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 40 筆
第 32 頁
... note first that if 0 ≤ x ≤ y ≤ L , then p ( y ) = p ( x ) + p ( y − x ) ≥ p ( x ) , since p ( y — x ) ≥ 0 by property ( 1 ) . Thus the function p ( x ) is monotone non - decreasing . Now if x / L is irrational , and ʼn is any ...
... note first that if 0 ≤ x ≤ y ≤ L , then p ( y ) = p ( x ) + p ( y − x ) ≥ p ( x ) , since p ( y — x ) ≥ 0 by property ( 1 ) . Thus the function p ( x ) is monotone non - decreasing . Now if x / L is irrational , and ʼn is any ...
第 59 頁
... ( Note that the divisor 1 is gotten by taking a = b = c = 0. ) Thus there are 5 possibilities for a ( namely 0 , 1 , 2 , 3 , or 4 ) , 3 possibilities for b , and 4 possibilities for c . Since these can be combined in all possible ways ...
... ( Note that the divisor 1 is gotten by taking a = b = c = 0. ) Thus there are 5 possibilities for a ( namely 0 , 1 , 2 , 3 , or 4 ) , 3 possibilities for b , and 4 possibilities for c . Since these can be combined in all possible ways ...
第 124 頁
... note that in part a N is an integer by its definition . Hence 2N is also an integer , which implies that [ ( p − 1 ) ! + 1 ] / p is an integer . This means that ( p 1 ) ! + 1 is divisible by p . - - Note that if p is not prime , then ...
... note that in part a N is an integer by its definition . Hence 2N is also an integer , which implies that [ ( p − 1 ) ! + 1 ] / p is an integer . This means that ( p 1 ) ! + 1 is divisible by p . - - Note that if p is not prime , then ...
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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 length mathematical induction maximum number multiple 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