Problem:
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?
Answer Choices:
A. 211
B. 191
C. 181
D. 21
E. 2111
Solution:
There are ⌊221⌋+⌊421⌋+⌊821⌋+⌊1621⌋=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 (B)191. In fact, 21!=218⋅39⋅54⋅73⋅11⋅13⋅17⋅19, so it has 19⋅10⋅5⋅4⋅2⋅2⋅2⋅2=60,800 positive integer divisors, of which 10⋅5⋅4⋅2⋅2⋅2⋅2=3,200 are odd.
The problems on this page are the property of the MAA's American Mathematics Competitions