(2^16)!/2^p maximum value of p? (result and p are integers )
- Thread starter locha
- Start date
- Joined
- Feb 4, 2004
- Messages
- 16,582
This exercise statement is somewhat cryptic. The following is what I think is meant:
Let p be an integer. Consider the following expression:
. . .\(\displaystyle \dfrac{(2^{16})!}{2^p}\)
...where "!" indicates "factorial". Find the maximum value of p such that the above expression is itself an integer.
Is the above how you interpret the exercise? If not, please reply with corrections. If so, please reply with a clear statement of your thoughts and efforts so far, so we can see where things are bogging down. Thank you!
