RUBIKS CUBE,  would someone please send me complete details and instructions?
473157
http://www.wrongway.org/cube/solve.html\n\nJust input the current colors of your cube into the site program, and hit solve. It gives you a STEP BY STEP solution to your cube. Just make sure you input the colors correctly.
254189
cube = a 3  \n\nrectangular prism = a b c  \n\nirregular prism = b h  \n\ncylinder = b h = pi r 2 h  \n\npyramid = (1/3) b h  \n\ncone = (1/3) b h = 1/3 pi r 2 h  \n\nsphere = (4/3) pi r 3  \n\nellipsoid = (4/3) pi r1 r2 r3
781998
27 cubes, then 81, then 243, then 729...
221698
A Rubik's Cube can have (8! × 38−1) × (12! × 212−1)/2 = 43,252,003,274,489,856,000 different positions (~4.3 × 1019), about 43 quintillion, but it is advertised only as having "billions" of positions, due to the general incomprehensibility of that number.
869806
213\n321\n132\n\nall rows, columns, diagonals = 6\nnice try rt11guru but that wasn't your original answer\nmirror image:\n312\n123\n231\nupside down:\n231\n123\n312\nmirror image of that:\n132\n321\n213\n\nfuntym888 what do you mean it cant be done? dont you read the other answers before you answer?  it certainly can be done, in 4 different ways no less, as I just showed!  you simply take a 1-9 magic square and replace 4,5,6 by 1,2,3 and 7,8,9 by 1,2,3 respectively.  And it is "really" a magic square, it just isnt a normal magic square.  You should read your own sources as well!
322239
Try Cube: http://www.cubeengine.com
977457
1 to 64k.. so there exists about 40 cubes (1^3, 2^3...  39^3, 40^3)..  and sum of 4 cubes. that's like. "40 choose 4" (combinatorics) so that comes out to be about 91390 combinations of 4 cubes and then you just find the cube roots of each. each should have its own cube root. and 91 thousand isn't too bad for an average computer to handle. i suggest you program a computer to do it for you. it wouldn't be hard, just basic programming language.\n\nif you cant do that, then i suggest you try to narrow down what you're looking for...
465982
It cannot end in the middle.\n\nLet's look at the cube.  \nIt's divided 3x3 to give you 27 subcubes.\nThe middle cube is therefore only one deep from any side.\nThere are eight subcubes on each corner of the cube.\nThere are six subcubes in the middle of each face of the cube.\nWe'll put all these subcubes into a group.\nCall it group A. 8+6=14 subcubes.\n\nAnd there are all the other cubes.\nThere are 27 subcubes.  14 are in group A.\nSo there are 27-14=13 of these.  \nWe'll group them together and call it group B.  13 subcubes.\n\nSome things become clear at this point.\nThe mouse could progress always ahead. So...\nEach time the mouse eats one of the group A subcubes, it must go to one of the group B subcubes.\n\nEach time it eats one of the group B subcubes, it must go to one of the group A subcubes.\n\nEven if it goes sideways, to a subcube in the same group (that's only possible for group B subcubes because they are the only subcubes where members of the group can be adjacent), what happens?\n\nThe number of subcubes in group B will reduce by two instead of one.\nThe number of subcubes in group A will remain the same as before.\n\nHowever the mouse moves and eats, there will always be more group A subcubes than there are group B subcubes.\n\nThe middle subcube belongs to group B.\nIt's one of the group of 13 subcubes.\n\nWhichever group the mouse eats from on the first day, he will always have one of group A subcubes left.\nBecause there are always more group A subcubes than group B subcubes.\n\nI'll repeat all that.\n27 subcubes, one per day = 27 days. \n13 subcubes in group B.\n14 subcubes in group A.\n\nIt is impossible for the mouse to have eaten the last group B subcube and there be no group A subcubes left.\nBecause there are always more group A subcubes than there are group B subcubes.\n\nSo the last subcube the mouse eats will always be one of group A.\nA corner subcube or a subcube that's in the middle of one of the faces.\nNot the subcube in the middle.\nQED, I think.
121959
I CRACKED MY HEAD OPEN LAST NIGHT TRYING TO SOLVE THIS ..... PUZZLE. THAT CRAZY 15 JUST WON'T BUDGE. I JUST COULDN'T FIND A WAY TO GET HIM IN THE RIGHT SPOT. \nsorry. and i did it on paper, too. try that for a change (desperate). i really like puzzles, but this one..... drove me crazy :((((((
557152
its easy....go to http://www.opencube.com and design the cool java menu. if you have little more knowledge you can go to http://www.milonic.com and design the menu. If you still find problem you can simnply send me the menu items and submenu items, i will design a menu for u.\n\nmail me at anshupp@gmail.com
4519
(4/3)*pi*r^3\n\n[four thirds pi r cubed]\n\ngood luck!
774727
try http://www.rube-goldberg.com/
686243
Email me; email on profile. I will give you complete instructions.
1459
Just randomly enter numbers according to the rules of the puzzle, but using a blank grid. Start with one or two per 9-square block. To make the puzzle difficult, only enter a few numbers, and add more to make it easier.\n\nIt should be quite simple for a computer program to make a Sudoku puzzle, since there are basic rules to follow. I'm sure the "made by hand" claim is for marketing purposes (though I'm sure it's true).
903014
What you do is take a really easy idea, and then complicate it.  So if you have some ideas, narrow it down to one.  Oh, and you should include a cat in there somehow.  It ain't Goldberg unless it's got wildlife in it!\n\nHere's a complicated way to look at some cool Rube Goldberg devices featured on a Japanese children's show.  At the bottom of the article on Pitagora Suicchi, there's a link to youtube which should let you see the things in action.  VERY cool!\n\nhttp://en.wikipedia.org/wiki/Pitagora_Suicchi
80762
Here are some code snippets that might be useful.\nNote: I wrote these down, did _not_ compile them, or test them.\n\nThe deck of cards is a array of 52 unsigned integers,\nThey can take a value of 0-52.\n\n//Create an array to hold the deck:\n\nunsigned int Deck[52];\n\nfor(unsigned int i = 0; 1 < 52; i++)\n{\n   Deck[i] = i;\n}\n\n// Shuffle the deck\nsrand(); // Initialize the random number generator\n\nunsigned int temp;\n\n// The shuffle works this way. First round, every card gets a chance\n// to be in slot 51. Second round, each of the remaining cards gets a \n// chance to be in slot 50, third round etc.\n\n\n// Note: This method allows an array value to be swapped with itself, i.e. not moved, which is what you want to get all possible shufflings of the deck.\nfor(unsigned int j = 52; 0< j; j--)\n{\n   // Generate an index in the range 0-j\n   unsigned int k = rand( void ) % (j);\n\n   temp = Deck[k];\n   Deck[k] = Deck[j-1];\n   Deck[j-1] = temp;\n}\n\n///////////////////////////////////\n\nunsigned int GetFaceValueOfCard(unsigned int n)\n{\n  return Deck[n] % 13 + 1;    // Returns a value 1-13\n}\n\n// If you want to get fancy:\n\nchar GetSuitOfCard(unsigned int n)\n{\n   char Suit[4] = "HDCS";\n   return Suit[n/13];\n}
2243
Unless you know about group theory, there is going to be no mathematics involved to solve the Magic Cube(better known as Rubik's Cube). To read about relation between group theory and Rubik's Cube, visit http://web.usna.navy.mil/~wdj/rubik_nts.htm\n\nThere are a lot of different algorithms to solve Rubik's cube. Layer by Layer is the most suitable for beginners. Fridrich method is popular among speedcubers. One that I use is mixture of these two algorithms and my average time is about 70 seconds. Good speedcubers have average of about 15 seconds.\nTo find out more about speedcubing visit http://www.speedcubing.com/\n\nIf you just want to solve the cube without learning how to solve it, you can visit http://www.roobik.com/cgi-bin/rubix/rubix.cgi\nto repair your cube.