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13 May 2019, 01:25
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Difficulty:
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Question Stats:
26% (02:35) correct
74%
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wrong
based on 211
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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?
A. 1/21
B. 1/19
C. 1/18
D. 1/2
E. 11/21
A. 1/21
B. 1/19
C. 1/18
D. 1/2
E. 11/21
13 May 2019, 18:51
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21!= 2^18 * 3^9*5^4 * 7^3 * 11* 13*17*19
Total number of divisors= 19*10*5*4*2*2*2*2
Odd divisors= 10*5*4*2*2*2*2
Probability of choosing odd divisors = (10*5*4*2*2*2*2)/(19*10*5*4*2*2*2*2) = 1/19
Total number of divisors= 19*10*5*4*2*2*2*2
Odd divisors= 10*5*4*2*2*2*2
Probability of choosing odd divisors = (10*5*4*2*2*2*2)/(19*10*5*4*2*2*2*2) = 1/19
13 May 2019, 08:06
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We can first work out the power on the '2' in the prime factorization of 21!. First, looking only at the even numbers in the product 21*20*19*18*..., we have
(20)(18)(16)(14)(12)(10)(8)(6)(4)(2)
and extracting only the 2's from each of the above (ignoring any odd prime divisors) the above gives us
(2^2)(2^1)(2^4)(2^1)(2^2)(2^1)(2^3)(2^1)(2^2)(2^1) = 2^18
So 21! = (2^18)(a lot of odd primes)
Notice then that for any one odd divisor d we have of 21!, we can make 18 even divisors, namely 2d, (2^2)d, (2^3)d, ... (2^17)d and (2^18)d. So the ratio of odd divisors to even divisors of 21! is 1 to 18, and picking one divisor at random, the probability it is odd is 1/19.
(20)(18)(16)(14)(12)(10)(8)(6)(4)(2)
and extracting only the 2's from each of the above (ignoring any odd prime divisors) the above gives us
(2^2)(2^1)(2^4)(2^1)(2^2)(2^1)(2^3)(2^1)(2^2)(2^1) = 2^18
So 21! = (2^18)(a lot of odd primes)
Notice then that for any one odd divisor d we have of 21!, we can make 18 even divisors, namely 2d, (2^2)d, (2^3)d, ... (2^17)d and (2^18)d. So the ratio of odd divisors to even divisors of 21! is 1 to 18, and picking one divisor at random, the probability it is odd is 1/19.
