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More generally, an a-by-b rectangle can be covered with square tiles of side length c only if c is a common divisor of a and b. For example, a 24-by-60 rectangular area can be divided into a grid of: 1-by-1 squares, 2-by-2 squares, 3-by-3 squares, 4-by-4 squares, 6-by-6 squares or 12-by-12 squares.
To do so, one goes outside the confines of the square area defined by the nine dots themselves. The phrase thinking outside the box, used by management consultants in the 1970s and 1980s, is a restatement of the solution strategy. According to Daniel Kies, the puzzle seems hard because we commonly imagine a boundary around the edge of the dot ...
The pieces to be moved may consist of simple shapes, or they may be imprinted with colours, patterns, sections of a larger picture (like a jigsaw puzzle), numbers, or letters. Sliding puzzles are essentially two-dimensional in nature, even if the sliding is facilitated by mechanically interlinked pieces (like partially encaged marbles) or three ...
For illustration, a 24×60 rectangular area can be divided into a grid of: 1×1 squares, 2×2 squares, 3×3 squares, 4×4 squares, 6×6 squares or 12×12 squares. Therefore, 12 is the GCD of 24 and 60. A 24×60 rectangular area can be divided into a grid of 12×12 squares, with two squares along one edge (24/12 = 2) and five squares along the ...
A domination (or covering) problem involves finding the minimum number of pieces of the given kind to place on a chessboard such that all vacant squares are attacked at least once. It is a special case of the vertex cover problem. The minimum number of dominating kings is 9, queens is 5, rooks is 8, bishops is 8, and knights is 12.
There are several ways to find the greatest common divisor of two polynomials. Two of them are: Factorization of polynomials, in which one finds the factors of each expression, then selects the set of common factors held by all from within each set of factors. This method may be useful only in simple cases, as factoring is usually more ...
The best calculation of symmetry is achieved by limiting one pawn to 24 squares in the rectangle a2-a7-d7-d2. All other pieces and pawns may be located in any of the 64 squares with respect to the pawn. Thus, an endgame with pawns has a complexity of 24/10 = 2.4 times a pawnless endgame with the same number of pieces.
However, there are three distinct ways of partitioning a square into three similar rectangles: [1] [2] The trivial solution given by three congruent rectangles with aspect ratio 3:1. The solution in which two of the three rectangles are congruent and the third one has twice the side length of the other two, where the rectangles have aspect ...