- #1
vagabond
- 6
- 0
Given that the 2005th digit in n! is zero, what is the possible value of n ?
Thanks
Thanks
Yes.Are you counting digits from the right? ie. the 3rd digit of 1234567 would be 5?
I would like to find all n where the 2005th digit of n! is 0.Are you looking for the least n where the 2005th digit of n! is 0?
If n is known, I can do it.Can you work out the highest power of 5 that divides n!? The highest power of 2? (the 2's won't really be an issue though)
Can we deduce if the possible number of such n is finite? Can we find the pattern?
I would like to find all n where the 2005th digit of n! is 0.
shmoe said:The digits are being counted from the right, not the left.
The 2005th digit being 0 indicates that the number n has at least 2005 digits. It also suggests that the number is likely very large.
Yes, n can have any combination of digits in the first 2004 digits. The only requirement is that the 2005th digit is 0.
Yes, there is a formula for determining the possible values of n in this scenario. It is: n = 10^(2004)k, where k is any positive integer.
There are infinitely many possible values of n in this scenario. As long as n follows the formula mentioned in question 3, it is a possible value.
This scenario can be useful in cryptography and data encryption, as large numbers with specific digit patterns are often used in these fields for added security. It can also be applied in number theory and mathematical research.