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  1. 26 Aug '06 21:25
    How many squares do you think are on a chess board?
  2. 26 Aug '06 21:33
    There are:
    8*8 squares of side 1.
    7*7 squares of side 2.
    6*6 squares of side 3.
    5*5 squares of side 4.
    4*4 squares of side 5.
    3*3 squares of side 6.
    2*2 squares of side 7.
    1 square of side 8.

    Making a total of 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 204.
  3. 26 Aug '06 21:44
    Originally posted by Fat Lady
    There are:
    8*8 squares of side 1.
    7*7 squares of side 2.
    6*6 squares of side 3.
    5*5 squares of side 4.
    4*4 squares of side 5.
    3*3 squares of side 6.
    2*2 squares of side 7.
    1 square of side 8.

    Making a total of 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 204.
    I didn't think that the first person to reply would get it right.
  4. 26 Aug '06 23:17
    Originally posted by tomtom232
    I didn't think that the first person to reply would get it right.
    Not the first time it's been asked here.
  5. 26 Aug '06 23:23
    64 if thtas what you mean
  6. 26 Aug '06 23:24 / 1 edit
    Originally posted by moon 111
    64 if thtas what you mean
    the first answer was correct.
  7. 27 Aug '06 12:55
    posser's and puzzles anyone?
  8. 27 Aug '06 13:04 / 2 edits
    Originally posted by Shinidoki
    posser's and puzzles anyone?
    Two posser's please.
  9. 27 Aug '06 19:28
    Here's a slightly trickier one, one of Martin Gardner's.

    Imagine you have an empty chess board in front of you and 32 2x1 rectangle shaped pieces of wood, each one composed of a black square glued onto a white square. The rectangles are such that each one will exactly cover two adjacent squares of the chess board.

    With your 32 rectangular pieces it is easy to see that you can cover the chess board so that every black square on the chess board has a black side of a rectangle over it, and every white square on the board has a white side of a rectangle over it.

    This is the puzzle - suppose I take away one of your rectangles, leaving you with 31, and obliterate the a1 and h8 squares on the chess board. Is it possible to cover the board (except for a1 and h8) in the same way as before?
  10. 27 Aug '06 19:33 / 2 edits
    Originally posted by Fat Lady
    Here's a slightly trickier one, one of Martin Gardner's.

    Sorry, this was posted just the other day (in the correct forum).
    http://www.timeforchess.com/board/showthread.php?threadid=50465

    So before this thread dies, how many rectangles are there on a chessboard?
  11. 29 Aug '06 13:35
    Originally posted by ThudanBlunder
    So before this thread dies, how many rectangles are there on a chessboard?
    1x2=2
    3x2=6
    4x2=8
    5x2=10
    6x2=12
    7x2=14
    8x2=16

    Total: 68

    1x3=3
    2x3=6
    4x3=12
    5x3=15
    6x3=18
    7x3=21
    8x3=24

    Total: 99

    1x4=4
    2x4=8
    3x4=12
    5x4=20
    6x4=24
    7x4=28
    8x4=32

    Total: 128

    1x5=5
    2x5=10
    3x5=15
    4x5=20
    6x5=30
    7x5=35
    8x5=40

    Total: 155

    1x6=6
    2x6=12
    3x6=18
    4x6=24
    5x6=30
    7x6=42
    8x6=48

    Total: 180

    1x7=7
    2x7=14
    3x7=21
    4x7=28
    5x7=35
    6x7=42
    8x7=56

    Total: 203

    1x8=8
    2x8=16
    3x8=24
    4x8=32
    5x8=40
    6x8=48
    7x8=56

    Total: 224

    224 + 203 + 180 + 155 + 128 + 99 + 68 = 1057
  12. Standard member XanthosNZ
    Cancerous Bus Crash
    29 Aug '06 14:08
    Originally posted by lausey
    1x2=2
    3x2=6
    4x2=8
    5x2=10
    6x2=12
    7x2=14
    8x2=16

    Total: 68

    1x3=3
    2x3=6
    4x3=12
    5x3=15
    6x3=18
    7x3=21
    8x3=24

    Total: 99

    1x4=4
    2x4=8
    3x4=12
    5x4=20
    6x4=24
    7x4=28
    8x4=32

    Total: 128

    1x5=5
    2x5=10
    3x5=15
    4x5=20
    6x5=30
    7x5=35
    8x5=40

    Total: 155

    1x6=6
    2x6=12
    3x6=18
    4x6=24
    5x6=30
    7x6=42
    8x6=48

    Total: 180

    1x7=7
    2x7=1 ...[text shortened]... 24
    4x8=32
    5x8=40
    6x8=48
    7x8=56

    Total: 224

    224 + 203 + 180 + 155 + 128 + 99 + 68 = 1057
    I have no idea what you are doing here. This problem can be bruteforced (there are 2*(9-m)*(9-n) mxn rectangles on a chessboard where m!=n [if so then remove the 2*]).

    However a more elegant solution is that a chessboard has 9 vertical boundaries and 9 horizontal boundaries and to form a rectangle we choose any two of each.

    There are 36 ways of choosing 2 from 9 (9C2) and therefore there are 36^2 rectangles.

    36^2 = 1296
  13. 29 Aug '06 15:27 / 2 edits
    Originally posted by XanthosNZ
    I have no idea what you are doing here. This problem can be bruteforced (there are 2*(9-m)*(9-n) mxn rectangles on a chessboard where m!=n [if so then remove the 2*]).

    However a more elegant solution is that a chessboard has 9 vertical boundaries and 9 horizontal boundaries and to form a rectangle we choose any two of each.

    There are 36 ways of choosing 2 from 9 (9C2) and therefore there are 36^2 rectangles.

    36^2 = 1296
    I realised I made a mistake in my bruteforce approach. Wrote an algorithm that did indeed get 1296. If you are talking about oblongs, it is 1092 (previously assumed that rectangles aren't squares).

    I like your elegant solution though.

    EDIT (short delphi code of bruteforce if anyone is interested):

    procedure TForm1.Button1Click(Sender: TObject);

    var
    n, m, total: integer;

    begin

    total := 0;

    for n := 1 to 8 do
    for m := 1 to 8 do
    if m n then // this is just for oblongs. For all rectangles, remove this line
    total := total + (m*n);

    showmessage(inttostr(total));

    end;

    EDIT2: meant to be an "m greater or less than n" in code, but did not render here, as well as indentation.