AMC10 2017 B

AMC10 2017 B · Q20

AMC10 2017 B · Q20. It mainly tests Probability (basic), Primes & prime factorization.

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21!=51,090,942,171,709,440,000有超过60,000个正整数因数。其中随机选一个。因数为奇数的概率是多少？

(A) \dfrac{1}{21} \dfrac{1}{21}

(B) \dfrac{1}{19} \dfrac{1}{19}

(D) \dfrac{1}{2} \dfrac{1}{2}

(E) \dfrac{11}{21} \dfrac{11}{21}

Answer

Correct choice: (B)

正确答案：（B）

Solution

There are \binom{21}{2} + \binom{21}{4} + \binom{21}{8} + \binom{21}{16} = 10 + 5 + 2 + 1 = 18 powers of 2 in the prime factorization of 21!. Thus 21! = 2¹⁸k, where k is odd. A divisor of 21! must be of the form 2ⁱb where 0 ≤ i ≤ 18 and b is a divisor of k. For each choice of b, there is one odd divisor of 21! and 18 even divisors. Therefore the probability that a randomly chosen divisor is odd is 1/19.

21!的质因数分解中2的幂为\binom{21}{2}+\binom{21}{4}+\binom{21}{8}+\binom{21}{16}=10+5+2+1=18。因此21!=2^{18}k，其中k为奇数。21!的因数形式为2^i b，其中0≤i≤18，b为k的因数。对每个b，有1个奇因数和18个偶因数。因此随机选因数为奇数的概率为1/19。

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