Challenging Mathematical Problems with Elementary Solutions, 第 1 卷Holden-Day, 1964 - 440 頁 |
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第 1 到 3 筆結果,共 26 筆
第 32 頁
... length of CD divided by the length of AB , i.e. the fraction of the total length contained in the interval CD . = To return to the original problem , we note that the length of the smaller of the pieces into which the rod is broken will ...
... length of CD divided by the length of AB , i.e. the fraction of the total length contained in the interval CD . = To return to the original problem , we note that the length of the smaller of the pieces into which the rod is broken will ...
第 213 頁
... length of each of these three pieces be less than the sum of the lengths of the other two pieces . Since the sum of the lengths of the pieces is 1 , this condition is equivalent to the requirement that each of the three pieces be less ...
... length of each of these three pieces be less than the sum of the lengths of the other two pieces . Since the sum of the lengths of the pieces is 1 , this condition is equivalent to the requirement that each of the three pieces be less ...
第 215 頁
... length is replaced by an arbitrary length a . As in the first solution of problem 96 , we will characterize all possible outcomes of the experiment in question in terms of two numbers AK = x and AL = y , where K and L are the break ...
... length is replaced by an arbitrary length a . As in the first solution of problem 96 , we will characterize all possible outcomes of the experiment in question in terms of two numbers AK = x and AL = y , where K and L are the break ...
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A₁ A₂ An+m arrangements b₁ B₂ binomial coefficients binomial theorem bishops black squares C₁ chessboard chord circle coefficient color column compute the number Consequently consider corresponding customers denote determine the number diagonals digits dihedral angle divided divisible draw equally likely possible equation equidistant equivalence classes exactly example experiment favorable outcomes follows formula given Hence inclusion and exclusion intersection k-gons knights L₁ length mathematical induction maximum number n-gon number of different number of favorable number of paths number of shortest obtain pairs partition passengers plane points A1 polygons positive integers possible outcomes Pr{E probability theory problem 54 prove queens rectangle relatively prime remaining required probability rooks segment selected at random sequence shortest paths side solution to problem solved sphere square controlled Suppose T₂ total number triangle unfavorable values vertex vertices