2017年AMC12竞赛B第16题 2017年AMC10竞赛B第20题
$$2017~ AMC10B$$ P20  The number $$21!=51,090,942,171,709,440,000$$, has over $$60,000$$ positive integer divisors. One of them is chosen at random. What is the probability that it is odd?  数字21！=51,090,942,171,709,440,000，有超过$60,000$ 个的正整数因数。随机选择其中一个，它是奇数的概率是多少？
美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Classical Models of Probabilities
Solution 1  We note that the only thing that affects the parity of the factor are the powers of $$2$$. There are $$10+5+2+1=18$$ factors of $$2$$ in the number. Thus, there are $$18$$ cases in which a factor of $$21!$$ would be even (have a factor of $$2$$ in its prime factorization), and $$1$$ case in which a factor of $$21!$$ would be odd. Therefore, the answer is $$\frac 1{19}$$.  Solution 2  Consider how to construct any divisor $$D$$ of $$21!$$. First by Legendre\textquotesingle s theorem for the divisors of a factorial , we have that there are a total of $$18$$ factors of $$2$$ in the number. $$D$$ can take up either $$0$$,$$1$$,$$2$$,$$3$$,$$\cdots$$, or all $$18$$ factors of $$2$$, for a total of $$19$$ possible cases. In order for $$D$$ to be odd, however, it must have $$0$$ factors of $$2$$, meaning that there is a probability of $$1$$ case$$/19$$ cases$$=\frac 1{19}$$. 