A random positive integer divisor of 20! is chosen. What is the probability that it is divisible by 100?
The prime factors of 20! are shown by:
2^18
3^ 8
5^ 4
7^ 2
11^ 1
13^ 1
17^ 1
19^ 1
The number of integer divisors is
19*9*5*3*2^4 = 41040
including 1 and 20! itself.
To be divisible by 100 is to have factors 2*2*5*5, so these divisors of 20! number
17*9*3*3*2^4,
where the possibilities for 2 and 5 have been reduced by 2 each, as they refer to the excess over 2*2*5*5 (or no excess at all, i.e. 100).
This product is 22032, so the requested probability is
22032/41040 = 51/95 =~ 0.536842105263158
|
|
Posted by Charlie
on 2025-04-30 08:50:33 |
solution
