Challenging Mathematical Problems with Elementary Solutions: Combinatorial analysis and probability theoryHolden-Day, 1964 |
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第 1 到 3 筆結果,共 44 筆
第 57 頁
... values of x , and the remaining four numbers contribute 2 values of x ( namely x = 9998 and x 10,000 ) . So the total is 2858 ; subtracting this from 10,000 we get 10,000 - 2858 = 7142 values of x for which 2 * x2 is not divisible by 7 ...
... values of x , and the remaining four numbers contribute 2 values of x ( namely x = 9998 and x 10,000 ) . So the total is 2858 ; subtracting this from 10,000 we get 10,000 - 2858 = 7142 values of x for which 2 * x2 is not divisible by 7 ...
第 65 頁
... values of m from 0 to 10 we get a total of N ( 16 ) N ( 196 ) + N N ( 126 ) N ( 169 ) + N N ( 576 ) N ( 144 ) + N N ( 1156 ) N ( 121 ) + N N ( 1936 ) 100 N 20 + N N ( 2216 ) N ( 31 ) + N ( 1996 ) N ( 54 ) = 49 -N ( 5476 ) N ( 28 ) + N ...
... values of m from 0 to 10 we get a total of N ( 16 ) N ( 196 ) + N N ( 126 ) N ( 169 ) + N N ( 576 ) N ( 144 ) + N N ( 1156 ) N ( 121 ) + N N ( 1936 ) 100 N 20 + N N ( 2216 ) N ( 31 ) + N ( 1996 ) N ( 54 ) = 49 -N ( 5476 ) N ( 28 ) + N ...
第 66 頁
... values 1 , 2 , ... , k any 1 ( y cannot take the value k , since then z would not be positive ) ; the corresponding value of z determined from the equation will also be a positive integer in each of these cases . Let us now turn to our ...
... values 1 , 2 , ... , k any 1 ( y cannot take the value k , since then z would not be positive ) ; the corresponding value of z determined from the equation will also be a positive integer in each of these cases . Let us now turn to our ...
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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 G₁ given Hence inclusion and exclusion intersection k-gons knights length mathematical induction maximum number n-gon number of different number of favorable number of paths number of shortest obtain pairs partition passengers plane polygons positive integers possible outcomes Pr{E probability theory problem 54 prove queens rectangle relatively prime remaining required probability rooks S₁ segment selected at random sequence shortest paths side solution to problem solved sphere square controlled Suppose T₂ total number triangle unfavorable values vertex vertices