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 16 筆
第 51 頁
A. M. I︠A︡glom, Isaak Moiseevich I︠A︡glom 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 ...
A. M. I︠A︡glom, Isaak Moiseevich I︠A︡glom 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 ...
第 54 頁
... multiples of 3. Then AC consists of the multiples of 53 15 , BOC of the multiples of 73 = 21 , and AOBOC of the multiples of 573 105 in this range . Reasoning as in part a we see that # ( C ) = [ 999 ] = = 333 , # ( ANC ) = [ 999 ] = 66 ...
... multiples of 3. Then AC consists of the multiples of 53 15 , BOC of the multiples of 73 = 21 , and AOBOC of the multiples of 573 105 in this range . Reasoning as in part a we see that # ( C ) = [ 999 ] = = 333 , # ( ANC ) = [ 999 ] = 66 ...
第 143 頁
... multiple of 5 , N is divisible by 5 if and only if e = 0 ore = 5 . Since N = ( 9999a + 999b + 99c + 9d ) + ( a + b + c + d + e ) , and since the first parenthesis is always a multiple of 9 , N is divisible by 9 if and only if a + b + c ...
... multiple of 5 , N is divisible by 5 if and only if e = 0 ore = 5 . Since N = ( 9999a + 999b + 99c + 9d ) + ( a + b + c + d + e ) , and since the first parenthesis is always a multiple of 9 , N is divisible by 9 if and only if a + b + c ...
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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