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 13 筆
第 197 頁
... powers of 2 , exactly one ends in a 2 ; this means that the probability that 2 " ends in a 2 is 1/4 = 0.25 . Now let us compute the last two digits of the successive powers of 2 : 02 , 04 , 08 , 16 , 32 , 64 , 28 , 56 , 12 , 24 , 48 ...
... powers of 2 , exactly one ends in a 2 ; this means that the probability that 2 " ends in a 2 is 1/4 = 0.25 . Now let us compute the last two digits of the successive powers of 2 : 02 , 04 , 08 , 16 , 32 , 64 , 28 , 56 , 12 , 24 , 48 ...
第 198 頁
... powers of 2 from 22 04 to 221 , only 29 ends in the digits 12. Thus , of any 20 consecutive powers of 2 , exactly one ends in the digits 12. Consequently , the probability that 2 " ends in the digits 12 is 1/20 = 0.05 = Remark . It can ...
... powers of 2 from 22 04 to 221 , only 29 ends in the digits 12. Thus , of any 20 consecutive powers of 2 , exactly one ends in the digits 12. Consequently , the probability that 2 " ends in the digits 12 is 1/20 = 0.05 = Remark . It can ...
第 202 頁
... power of 10 ) this same expression gives the probability that a power An picked at random will begin with the digits which represent the number M. The proof of this does not differ essentially from the proof for the case of powers of 2 ...
... power of 10 ) this same expression gives the probability that a power An picked at random will begin with the digits which represent the number M. The proof of this does not differ essentially from the proof for the case of powers of 2 ...
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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