

Jason C LMGTFY 14 For the use of LMGTFY (Let Me Google That For You)

Trimsty Python 12 For the clever use of an error message to output 42.

Mukul Kumar C++ 7 For the nice ASCII output.

If you think that there is another answer worth putting on the list, please comment it!

Answer 1

Double Brainfuck

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

which outputs...

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

which outputs...

6*7=42

Answer 2

C

Here's an oldie but goodie...

#include <stdio.h>

#define six  1+5
#define nine 8+1

int main()
{
    printf("what do you get when you multiply six by nine?\n");
    printf("%i x %i = %i\n", six, nine, six*nine);
}

This program contains 42 different ASCII characters.

Answer 3

Brainfuck

Took a while to get there, but I like the result:

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

When run, it will print 42, of course.

Answer 4

JavaScript:

var ________ = 0.023809523809523808, ____ = 1, ___ = 0, __ = 0, _ = 1;

       __ -           ___
     /_  |0        //     \\
    /_/   0     //          \\
   /_/_  |0                //
  /_/_   |0              //
 /_/____ |_           //
/________|0        //
         |0     //______________

The output is:

42

Not bad, eh? :)

For the people who don't understand, it actually evaluates the following:

__ - ___ / _ | 0 / _ / 0 / _ / _ | 0 / _ / _ | 0 / _ / ____ | _ / ________ | 0 | 0

Answer 5

C, Twelve Days of Xmas Style

New version:

main(Z,_){Z?(_=Z[" $X,X3Y<X@Z@[<XHZHX,"
"` \\(Z(X0Z0Z8[@X@^8ZHZHX(Z(`#Y(Z(X3[8"
"\\@_8ZHXHXHX(Z(` \\(Z(X0Z0Z8\\@_8ZIXI"
"X(Z(` \\,X0Z0Z8\\@_8ZHZHX,"])?main(0,_
-32),main(Z+1,_):0:(putchar((_>>3)["kt"
"wy~|tE/42"]-37),(_&7)?main(0,_-1):0);}

Output:

FFFFF OOOOO RRRR  TTTTT Y   Y    TTTTT W   W OOOOO
F     O   O R   R   T    Y Y       T   W   W O   O
FFFF  O   O RRRR    T     Y        T   W W W O   O
F     O   O R   R   T     Y        T   WW WW O   O
F     OOOOO R   R   T     Y        T   W   W OOOOO

By the way, also check out my text-to-speech answer.

---

Original Version:

main(c,z,_){c==1?main(c+1,0,c^c):c==2?
z=_["##$#%&#%#x'%%()&(%%x$%$((&(*%x'%"
"%((&(+%x'#%((&(%#x"],z?z=='x'?main(4,
_,c*5):main(c+1,z,0),main(c,z,_+1):0:c
==3?(_-2)==3?main(_-1,_,32):(main(c+1,
c,((2+c)*(z-35)+_)["six*nine= {   }  "
"   ;      _   ( ) [ 3 ]do {;=0xDA"]==
32?32:043),main(c,z,_+1)):putchar(_);}

The output is:

##### ##### ####  ##### #   #       ##### #   # #####
#     #   # #   #   #    # #          #   #   # #   #
####  #   # ####    #     #           #   # # # #   #
#     #   # #   #   #     #           #   ## ## #   #
#     ##### #   #   #     #           #   #   # #####

Alternate spacing, if you're feeling tacky:

        main(c     ,z,_){c==01?
       main(c+     1,0,c^c):c==2
      ?z=_["#"     "#$#%&#%#x'%%"
     "()&(%%x"             "$%$("
    "(&(""*%x"             "'%%("
   "(&(" "+%x"             "'#%("
  "(&("  "%#x"             ],z ?z
 =='x'?main(4,_     ,c*5):main(c
 +1,z,0),main(c    ,z,_+1):00:c
 ==3?(_+-2)==3?    main(_-1,_,
         32):(     main(
         c+1,c     ,((2+
         c)*(z     -35)+
         _)[""     "six"
         "*ni"     "ne= {   }   "
         "  ;"     "      _   ( "
         ") ["     " 3 ]do {;"]==
         32?32     :043),main(c,z
         ,_+1)     ):putchar(_);}

The program is a single recursive statement. I made it in the style of my favorite obfuscated C program ever, Twelve Days of Christmas (compile, prepare mind to be blown, run).

---

HOW TO

Also, since this seems as good a place as any, here is a guide describing how to make this type of program. This guide uses the original version above as an example. Aside from the first bit with the block letters, they are general steps:

INITIAL: First, I started by making the block letters:

##### ##### ####  ##### #   #       ##### #   # #####
#     #   # #   #   #    # #          #   #   # #   #
####  #   # ####    #     #           #   # # # #   #
#     #   # #   #   #     #           #   ## ## #   #
#     ##### #   #   #     #           #   #   # #####

I then made a numbered list of the unique patterns in each 5-column character row:

0: *****  
1: **** 
2: *   * 
3:       
4: *     
5:   *   
6:  * *  
7: * * * 
8: ** ** 

And so each of the 5 pixel rows of text becomes a series of 9 numbers:

00000 00000 11111 00000 22222 33333 00000 22222 00000
44444 22222 22222 55555 66666 33333 55555 22222 22222
11111 22222 11111 55555 55555 33333 55555 77777 22222
44444 22222 22222 55555 55555 33333 55555 88888 22222
44444 00000 22222 55555 55555 33333 55555 22222 00000

For obfuscation (and ease of programming) we add the '#' character to the numbers. In the program below, patterns is the array of pixel patterns, and lines is the obfuscated array of pattern codes for each line, terminated by an 'x'. For further obfuscation we define "on" pixels in patterns to be any character that isn't a space; this lets us put more misleading text in pattern:

#include <stdio.h>

char pattern[] = 
  "six*n"
  "ine= "
  "{   }"
  "     "
  ";    "
  "  _  "
  " ( ) "
  "[ 3 ]"
  "do {;";

char lines[] =
  "##$#%&#%#x"
  "'%%()&(%%x"
  "$%$((&(*%x"
  "'%%((&(+%x"
  "'#%((&(%#x";

void printpattern (char c) {
  int n;
  for (n = 0; n < 5; ++ n)
    putchar(pattern[5*(c-'#') + n]==32?32:'#');
  putchar(' ');
}

int main () {
  char *ptr = lines;
  while (*ptr) {
    while (*ptr != 'x')
      printpattern(*(ptr++));
    putchar('\n');
    ++ ptr;
  }
}

STEP 1: The next step involves a few tasks:

• Remove all loops and use recursion.
• Change all functions (except main) to the form int function (int, int) and use the same parameter names for each. The reasons will become clear later.
• Change main to the form int main (int, int, int) and name the last two parameters the same as your function parameter names.
• Replace all references to string constants with the strings themselves; and use each string only once if possible.
• The include can be removed; it's unnecessary for int putchar (int).

We can also take advantage of the weird C feature where a[b] is equivalent to b[a] to obfuscate further.

int printpattern (int z, int _) {
  if (_==5)
    putchar(' ');
  else{
    putchar((5*(z-'#') + _)["six*nine= {   }     ;      _   ( ) [ 3 ]do {;"]==32?32:'#');
    printpattern(z, _+1);
  }
  return 0;
}

// z ignored, _ is index
int printtext (int z, int _) {
  z = _["##$#%&#%#x'%%()&(%%x$%$((&(*%x'%%((&(+%x'#%((&(%#x"];
  if (z) {
    if (z == 'x')
      putchar('\n');
    else
      printpattern(z, 0);
    printtext(z, _ + 1); // first parameter arbitrary
  }
  return 0;
}

int main (int c, int z, int _) {
  printtext(0, 0);
}

STEP 2: Next, make use of the ?: and , operators to transform each function into a single return statement. I'm illustrating this separately from the above because this is where things start getting confusing to look at. Remember that putchar() returns an int, and ?: takes precedence over ,:

int printpattern (int z, int _) {
  return _==5 ?
    putchar(' ')
  :
    (putchar((5*(z-'#') + _)["six*nine= {   }     ;      _   ( ) [ 3 ]do {;"]==32?32:'#'),
     printpattern(z, _+1));
}

// z ignored, _ is index
int printtext (int z, int _) {
  return
    z = _["##$#%&#%#x'%%()&(%%x$%$((&(*%x'%%((&(+%x'#%((&(%#x"],
    z ?
      z == 'x' ?
        putchar('\n')
      :
        printpattern(z, 0)
      ,
      printtext(z, _ + 1)
    :
    0;
}

int main (int c, int z, int _) {
  printtext(0, 0);
}

STEP 3: Ok. The next step is a big one. All of the functions are now a single statement of the same form. We can now combine them all into a single function, identifying each one by a number -- essentially turning the entire program into a single recursive function. Note that the first parameter to main will be 1 when the program is run with no arguments, so that should be our initial state.

Also, since our parameter c to main is our state variable, we know its value at all times, and we can obfuscate a little further by replacing integer constants with their values in terms of c (for example, when we know c is 2, we can replace 5 with c+3). Other little obfuscations can be done too (e.g. I replaced '#' with 35 and 043):

int main (int c, int z, int _) {
  switch (c) {
  case 1: // main
    return main(c+1, 0, c^c); // (2, 0, 0)
  case 2: // printtext
    return
    z = _["##$#%&#%#x'%%()&(%%x$%$((&(*%x'%%((&(+%x'#%((&(%#x"],
    z ?
      z == 'x' ?
        putchar('\n')
      :
        main(c+1, z, 0) // c+1==3
      ,
      main(c, z, _ + 1)
    :
    0;
  case 3: // printpattern
    return (_-2)==3 ? // _==5
    putchar(' ')
    :
     (putchar(((2+c)*(z-35) + _)["six*nine= {   }     ;      _   ( ) [ 3 ]do {;"]==32?32:043),
     main(c, z, _+1));
  }
}

STEP 4: Finally, remove the switch block by using a series of ?: operators. We can also remove the int declarations, since C will use them by default, as well as the return itself (which will generate a warning at worst). After this, our program is a single, recursive function with one statement. Pretty cool, right?

Edit: I replaced putchar() with a main and c==4 below; because I just thought of it at the last minute:

main (c, z, _) {
  c == 1 ?
     main(c+1, 0, c^c)
  : c == 2 ?
    z = _["##$#%&#%#x'%%()&(%%x$%$((&(*%x'%%((&(+%x'#%((&(%#x"],
    z ?
      z == 'x' ?
        main(4,_,c*5)
      :
        main(c+1, z, 0) 
      ,
      main(c, z, _ + 1)
    :
    0
  : c==3 ?
    (_-2)==3 ? 
    main(_-1,_,32)
    :
    (main(c+1,c,((2+c)*(z-35) + _)["six*nine= {   }     ;      _   ( ) [ 3 ]do {;"]==32?32:043),
      main(c, z, _+1))
  : // c==4
    putchar(_);
}

If you want to add a little flair, you can use more interesting numbers for c and even base the checks off of other numbers (e.g. for the c==2 case, z is ignored and available, so instead of calling main(2,z,_) you could call main(-97,_,_) and replace c==2 with c<-z). Be creative; the possibilities are endless.

FINISH: The final step, then, is to arrange the text in some creative pattern, and voila! You can adjust the code a little to help with formatting (e.g. I added some extra data at the end of the patterns string in the posted program to help get the line length right). The ladies are sure to be all up ons.

Answer 6

I'm feeling lazy.

Python

t
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e
a
n
s
w
e
r
t
o

l
i
f
e

t
h
e
u
n
i
v
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r
s
e

a
n
d
e
v
e
r
y
t
h
i
n
g:

Output:

  File "golf.py", line 42
    g:
     ^
SyntaxError: invalid syntax

Answer 7

Java

(or C++, the code's almost similar)
Using String functions, so don't forget to include your library!

P.S. I know it's lengthy, but it's supposed to be creative, right? And anyway, it isn't a "lowest-byte-wins".

String s = "Hitchhiker's Guide to the Galaxy";
String s2 = "Don'tPanic";
String s3 = "The Restaurant at the End of the Universe.";

int arthur_dent = s.length();
int ford_prefect = s2.length();
int zooey_deschanel = s3.length();
int vogon_poetry = arthur_dent + ford_prefect;

System.out.println("         " + vogon_poetry + "       " + zooey_deschanel + " " + zooey_deschanel); //in case you're confused, I'm using Zooey to print the big '2', and Vogons to print the big '4'.
System.out.println("       " + vogon_poetry + vogon_poetry + "     " + zooey_deschanel + "     " + zooey_deschanel);
System.out.println("     " + vogon_poetry + "  " + vogon_poetry + "    " + zooey_deschanel + "       " + zooey_deschanel);
System.out.println("   " + vogon_poetry + "    " + vogon_poetry + "            " + zooey_deschanel);
System.out.println(" " + vogon_poetry + "      " + vogon_poetry + "          " + zooey_deschanel);
System.out.println(vogon_poetry + " " + vogon_poetry + " " + vogon_poetry + " DA " + vogon_poetry + "     " + zooey_deschanel);
System.out.println("         " + vogon_poetry + "     " + zooey_deschanel);
System.out.println("         " + vogon_poetry + "    " + zooey_deschanel + " " + zooey_deschanel + " " + zooey_deschanel + " " + zooey_deschanel);

Here's the output:

         42       42 42
       4242     42     42
     42  42    42       42
   42    42            42
 42      42          42
42 42 42 DA 42     42
         42     42
         42    42 42 42 42 

Imagine my misery when I counted and found out that "The Restaurant at the End of the Universe" had 41 characters! :/ Sigh.

Answer 8

Mathematica

Take 1

With some work, I ought be able to golf this down a bit. :)

In InputForm:

answer[ultimateQuestion[Life,theUniverse,Everything]] =
  Times[Plus[-1,Limit[Power[Plus[1,Times[Complex[0,1],
  Power[n,-1],Pi]],n],Rule[n,DirectedInfinity[1]]]],Sqrt[-1]^2,
  Times[Rational[1,2],Plus[-1,Fibonacci[4]],Fibonacci[2]],
  Binomial[3,2],LucasL[4]]

In TraditionalForm:

Testing:

answer[ultimateQuestion[Life,theUniverse,Everything]] 

42

---

Take 2

Note: The numerals were made as follows.

• "42" was first printed on the screen in very large font, axes displayed, so that the coordinates of the key points could be identified.
• Another "4" was drawn a broad straight lines connecting the respective key points. It was superimposed on the previously drawn "4" to check for accuracy. The "2" was drawn as a BSpline curve. Some of the key points, which were now control points, had to be set in position by trial and error to get the desired curves.
• An third coordinate (always zero) was added to the line and BSplineCurve to enable 3D display.

---

answer[ultimateQuestion[Life,theUniverse,Everything]]  = 
 Table[With[{v = RotationTransform[θ, {0, 0, 1}][{3, 0, -.2}]},
   Graphics3D[{Thickness[.06], CapForm["Round"],
     Tube[Line[{{-67, 0, -44}, {-30, 0, -44}}], 10],
     Tube[
      Line[{{-25, 0, -12}, {-100, 0, -12}, {-52, 0, 70}, {-45, 0, 70}, {-45, 0, -43}}], 10], 
     Tube[BSplineCurve[l = {{27, 0, 52}, {27, 0, 57}, {57, 0, 85}, {113, 0, 40}, 
     {12, 0, -45}, {35, 0, -45}, {90, 0, -45}, {90, 0, -45}, {92, 0, -35}}], 10]},
     Boxed -> False, PlotRange -> 100, ImageSize -> 250, SphericalRegion -> True, 
     ViewPoint :> v, 
     PlotRangePadding -> 10]],(*{θ,0,2Pi,Pi/24},*){θ, -Pi/2, -Pi/2 + 2 Pi, Pi/24}]

Export["theMeaningOfLife.gif", answer[ultimateQuestion[Life,theUniverse,Everything]] ]

Answer 9

Seems appropriate:

grep -i "DON'T" /bin/lesspipe | wc -l ; grep -i "PANIC" /usr/share/pyshared/mx/Log.py | head -n 1 | cut -d '=' -f 2 | tr -d ' '

Output:

4
2

Ubuntu 12.04, 64-bit Desktop

Answer 10

Windows calculator

Multiplying Pi with 13.37 and ignoring the decimal :P

Answer 11

Python

I guess it only works on Windows 7.

import win32api, win32con, win32gui
from time import time, sleep
import os

w = { 1:[(358, 263), (358, 262), (358, 261), (359, 261), (359, 262), (359, 264), (359, 266), (359, 270), (359, 282),
     (358, 289), (357, 308), (356, 319), (355, 341), (355, 351), (355, 360), (355, 378), (355, 388), (354, 397),
     (354, 406), (354, 422), (354, 428), (354, 436), (354, 438), (354, 439), (354, 440), (355, 440), (356, 439),
     (357, 439), (358, 438), (360, 438), (362, 437), (369, 437), (372, 437), (381, 437), (386, 437), (391, 437),
     (397, 436), (411, 436), (419, 435), (434, 435), (442, 435), (449, 434), (456, 434), (468, 434), (473, 435),
     (480, 436), (483, 436), (485, 436), (487, 437), (488, 437), (488, 438), (488, 439), (487, 440), (486, 440),
     (485, 440), (484, 440), (483, 439), (483, 437), (481, 431), (481, 427), (481, 420), (481, 413), (483, 396),
     (485, 387), (488, 367), (491, 356), (493, 345), (500, 321), (503, 310), (507, 299), (514, 280), (517, 272),
     (520, 266), (523, 260), (524, 258), (524, 259), (524, 261), (524, 265), (524, 269), (523, 275), (522, 289),
     (521, 297), (518, 315), (516, 324), (515, 334), (513, 345), (509, 368), (507, 382), (502, 411), (500, 426),
     (498, 440), (495, 453), (491, 478), (489, 491), (485, 517), (483, 530), (481, 542), (479, 552), (476, 570),
     (475, 577), (474, 588), (473, 592), (473, 595), (473, 597), (473, 600), (473, 601), (473, 602), (473, 601),
     (474, 599), (475, 597), (476, 594), (478, 587)],
  2:[(632, 305), (634, 306), (636, 309), (639, 314), (641, 319), (645, 330), (647, 337), (649, 353), (649, 362),
     (649, 372), (649, 384), (645, 409), (639, 436), (636, 448), (632, 459), (627, 470), (623, 479), (613, 497),
     (608, 503), (599, 512), (595, 514), (591, 514), (587, 513), (581, 504), (578, 498), (576, 483), (575, 476),
     (575, 469), (579, 454), (582, 447), (591, 436), (595, 432), (600, 430), (605, 429), (617, 432), (624, 437),
     (639, 448), (646, 455), (654, 461), (662, 469), (679, 484), (686, 491), (702, 504), (710, 509), (718, 512),
     (727, 514), (744, 515), (752, 515), (767, 512), (774, 510), (779, 508), (783, 505), (788, 499), (789, 495),
     (789, 486)] }

def d( x1, y1, x2, y2 ):
    win32api.SetCursorPos((x1, y1))
    win32api.mouse_event(win32con.MOUSEEVENTF_LEFTDOWN, 0, 0, 0, 0)
    win32api.SetCursorPos((x2, y2))
    win32api.mouse_event(win32con.MOUSEEVENTF_LEFTUP, 0, 0, 0, 0)
    sleep(0.01)

def p( l1 ):
    l2 = [""]
    l2.extend(l1)
    l1.append("")
    l3 = zip(l2, l1)
    l3.pop(0)
    l3.pop(-1)
    for n in l3:
        d(n[0][0], n[0][1], n[1][0], n[1][1])

os.startfile("C:\Windows\system32\mspaint.exe")
sleep(0.5)
win32gui.ShowWindow(win32gui.GetForegroundWindow(), win32con.SW_MAXIMIZE)
sleep(0.5)

for n in w:
    p(w[n])

The result is opening Paint and painting 42 as free hand.

Answer 12

Java (Swing)

This will display a frame drawing the answer. It only uses 42 for values.

public        class         FourtyTwo{ public
static         void         main(String[]args)
{  new        javax                    .swing.
JFrame        () {{                    setSize
(42 /(        42/42                    +42/42)
*42/ (        42/42                    +42/42)
,42/(42/ 42+42/42)*         42/(42/42+42/42));
}public void paint(         java.awt .Graphics
  g){g.drawPolygon(         new int[]{42,42,42
              + 42+         42,42+
              42+42         ,42+42
              +42 +         42,42+
              42+42         +42,42
              + 42+         42,42+42+42,42+42,
              42+42         },new int[]{42,42+
              42+42         +42,42+42+42+42,42

+42+42+42+42+42,                  42+42+
42+42+42+42,42,42,               42+42+42
,42 +        42+42              ,42}, (42/
42+42        /42)*              (42/  42 +
42/42        + 42/             42 +    42 /
42+42        /42))            ;g.drawPolygon
( new        int[]           {42+42+42+42+42,
42+42        +42 +           42+42      , 42+
42+42        + 42+          42+42        + 42,
42+42        +42 +          42+42        +42 +
42,42+42+42+42+42,         42+42          + 42+
42+42,42+ 42+42+           42+42          +42 +

42+42,42+42+42+42+42+42+42+42,42+42+42+42+42+42,
42+42+42+42+42+42,42+42+42+42+42+42+42+42,42+42+
42+42+42+42+42+42},new int[]{42,42 +42,42+42,42+
42+42,42+42+42,42+42+42+42+42+42,42+42+42+42+42+
42,42+42+42+42+42,42+42+42+42+42,42+42+42+42,42+
42+42+42,42},(42/42+42/42+42/42)*((42/42+42/42)*
(42/42+42/ 42)));};}.setVisible(42*42*42!=42);}}

Answer 13

Mathematica

WolframAlpha["meaning of life", {{"Result", 1}, "Content"}]

42

though I think it's cheating, really, since it's hard-coded. And not very creative, on my part... :)

Answer 14

Ruby

It is well known what you get if you multiply six by nine. This gives one solution:

puts (6 * 9).to_s(13)

---

Python

A variant of Tupper's self-referential formula:

# Based loosely on http://www.pypedia.com/index.php/Tupper_self_referential_formula
k = 17 * (
    (2**17)**0 * 0b11100000000000000 +
    (2**17)**1 * 0b00100000000000000 +
    (2**17)**2 * 0b00100000000000000 +
    (2**17)**3 * 0b11111000000000000 +
    (2**17)**4 * 0b00100000000000000 +
    (2**17)**5 * 0b00000000000000000 +
    (2**17)**6 * 0b01001000000000000 +
    (2**17)**7 * 0b10011000000000000 +
    (2**17)**8 * 0b10011000000000000 +
    (2**17)**9 * 0b01101000000000000 +
0)
# or if you prefer, k=int('4j6h0e8x4fl0deshova5fsap4gq0glw0lc',36)

def f(x,y):
    return y // 17 // 2**(x * 17 + y % 17) % 2 > 0.5
for y in range(k + 16, k + 11, -1):
    print("".join(" @"[f(x, y)] for x in range(10)))

Output:

@  @   @@ 
@  @  @  @
@@@@@    @
   @   @@ 
   @  @@@@

Answer 15

Javascript

alert((!![]+ -~[])*(!![]+ -~[])+""+(!![]+ -~[]))

Answer 16

LMGTFY

http://bit.ly/1ldqJ8w

Short enough that I had to type this to reach the minimum character count...

Answer 17

Forth:

SCR # 1
 0 ( FORTY-TWO @ ES-FORTH )
 1 HEX 0 24 -31 21 -31 31 -31
 2 31 -14 51 11 -11 51 11 -11 23 31
 3 : T SWAP 0 DO DUP EMIT LOOP DROP ;
 4 : K BEGIN DUP WHILE DUP 0< IF CR
 5  ABS THEN 10 /MOD 20 T A0 T
 6  REPEAT DROP ;
 7 K CR

That 1 LOAD outputs:

   █  ███
 █ █     █
 █ █     █
 ████   █
   █   █
   █  █
   █  ████

Answer 18

C++

cout<<"....-"<<" "<<"..---";

Morse code ;)

Answer 19

R

sum(as.numeric(factor(unlist(strsplit(gsub(" |[.]","","D. ADAMS"),"")),levels=LETTERS)))

Result:

42

Answer 20

Java

public class MainProgram {    
    public static void main(String[] args) {
        int[] the      = { 'T', 'h', 'e' };
        int[] most     = { 'M', 'o', 's', 't' };
        int[] creative = { 'C', 'r', 'e', 'a', 't', 'i', 'v', 'e' };
        int[] way      = { 'W', 'a', 'y' };
        int question   = '?';

        double x = -3.18906605923E-2;

        int The      = 0;
        int Most     = 0;
        int Creative = 0;
        int Way      = 0;

        for(int i : the) {
            The += i;
        }
        for(int i : most) {
            Most += i;
        }
        for(int i : creative) {
            Creative += i;
        }
        for(int i : way) {
            Way += i;
        }
        System.out.println((int)((The*x)-(Most*x)-(Creative*x)-(Way*x)-(question*x)));      
    }//SSCE
}//good1

Output:

42

Answer 21

SWI-Prolog, anyone?

?- X.

Output:

% ... 1,000,000 ............ 10,000,000 years later
% 
%       >> 42 << (last release gives the question)

This is even lazier than the Mathematica-calling-Wolfram-Alpha one, but hey!

Answer 22

Linux shell

Here’s something I wrote in 1999 and used as my Usenet signature back then.

echo "what is the universe"|tr "a-z " 0-7-0-729|sed 's/9.//g;s/-/+/'|bc

Edit: Ha! This was the 42nd answer.

Answer 23

PHP version:

 echo strlen("Douglas Adams")+strlen("born on")+array_sum(array(1,1,0,3,1,9,5,2));
 /* array(1,1,0,3,1,9,5,2) => March 11, 1952 */

JavaScript version:

 console.log("Douglas Adams".length + "born on".length + [1,1,0,3,1,9,5,2].reduce(function(previousValue, currentValue, index, array){return previousValue + currentValue;}));

 /* [1,1,0,3,1,9,5,2] => March 11, 1952 */

Output:

 42

Answer 24

dc

$ dc <<< "1 8 sc 1 5 lc *++p"
42

Trying to multiply 1+8 and 5+1 to get 42. It looks like that ignorance of operator precedence led to 42.

---

Python

>>> p = lambda x: x%2!=0 and True<<x
>>> sum(p(i) for i in range(0,6))

Output: 42

---

bash

(Multiplying 6 by 9)

$ echo "obase=13;6*9"|bc|figlet
 _  _  ____
| || ||___ \
| || |_ __) |
|__   _/ __/
   |_||_____|

---

C

#include <stdio.h>
int main(){printf("%d",fprintf(fopen("/dev/null","w"), "so-popularity-contest\b\b\b\b\b\b\b\b\b\b\b\b\b\b\b\b\b\b\b\b\b"));}

Output: 42

• Did you notice the use of backspace characters?
• Do you know what do those result in?

Answer 25

Brainf**k

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

Answer 26

C++

#include<iostream>
#include<conio.h>
using namespace std;
int main()
{
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)66<<(char)73<<(char)82;
    cout<<(char)84<<(char)72<<(char)32;
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)32<<(char)68<<(char)69;
    cout<<(char)65<<(char)84<<(char)72;
    cout<<(char)32<<(char)32<<'\n';
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)32<<(char)32<<(char)95;
    cout<<(char)95<<(char)95<<(char)32;
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)32<<(char)95<<(char)95;
    cout<<(char)95<<(char)95<<(char)95;
    cout<<(char)95<<(char)32<<'\n';
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)32<<(char)47<<(char)32;
    cout<<(char)32<<(char)32<<(char)124;
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)124<<(char)32<<(char)32;
    cout<<(char)95<<(char)95<<(char)32;
    cout<<(char)32<<(char)124<<'\n';
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)47<<(char)32<<(char)47;
    cout<<(char)124<<(char)32<<(char)124;
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)124<<(char)95<<(char)124;
    cout<<(char)32<<(char)32<<(char)124;
    cout<<(char)32<<(char)124<<'\n';
    cout<<(char)32<<(char)32<<(char)47;
    cout<<(char)32<<(char)47<<(char)32;
    cout<<(char)124<<(char)49<<(char)124;
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)32<<(char)32<<(char)47;
    cout<<(char)50<<(char)124<<'\n';
    cout<<(char)32<<(char)47<<(char)32;
    cout<<(char)47<<(char)32<<(char)32;
    cout<<(char)124<<(char)57<<(char)124;
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)84<<(char)79<<(char)32;
    cout<<(char)32<<(char)47<<(char)48;
    cout<<(char)47<<(char)32<<'\n';
    cout<<(char)47<<(char)32<<(char)47;
    cout<<(char)95<<(char)95<<(char)95;
    cout<<(char)124<<(char)53<<(char)124;
    cout<<(char)95<<(char)95<<(char)32;
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)47<<(char)48<<(char)47;
    cout<<(char)32<<(char)32<<'\n';
    cout<<(char)124<<(char)95<<(char)95;
    cout<<(char)95<<(char)95<<(char)95;
    cout<<(char)124<<(char)50<<(char)124;
    cout<<(char)95<<(char)95<<(char)124;
    cout<<(char)32<<(char)32<<(char)47;
    cout<<(char)49<<(char)47<<(char)32;
    cout<<(char)32<<(char)32<<'\n';
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)124<<(char)32<<(char)124;
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)32<<(char)47<<(char)32;
    cout<<(char)47<<(char)32<<(char)32;
    cout<<(char)32<<(char)32<<'\n';
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)124<<(char)32<<(char)124;
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)47<<(char)32<<(char)47;
    cout<<(char)95<<(char)95<<(char)95;
    cout<<(char)95<<(char)32<<'\n';
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)32<<(char)32<<(char)32;
    cout<<(char)124<<(char)95<<(char)124;
    cout<<(char)32<<(char)32<<(char)124;
    cout<<(char)95<<(char)95<<(char)95;
    cout<<(char)95<<(char)95<<(char)95;
    cout<<(char)95<<(char)124<<'\n';
    getch();
    return 0;
}  

output

Answer 27

JavaScript

window.location = "https://www.google.nl/search?q=the+answer+to+life+the+universe+and+everything";

Outputs 42.

Answer 28

J

Symmetric one-liner without alphanumeric chars.

   _<.>.>_ (=(+^:]) ~=(-*-)=~ ([:^+)=) _<.<.>_

Outputs 42.

The main computation is:

ceiling( 1 + ( 1 - e ^ 2 ) ^ 2 ) = 42

Answer 29

JavaScript

The ASCII code for *, which for most programmers stands for "everything", is 42. +!"The End of the Universe" evaluates to 0.

String.prototype.answer = function() { alert(this.charCodeAt(+!"The End of the Universe")); };
'*'.answer();

Answer 30

PHP

Ask WolframAlpha. Here's some code that uses the WolframAlpha API to retrieve the result of a specific search query:

<?php

$searchTerm = "What's the answer to life, universe and everything?";

$url = 'http://api.wolframalpha.com/v2/query?appid=APLTT9-9WG78GYE65&input='.urlencode($searchTerm);
$xml = file_get_contents($url);

$xmlObj = simplexml_load_string($xml);
$plaintext = $xmlObj->xpath('//plaintext')[1];
$answer = preg_replace('/\D/', '', $plaintext);

echo $answer;

Output:

42

Working demo