Challenging Mathematical Problems with Elementary Solutions, 第 1 卷Holden-Day, 1964 - 440 頁 |
搜尋書籍內容
第 1 到 3 筆結果,共 42 筆
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 . - - 1 ) ! is Note that if p is not ...
... 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 . - - 1 ) ! is Note that if p is not ...
第 198 頁
... Note that there cannot be two different powers of 2 which have the same number of digits and which both begin with a 1. If two distinct powers of 2 are given , the larger must be at least twice the smaller ; hence if both have the same ...
... Note that there cannot be two different powers of 2 which have the same number of digits and which both begin with a 1. If two distinct powers of 2 are given , the larger must be at least twice the smaller ; hence if both have the same ...
其他版本 - 查看全部
常見字詞
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 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 total number triangle unfavorable values vertex vertices