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Lepton
Posted: Fri Feb 08, 2002 2:35 am    Post subject: 1

Can't we just round it up to 250000?
Tahnan
Posted: Thu Feb 07, 2002 8:25 am    Post subject: 0

Oh, hey, empirical tests!

Sure enough, a spreadsheet gives me the progression:

code:
1	2
2 24
3 249
4 2499
5 24999
6 249998
7 2499999
8 24999999
9 249999998
10 2499999997
11 24999999994

So there you are.
Mikko
Posted: Wed Feb 06, 2002 9:09 pm    Post subject: -1

N=6 is a quite simple case that proves the theory wrong.
borschevsky
Posted: Tue Feb 05, 2002 10:24 pm    Post subject: -2

I was thinking along those lines, Eykir (hey, in base 5, does 0.1111... = 1/4 or is it just really close? ) but I'm not sure if that proves we will always get .25*10^N - 1.

Going from 10,000 to 100,000 (for example) we must end up with 10 times as many zeros plus "a few", because any 5 we counted in 10,000 gets counted 9 more times in 100,000 and there will be a few from a new power of 5.

Also, your logic is right to show that we'll never get to .25*10^N.

But doesn't this prove only that the number of zeros Z is:
.25*10^N - 10 <= Z < .25*10^N

Might we not find somewhere in the sequence a number like 249,999,998? This would fit into the increasing sequence ok, because:
24,999,999/100,000,000 < 249,999,998/1,000,000,000 < .25

Am I making any sense? I don't know; I'm thinking as I go.

[This message has been edited by borschevsky (edited 02-05-2002 05:25 PM).]
Eykir
Posted: Tue Feb 05, 2002 6:54 pm    Post subject: -3

you have a series that is for any number x,

x/5 + x/5[sup]2[/sup]+ x/5[sup]3[/sup]+ x/5[sup]4[/sup]+ x/5[sup]5[/sup]...

This is the Sum(n=1 to inf) x/5[sup]n[/sup] which reaches a limit at x/4

10^x/4 = 2.5 * 10^(x-1) which is one more than you are getting... because it never gets that last one.

*hopes he makes sense*
borschevsky
Posted: Tue Feb 05, 2002 6:37 pm    Post subject: -4

And you guys better answer me, I'm the King of Windsor Castle. They don't just give these titles to any old chump.
borschevsky
Posted: Tue Feb 05, 2002 6:31 pm    Post subject: -5

So there's a type of math question that I've seen a few times in math contests: How many zeros are at the end of X factorial (for some given X)? The idea is that you have to count up all the factors of five that get multiplied into X!.

For example, 100! has 24 zeros at the end, because from 1 to 100 there are 20 numbers divisible by 5, and 4 numbers divisible by 25.

I was thinking about this today for no good reason, and I tried 1000! - it has 249 zeros: (200 for 5s + 40 25s + 8 125s + 1 625)

And 10000! has 2499 zeros, and 100000! has 24999 zeros at the end.

So, question: for N >= 2, is it true in general that (10^N)! has (10^N)/4 - 1 zeros at the end?

[This message has been edited by borschevsky (edited 02-05-2002 01:34 PM).]