Question

Find last 5 significant digits of 2017!

Answer 1

So, since there are less powers of 5 than 2 and since \(10 = 2 \cdot 5\), we need to count powers of 5 to obtain the number of 0's in the factorial:
2017/5 + 2017/25 + 2017/125+2017/625 = 502

Now, using Chinese remainder theorem we find that \(\frac{2017!}{10^{502}} \equiv 0 \pmod{2^5}\) (there are a lot more 2's left), so we need to find \(\frac{2017!}{10^{502}} \equiv ~? \pmod{5^5}\).

And that's where I get stuck. How would I compute that?

Answer 2

May not be useful, just an observation

2017 is a prime number, so we have by Wilson :

2016! = -1 mod 2017

2016! = 2017u - 1

Answer 3

@ganeshie8 Can we do this, to simplify the fraction \(10^{502} \pmod{5^5}\)? I think it might not be valid, as the fractional values aren't(?) valid in modulus operations.

Answer 4

Since \(a * b \pmod{c} = a\pmod{c} * b\pmod{c}\), I thought perhaps simplifying \(\frac{2017!}{10^{502}}\pmod{5^5}\) would work.
(Basically, could we do this: \(\frac{2017!}{10^{502}}\pmod{5^5} = 2017!\pmod{5^5}\cdot\frac{1}{10^{502}}\pmod{5^5}\)?)

Answer 5

Only thing that would be left to do, would be getting 2017! out of the way with the right u (from Wilson, not sure how to find the right u, satisfying the mod \(5^5\)).

Answer 6

We can't do that here because 10 has no inverse in modulus 5

Answer 7

Ahh.. too bad then.

Answer 8

1/10^a is meaningless in modulus 5^b

Answer 9

Yes, you are right. Didn't notice that earlier :/

Answer 10

10^a * x = 1 mod 5^b

must have a solution in order to divide 10^a in mod 5^b

Answer 11

OK let's get back to the problem

Answer 12

http://m.wolframalpha.com/input/?i=2017%21+mod+10%5E507&x=0&y=0

Answer 13

Why "2017! mod 10^507"? And is it not possible to solve it by hand?

Answer 14

It should be possible, I'm just not getting ideas...

Answer 15

10^507 because you've said earlier there were 502 zeroed at the end

Answer 16

and we are looking for 5 significant digits

Answer 17

Oh, you are right, sorry.

Answer 18

Hey I have to go now. Do post here if you find any interesting way to find the last nonzero digits of these factorials. I'll check when I'm back. good luck !

Answer 19

@ganeshie8 I will. Thank you for your help!

Answer 20

Posted the question in MSE, so far no answers. Comments gave some interesting stuff about finding the last digit, but I can't find a way to get the last 5 from any of those methods.
http://math.stackexchange.com/questions/1841331/find-last-5-significant-digits-of-2017/1841384#1841384

Answer 21

Hey nice
Could you explain how you're using the fact that 1.2.3.4 = -1 mod 5 ?

Answer 22

@ganeshie8, the problem is still unsolved.
I think that the person who suggested looking at this meant to suggest searching for some kind of pattern and group the numbers in such way.
Something like this: http://www.campusgate.co.in/2013/10/finding-right-most-non-zero-digit-of.html
However, I hadn't figured out how would I apply this for \(5^{5}\)...

Answer 23

The last 5 non zero consecutive digits is 15968 .

991033159680000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

Answer 24

@robtobey7, well I am well aware of the result using calculator, however, what i am interested into is solving this problem without use of calculator.