Quant Question Of The Day: 168
Numbers/Counting
The number 21! = 51090942171709440000 has over 60,000 positive integer divisors. One of them is chosen at random. What is the probability that it is odd ?
1. 1/21
2. 1/18
3. 1/19
4. 1/2
The number 21! = 51090942171709440000 has over 60,000 positive integer divisors. One of them is chosen at random. What is the probability that it is odd ?
1. 1/21 21! = 218 × 39 × 54 × 73 × 11 × 13 × 17 × 19 Total factors = 19*10*5*4*2*2*2*2 since first multiple in above equation that is 19 is all the powers of 2, and out of 19 only 1 that is 20 is odd. In all other cases it is even, hence resultant factor will be even. The only thing that affects the parity of factors are the powers of 2. Therefore the answer is 1/ 19.
2. 1/18
3. 1/19
4. 1/2
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