Challenging Mathematical Problems with Elementary Solutions, 第 1 卷Holden-Day, 1964 - 440 頁 |
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第 57 頁
... divisible by 49 , then x2 + y2 is divisible by 7. But x2 can only have 0 , 1 , 2 , or 4 as its remainder upon division by 7. ( See the solution to problem 15. ) The same is true of y2 . Now it is easy to verify that of all sums of two ...
... divisible by 49 , then x2 + y2 is divisible by 7. But x2 can only have 0 , 1 , 2 , or 4 as its remainder upon division by 7. ( See the solution to problem 15. ) The same is true of y2 . Now it is easy to verify that of all sums of two ...
第 143 頁
... divisible by 495 5.911 , it is necessary and sufficient that it be divisible by 5 , by 9 , and by 11. Since 10,000a + 1,000b + 100c + 10d is always a multiple of 5 , N is divisible by 5 if and only if e = 0 ore = 5 . Since N = ( 9999a + ...
... divisible by 495 5.911 , it is necessary and sufficient that it be divisible by 5 , by 9 , and by 11. Since 10,000a + 1,000b + 100c + 10d is always a multiple of 5 , N is divisible by 5 if and only if e = 0 ore = 5 . Since N = ( 9999a + ...
第 196 頁
... divisible by 49. But the latter can occur only if one of the seven factors is divisible by 49 , that is , if n gives a remainder of 0 , 1 , 2 , 3 , 4 , 5 , or 6 upon division by 49. Of any 49 consecutive integers , exactly seven satisfy ...
... divisible by 49. But the latter can occur only if one of the seven factors is divisible by 49 , that is , if n gives a remainder of 0 , 1 , 2 , 3 , 4 , 5 , or 6 upon division by 49. Of any 49 consecutive integers , exactly seven satisfy ...
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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 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