i know it's made up of 8*8=64 squares but that's not quite the answer i'm looking 4 here. How many total squares are there and is there a formula to calculate that number for any number of given squares?
132148
it's 8*8 + 7*7 + ... +2*2 + 1*1. \n\nso for the chessboard it would be 204. :]
664297
2[w+ {2w+8}]= 64\n3w+8 = 32\nw= 8 \nyour are right
376863
It does look like squares of 1,2,3,4,5 so the square of 1 would be 1.
14470
Let No of square is N\nPerimeter = 2xN + 2 \n\nAssumptions:\nThey are arranged in a row.\nThey are of same size.\nThey are of UNIT size.
187083
ANSWER : 25 square units.\n\n---------------------------------\n\nFirst off, draw a picture. \nSo, you have one point on the (0,0), one on the x-axis, one on the y-axis, and one on the diagonal line.\n\nLet's call the length of the horizontal line segment of the rectangle w.\n\nThe other side will then be (10-w).\n\nSo, the area of the rectangle is A = w*(10-w).\n\nIf you graph this, there is a maximum at the point (5,25), so 25 is the max area.\n\nTo verify this, take the derivative of A with respect to w, and set equal to zero.\nA' = 10 - 2w = 0\n2w = 10\nw = 5\nA = 5(5) = 25
5659
According to this page..\nhttp://www.shef.ac.uk/~pm1afj/sudoku/\n..there are 6,670,903,752,021,072,936,960 different completed 9x9 Sudokus.
157023
8 X 11=88\n\n88 X 154 = 13552\n\n13552 +76= 13628\n\n13628 -3 = 13625\n\n13625 + 7= 13632\n\nTHERE THERE ALL BETTER
175516
This problem is certainly tractable, you just need to\nfind a consistent/methodical way of counting the\nrectangles -- without missing any and without counting\nany of them multiple times.\n\nThe way I chose is to define each rectangle as being\ndefined by the square in the upper left corner and the\nsquare in the lower right corner.  Consider the following\nsimplified 2x3 case (where each square is labelled by\nits X and Y coordinates:\n\n0,0     0,1   0,2\n1,0     1,1   1,2\n\n\nThis 2X3 example has the following rectangles starting\nat 0,0.  Remember, each rectangle is defined by the\nsquare in its upper left and the square in the lower right:\n0,0   0,0\n0,0   0,1\n0,0   0,2\n0,0   1,0\n0,0   1,1\n0,0   1,2\n\nThen it also has the set anchored in the upper left by 0,1:\n0,1   0,1\n0,1   0,2\n0,1   1,1\n0,1   1,2\n\nThen it has the set anchored in the upper left by 0,2:\n0,2   0,2\n0,2   1,2\n\nThen those anchored by 1,0:\n1,0   1,0\n1,0   1,1\n1,0   1,2\n\nThose anchored by 1,1:\n1,1   1,1\n1,1   1,2\n\nAnd finally those anchored by 1,2:\n1,2   1,2\n\n\nRewriting the matrix another way, showing how many\nrectangles are anchored by each location in the matrix:\n\n2x3    2x2   2,1\n1x3    1x2   1x1\n\n\nand then doing the math:\n\n6      4     2\n3      2     1\n\nfor a total of 6+4+2+3+2+1 = 18\n\n\nDoing the same analysys for a 4x5 gives:\n\n4x5   4x4   4x3   4x2  4x1\n3x5   3x4   3x3   3x2  3x1\n2x5   2x4   2x3   2x2  2x1\n1x5   1x4   1x3   1x2  1x1\n\nor rewriting after doing the multiplication:\n\n20    16    12    8    4\n15    12     9    6    3\n10     8     6    4    2\n 5     4     3    2    1\n\nAdding up the columns gives:\n\n50    40    30   20   10\n\nFor a total of 50+40+30+20+10 = 150 total rectangles
550054
OK, I think what you're saying is that all the corners have to be colored and that if you can rotate a square to superimpose it on another square, or if they're mirror images, then they're the same.  You didn't say that all colors have to be used, so I'll assume they don't.  That said...\n\nStart with squares in which all four corners are the same color.  There are obviously 4 of these.\n\nNow take squares in which three corners are one color and the fourth is different.  There are 12 of these (three squares can be any of four colors, and the odd one can be any of the other three.)\n\nNext, take squares in which two corners are one color and two are another.  There are six color combinations (red-white, red-blue, red-green, white-blue, white-green, blue-green) and each one can have two patterns:  either the same-colored corners are next to each other, or they're diagonally opposite.  So there are 12 of these.\n\nNext, consider square in which two corners are the same and the others are two different colors.  For each combination of a major (2-corner) color and two minor colors, there are only two patterns:  same colors adjacent and same colors opposite.  Whichever way you arrange the odd colors, it will be the mirror image of the other way and therefore identical.  So for each major color, you can pick any two of three minor colors.  This is the same as leaving out any one, so there are three options for minor colors.  4*3*2 patterns is 24 different squares.\n\nFinally, consider squares in which each corner is a different color.  Now, pick a color and a corner.  Two of the other colors will be adjacent to that corner, and one will be diagonally opposite.  For any choice of opposite colors, either arrangement of adjacent colors will just make another mirror image.  So the only choice here is which color will be opposite.  That means there are three different 4-color squares.\n\nNow, add it all up: 4 one-color squares, (12 + 12 =) 24 two-color squares, 24 three-color squares, and 3 four-color squares, equals 55 unique squares.
585231
To solve this problem, you need to complete the square.\nx^2 can stay by itself.\ny^2-16y+64 since this qequals to (y-8)^2\nTherefore, since 64 is added to one side, 64 must be added to the other. 2004+64=2068\nx^2+(y-8)^2=2068\nNow divide both sides by 2068\n(x^2)/2068+(y-8)^2/2068=1\n\nNow, this is a circle. You should be fine from now on.
391900
28! I just did it on paper putting 8 dots and connecting each one with the rest. There is a math formula for it but I don't remember it!
10895
There are 65 Rectangles on a regular checkers checkerboard.  (64 playing squares plus the board itself)\n\nFor the poor soul that told you that a square is not a rectangle... A Square is defined as "A Rectangle with 4 equal sides"
514452
(a^2 - 4^2)^2 - 4(a^2 - 4) - 32 = 0\n(a^2 - 16)^2 - 4(a^2 - 4) - 32 = 0\na^4 - 32a^2 + 256 - 4a^2 + 16 - 32 = 0\na^4 - 36a^2 + 240 = 0\na^4 - 36a^2 = -240\na^4 - 36a^2 + 324 = 84\n(a^2 - 18) = +/- 84^(1/2)\na^2 = 18 +/- 84^(1/2)\na = +/- (18 +/- 84^(1/2))\n\nSo, there'll be four answers. Also, of course, x^(1/2) means the square root of x. Hope this helps, and good luck.\nP.S. Completing the square is always fun.
108763
False, a square is a regular polygon.  Say each side is 4 units long.  Use the pythagorean theorem to solve for the diagonal.  So 4^2+4^2=x^2, or the square root of 32 which is between 5 and 6.  The diagonal would have to be about 5.7 (mental math) which is not 4.
127108
9^8 is 9 times itself 8 times\n9^8\n= 9*9*9*9*9*9*9*9*9\n= 43046721
378216
The answer is 78:\n\n1² + 2² + 3² + 8² = 1 + 4 + 9 + 64 = 78\n2² + 3² + 4² + 7² = 4 + 9 + 16 + 49 = 78\n1² + 4² + 5² + 6² = 1 + 16 + 25 + 36 = 78\n\nI did this creating a spreadsheet with the first 9 squares.  There were 126 ways (9 choose 4) ways to pick the four squares which then resulted in a sum.  I sorted the sums and found the one that repeated itself 3 times (78).\n\nThe answer is 78.
479157
I think you are referring to magic numbers making up a magic square wherein you arrange such numbers in a grid and get the same sum for all the numbers in the same row, same column and diagonal.\n\nFor example, take the numbers 1 -9 and arrange them in a 3 x 3 grid (3 rows and 3 columns).  Set the 5 in the middle and arrange the rest around it so as to get a total of 15.  You should get something like this arrangement.\n\n     6     1     8     =     15\n     7     5     3     =     15\n     2     9     4     =     15\n\n     =     =     =      two diagonals each add up to 15 as well\n    15    15   15