A Sudoku-2 puzzle is similar to the usual Sudoku puzzle, except for the fact that it includes 16 squares on a 2*2 board, rather than the 9*9 one, and it's played in the same way (of course you can only use numbers ranging from 1 to 4). I have estimated the possible results of a solved Sudoku-2 into 4*3*2*3*2*2*2 = 4!*3!*2!*2 = 576 but I have followed a heuristic method and I have not been able to find a way to prove it mathmatically. Is the result correct? And if it is, is there a way to prove it?
210029
I normally hate referring people to wikipedia articles, since it is so often inaccurate, but this one is quite good (link is below).  If you scroll down, there is a table that gives just what you're looking for.\n\n2x2 grid: 288 possible solutions\n3x3 grid: 6.7*10^21 possible solutions!
621993
The modular (and clock) arithmetic answers are correct, but I think they miss the point a bit, since in any modular arithmetic you still have 2+2 = 4 mod n. Let's suppose we are using the natural numbers with the usual addition.\n\nThe recurring decimal answers are just stupid, 1 = 0.9999... recurring (as shown in fact by the first answer here), so all of these answers are either irrelevant or wrong.\n\nThe '2' is just a symbol arguments possibly have some merit, but you should be careful. In Indianna in 1872 (or thereabouts) it was ruled by State legislature, that the value of pi is 4. Obviously this creates problems, if you go to Indianna you won't find many buildings that were designed in 1872. (Their mistake of course was that they didn't alter the representation of all the other numbers they where using accordingly).\n\nI'm not really a logician so I don't know the answer to this off the top of my head, I'm pretty sure that we don't have a satisfying answer here (sorry if we have but I missed it). \n\nPerhaps a good place to start looking for a serious discussion on this kind of topic would be to understand what a Peano system is (wiki it, I'm not explaining that here, 'tis an axiomatic description of the natural numbers). I've also heard that Bertrand Russell gave a proof that 1+1 =2 in Principa Mathematica, although I haven't read that and I don't know precisely where you can find it.
56266
Definitions of  sudoku on the Web:\n\n    * Sudoku (: 数独, sūdoku), sometimes spelled Su Doku, is a placement puzzle, also known as Number Place in the United States. The aim of the puzzle is to enter a number from 1 through 9 in each cell of a grid, most frequently a 9×9 grid made up of 3×3 subgrids (called "regions"), starting with various numbers given in some cells (the "givens"). Each row, column and region must contain only one instance of each number. ...\n      en.wikipedia.org/wiki/Sudoku
19889
First we calculate all squares:\nWe have 9*9 one by one squares, 8*8 two by two squares and\nSo total squares are: 9^2+8^2+7^2+...+1^2=285\n Where 1^2 stands for 9 by 9 square.\n\nNow we must calculate all squares containing the red spot:\n\nWe have one "one by one" square, 4 " two by two" squares,\n9 "for 3*3", 9 for "4*4", 9 for "5*5", 9 for "6*6", 9 for "7*7", 4 for "8*8" and 1 for "9*9";\nWhich is count to 55. \n\nSo the answer is 285-55=230
54139
That's an extremely good question. \n\nThe answer is simple. Before that, i will explain what your question means to those who don't get. Well, it means that we must get 24 using 1,3,4 and 6 any number of times however involving operations +,-,* and / only once. \n\n  Then, the answer is very simple :\n\n  {[(6*6)/(3-1)]+[6]}  =  24\n\n---------------------------------------------------------\nWOKRING :\n\n {[(6*6)/(3-1)]+[6]}   =  {[(36)/(2)]+[6]}\n                       =  18 + 6\n                       =  24\n\n----------------------------------------------------------\n\nI hope you post more questions like these. They evoke a lot of thought. I hope the solution serves th purpose.
99914
im a geo student, but i havent done proofs for a long time sorry. ok\n2. false, because squares are parallelograms with 4 congruent sides and 4 right angles, while kites have 2 pairs of congruent sides and only 1 pair of congruent angles that are NOT right angles\n3. false, because a rhombus is a parallelogram with 4 congruent sides, while a kite is any thing BUT a parallelogram. enuff said\n4. false, because kites are quadrilaterals with only two pairs of congruent sides, while the quadrilateral mentioned has 3 congruent sides\n5. false, for example- do trapezoids have 4 congruent sides, and 4 right angles, and is a parallelogram (characteristic of squares)? i think NOT\n6. read above- trapezoids are not parallelograms.
297900
no. square root 32 = square root 4*4*2 = 4 square root 2\n\nbreak down the 32 into parts that will come out of the root. Then if there are two numbers inside, they become 1 number outside. If there is a loner number inside, then it stays there.
665102
6√2\n\nRemember it's a square so the sides are all the same length, and the area is s^2\n\nso s^2=36==>s^2-36=(s-6)(s+6)=0 so s is either 6 or -6.  -6 makes no sense here, so it is obviously 6.\n\nNow, since we have a square we can "draw" (imagine) a line at the diagonal, and that makes a right triangle (since the corners of a square are all 90º) with legs 6 and 6.  using the Pythagorean relation, we have d^2=s^2+s^2=6^2+6^2=72.  Thus d^2-72=(d-6√2)(d+6√2)=0.  Thus d is either 6√2 or -6√2, and again is obviously the positive one :)
540250
I was able to draw 16 different forms (8 with their symmetrical reversed forms).  There may be more but I don't have the time to work on it.\n\nI took another look at the problem and realized that in addition to the left-right symmetry there is also a top-bottom symmetry that adds at least another 14 answers to the 16 that I came up with earlier.  So there are at least 30.\n\nCould be more, but it is my bed time.
556966
Yes, in that division process you are writing the binary expression of the first number, so when you add the elements of the second column, this is the result (as in your example N=64):\n\n27*N=(2^4+2^3+0*2^1+2^0)*N\n=N*2^4+N*2^3+ (0*N*2^2) +N*2^1+N*1\n\nand this is the reason why you have to duplicate N each time and neglect some terms (those corresponding to 0 in the binary form for your first number).\n\nVery good trick!
575781
You can try http://www.web4sudoku.com they offer a free sudoku puzzle script as long as you leave the link at the bottom.
132148
it's 8*8 + 7*7 + ... +2*2 + 1*1. \n\nso for the chessboard it would be 204. :]
420160
Let us call the first 2 "2 sub 1", the second "2 sub 2" ... same thing for the 6's.\n\nLet us say that "2 sub 1" is distinct from "2 sub 2", and so on.\n\nWe have 8 choices for the first number, seven choices for the second number, etc. This means that there are 8*7*6*5*4*3*2*1 choices, or 8!. \n\nNow if you look at any number, take all of the 2's out. You see that you can rearrange the 2's in 3! different ways (including the original arrangement) using the procedure in the paragraph above. If we take out the 2's rearrange them, and put them back in we have created the a new arrangement. So for each arrangement of numbers, we have counted it 3! times. So we divide by 3!. The same thing for the sixes, we divide by 2!.\n\nThe result it (8!)/((3!)(2!)) = 3360 choices.
103758
I've seen these before, I think.  You're questions is a little vague but I'll give it a shot.\n\nFor instance, 81 K on a P is actually 81 Keys on a Piano?\n\nIf this is what you're talking about there is no real method that I'm aware of to solve them, except to think about the distinct number at the beginning, as this is what yields the greatest clue.  \n\nFor instance, 24 hours, 7 days, 60 minutes/seconds, 100 cents, 365 days, etc.  I hope this helps!
408290
Firstly, it has to start with either a 1 or a 2, and end with a 2 or a 6. So we just take each case one at a time.\n1..2: that leaves 23556, so there are 5!/2! ways, ie 60 ways.\n1..6: that leaves 22355, so there are 5!/(2!2!) ways, ie 30 ways.\n2..2: that leaves 13556, so there are 5!/2! ways, ie 60 ways.\n2..6: that leaves 12355, so there are 5!/2! ways, ie 60 ways.\n\nNote that in each case there are 5! ways of arranging the other 5 numbers, but we need to divide by 2 if there is a pair of repeated numbers.\nThat gives a total of 210.\n\n(note that both of the other answers given so far are wrong - its not 360, because that person has forgotten they have to be even; and its not 120. They counted 2x2 possibilities for the first and last digits which is correct, but one of those cases already uses both 2s, so you *don't* want to divide by 2 in the end for that particular case (you're not counting them twice are you!)\nTheres really no other way of doing it than splitting them into cases like I did.
559964
You place all 9 squares. There are 9! ways to place the pieces(without being concerned about each pieces orientation)\nFor each of the 9!, each piece can be rotated into 9 positions. So there are going to be 4^9 possiblities.\n\nOnce you find the correct solution, you can rotate the entire puzzle 90 degrees 4 times.\n\nso the correct number is 9!* 4^9 /4\n9!*4^8\n= 23,781,703,680\n\nNOTE--To the people who are answering 9!*4^9\nWhat would the possible solutions be if there was only one piece?\nWould it be 1!*4^1 =4\nNo, it would be 1/4 of that.\n\nIf there is a similar puzzle of NxN size\nthe solution would be (N^2)! *4^(N^2 -1)
47518
While it isn't possible, and all these "proofs" are fake (i.e. they contain a methematical error), I'll provide you one just for fun.\n\nLet a and b be two identical numbers (a = b)\n\na^2 = ab\na^2 + (a^2 - 2ab) = ab + (a^2 - 2ab)\n2(a^2 - ab) = a^2 - ab\n\nDividing both sides by (a^2 - ab) gives you\n\n2 = 1\n\n2 = 1 + 1 (common mathematical knowledge)\n\n2 = 2 + 1 (since 1 = 2)\n2 = 2 + 2 (again)\n1 + 1 = 2 + 2\n2 + 2 = 2 + 2\n1 + 1 + 2 = 2 + 2\n2 + 1 + 2 = 2 + 2\n5 = 2 + 2\n\n2 + 2 = 5