Possible Value of n When 2005th Digit of n is 0

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In summary, there are an infinite number of integers n such that the 2005th (from the right) digit of n! is 0. However, if we want to find all values of n where this is true, we can work out the first integer m such that 5^2005 divides m!, and then all n greater than m will have the same property. To find this m, we can use the formula: m = 1 + \sum_{n=1}^{\infty} \biggr \lfloor \frac{5^n}{5^n} \biggr \rfloor where \lfloor x \rfloor represents the floor function. This formula counts the number of
  • #1
vagabond
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Given that the 2005th digit in n! is zero, what is the possible value of n ?
Thanks :smile:
 
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  • #2
Are you counting digits from the right? ie. the 3rd digit of 1234567 would be 5? Are you looking for the least n where the 2005th digit of n! is 0?

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)
 
  • #3
Are you counting digits from the right? ie. the 3rd digit of 1234567 would be 5?
Yes.

Are you looking for the least n where the 2005th digit of n! is 0?
I would like to find all n where the 2005th digit of n! is 0.
Can we deduce if the possible number of such n is finite? Can we find the pattern?

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)
If n is known, I can do it.
For instance, if n=100, then in n!, the factors 5, 10, 15, 20, ..., 95, 100 involve 5.
So the highest power of 5 that divides n! is 1+1+1+1+2+1+1+1+1+2+1+1+1+1+2+1+1+1+1+2 = 24.
But if n is unknown, I do not know how to do.

Also, this method is used to count the number of '0' appeared in n!.
But how do we know the location of the '0' is at 2005th position or not?
 
  • #4
Can we deduce if the possible number of such n is finite? Can we find the pattern?

This is easy. No, there are an infinite number of integers n such that the 2005th (from the right) digit of n! is 0, since letting m be the first integer such that [itex]10^{2005}|m![/itex] we find that the 2005th digit of m! is 0 and so is the 2005th digit of n! for every [itex]\mathbb{Z} \ni n > m[/itex].

I would like to find all n where the 2005th digit of n! is 0.

This is not so easy. What we can do is work out the m that I mentioned above (so that every n>m has the desired property also). This is equivalent to finding the first [itex]m[/itex] s.t. [itex]5^{2005}|m![/itex] (since there are clearly more factors of 2 than there are of 5 in m!), so let's see what we can do. I claim that the number of factors of 5 in m! for any positive integer m is

[tex]\sum_{n=1}^{\infty} \biggr \lfloor \frac{m}{5^n} \biggr \rfloor.[/tex]

Now, before you use this, you should definitely prove it on your own (I haven't written out a rigorous proof - so maybe I'm wrong! You shouldn't risk it!).

Once you show that the statement above is true, you're almost done - you just have to solve for the first [itex]m[/itex] such that

[tex]\sum_{n=1}^{\infty} \biggr \lfloor \frac{m}{5^n} \biggr \rfloor \geq 2005.[/tex]
 
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  • #5
Here's the set between 800! and 1000! that have the 2005'th digit equal to 0:

{810, 811, 812, 813, 814, 815, 816, 817, 818, 819, 820, 821, 822, 823, 824,
825, 826, 827, 828, 829, 830, 831, 832, 833, 834, 835, 836, 837, 838, 839,
840, 841, 842, 843, 844, 845, 846, 847, 848, 849, 850, 851, 852, 853, 854,
855, 856, 857, 858, 859, 860, 861, 862, 863, 864, 865, 866, 867, 868, 869,
870, 871, 872, 873, 874, 875, 876, 877, 878, 879, 880, 881, 882, 883, 884,
886, 905, 922, 934, 948, 955, 965, 969, 985, 986, 992}
 
  • #6
The digits are being counted from the right, not the left.
 
  • #7
i guess it should be 2008.i mean n should be 2008.just a guess you might give it a try
 
  • #8
shmoe said:
The digits are being counted from the right, not the left.

Ok, thanks. These then:

{811, 821, 828, 846, 850, 867, 872, 878, 880, 893, 896, 924, 925, 926, 932,
937, 938, 944, 978, 985}
 

1. What is the significance of the 2005th digit being 0 in this scenario?

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.

2. Can n have any other digits besides 0 in the first 2004 digits?

Yes, n can have any combination of digits in the first 2004 digits. The only requirement is that the 2005th digit is 0.

3. Is there a specific formula or equation to determine the possible values of n?

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.

4. How many possible values of n are there?

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.

5. What practical applications does this scenario have in the scientific field?

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.

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