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 17 筆
第 51 頁
А. М Яглом, Исаак Моисеевич Яглом Basil Gordon. multiple of 1000 and the next 999 integers after it , etc. Note that 0 , being a multiple of any number , is a multiple of 10 , of 100 , of 1000 , etc .; hence { 0 , 1 , . . . , 9 } , { 0 ...
А. М Яглом, Исаак Моисеевич Яглом Basil Gordon. multiple of 1000 and the next 999 integers after it , etc. Note that 0 , being a multiple of any number , is a multiple of 10 , of 100 , of 1000 , etc .; hence { 0 , 1 , . . . , 9 } , { 0 ...
第 54 頁
... multiples of 3. Then AC consists of the multiples of 53 15 , BC of the multiples of 73 = 21 , and AOBOC of the multiples of 57.3 105 in this range . Reasoning as in part a we see that # ( C ) = = = 66 , 9997 [ 999 3 = 333 , # ( ANC ) ...
... multiples of 3. Then AC consists of the multiples of 53 15 , BC of the multiples of 73 = 21 , and AOBOC of the multiples of 57.3 105 in this range . Reasoning as in part a we see that # ( C ) = = = 66 , 9997 [ 999 3 = 333 , # ( ANC ) ...
第 196 頁
... multiples of 3 , one has to be a multiple of 9. Consequently , the product n ( n - 1 ) ( n - 2 ) ( n − 3 ) ( n — 4 ) ( n − 5 ) ( n − 6 ) is divisible by 27 if and only if n has the form 9k + r , where r = 0 , 1 , 2 , 3 , 4 , 5 , or 6 ...
... multiples of 3 , one has to be a multiple of 9. Consequently , the product n ( n - 1 ) ( n - 2 ) ( n − 3 ) ( n — 4 ) ( n − 5 ) ( n − 6 ) is divisible by 27 if and only if n has the form 9k + r , where r = 0 , 1 , 2 , 3 , 4 , 5 , or 6 ...
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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