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 31 筆
第 53 頁
... Similarly e will be counted by ( 3 ) The total number of times e is counted is therefore - ( 1 ) - ( 2 ) + ( 3 ) of the terms # ( A ; ~ A , ~ Ax ) , etc. - + ( - 1 ) -1 ( 2 ) . We must show that this expression is equal to 1. To see ...
... Similarly e will be counted by ( 3 ) The total number of times e is counted is therefore - ( 1 ) - ( 2 ) + ( 3 ) of the terms # ( A ; ~ A , ~ Ax ) , etc. - + ( - 1 ) -1 ( 2 ) . We must show that this expression is equal to 1. To see ...
第 60 頁
... Similarly we see that o ( 18,000 ) o ( 24 ) o ( 32 ) o ( 53 ) . Since = o ( p ' ) = 1 + p + p2 + ··· + pr • - = pr + 1 1 p - 1 we have σ ( 18,000 ) 31 13 156 • = · = 62,868 . Remark . The same reasoning leads immediately to the fact ...
... Similarly we see that o ( 18,000 ) o ( 24 ) o ( 32 ) o ( 53 ) . Since = o ( p ' ) = 1 + p + p2 + ··· + pr • - = pr + 1 1 p - 1 we have σ ( 18,000 ) 31 13 156 • = · = 62,868 . Remark . The same reasoning leads immediately to the fact ...
第 166 頁
... Similarly all the terms # ( A ; ~ A ; ) have the same value , which we will call aą . We similarly define a ̧ , ¤ , Then the principle of inclusion and exclusion gives 10 # ( A1 UU A10 ) = 10a1 - ( 19 ) a2 + 2 ( 19 ) ... 9 ( 10 ) a2 a10 ...
... Similarly all the terms # ( A ; ~ A ; ) have the same value , which we will call aą . We similarly define a ̧ , ¤ , Then the principle of inclusion and exclusion gives 10 # ( A1 UU A10 ) = 10a1 - ( 19 ) a2 + 2 ( 19 ) ... 9 ( 10 ) a2 a10 ...
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