How many zeros ????

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Moonbeam
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How many zeros ????

Postby Moonbeam » Sun Jun 07, 2009 9:44 pm UTC

I wasn't sure whether to post this in maths, or in here, so ..... well, it's here now (unless a mod decides to move it :? ).

Consider the number 100!

How many zeros are there at the end of 100! ????

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Wednesday
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Re: How many zeros ????

Postby Wednesday » Sun Jun 07, 2009 9:53 pm UTC

... There are two zeros in the number 100. And I suppose an infinate number after a decimal point.
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Re: How many zeros ????

Postby Moonbeam » Sun Jun 07, 2009 10:02 pm UTC

horisustar wrote:... There are two zeros in the number 100. And I suppose an infinate number after a decimal point.


I think you missed the ! :?

The question asked "How many zeros are there at the end of 100! ????" (Note the ! - denoting "factorial").

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Re: How many zeros ????

Postby Wednesday » Sun Jun 07, 2009 10:03 pm UTC

Oh. Meh. Im not really in the mood to think anything maths-y. Sorry. Ill come back and contribute later.
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Re: How many zeros ????

Postby CueBall » Sun Jun 07, 2009 11:37 pm UTC

Numbers which add a 0 to the end:

Anything with 5 as a factor

(There is an abundance of things with 2 as a factor)

so, 5, 10, 15... 100

that's 20.

However, 25, 50, 75 and 100 all have 5 as a factor twice.

So I make it 24 0's.
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Re: How many zeros ????

Postby Poohblah » Mon Jun 08, 2009 6:03 am UTC

Spoiler:
like the above poster said, this is determined by the number of factors of 10. We have 10, 20, 30... 100 to make 11 factors of 10. There are an abundance of factors of 2, far more then there are of 5, so we can just add all the factors of 5 up. There are 20 different numbers no more than 100 with a factor of 5, minus the 11 we have already counted to make 20. Then we can add the factors of 5 that appear twice in a number - 25, 50, 75, 100. That's 24. Exactly the same as the above poster got.

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Re: How many zeros ????

Postby Macbi » Mon Jun 08, 2009 6:24 am UTC

A good rough estimation is just to divide by four. In fact it's always an overestimation, so a better one is to divide by 4, subtract one, and round up. Here that gives the exact answer. Puzzle: Why does this estimation work?
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Re: How many zeros ????

Postby silvermace » Tue Jun 09, 2009 12:56 am UTC

Spoiler:
100! =

9332621544394415268169923885626670049071596826438162146859
29638952175999932299156089414639761565182862536979208272237582511852109168
64000000000000000000000000

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Re: How many zeros ????

Postby skeptical scientist » Tue Jun 09, 2009 2:02 am UTC

Macbi wrote:A good rough estimation is just to divide by four. In fact it's always an overestimation, so a better one is to divide by 4, subtract one, and round up. Here that gives the exact answer. Puzzle: Why does this estimation work?

Spoiler:
The number of zeros at the end of the decimal representation to n!, is, as previously noted, determined entirely by the largest power of 5 which divides n!. Of every number below n, exactly n/5 of them are divisible by 5 (rounding down). Also, n/25 of them (again round down) are divisible by 5, n/125 of them are divisible by 125, and so on. So the total number of zeros at the end of n! is about n/5+n/25+n/125+n/625+...=n/4. This is an overestimation because you should round all of the terms of the sum down before adding, and they can't all be integers. Since the number of zeros must be an integer, if you round this up you are overestimating by at least one, so you can round up and then subtract 1. This is still not exact, however; for example, 24/4-1=5, but 24! has 4 zeros at the end of it. In fact, this can overestimate by a very large amount: consider the case when n=5k-1, and k is a big number.
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Re: How many zeros ????

Postby Nitrodon » Tue Jun 09, 2009 4:24 am UTC

The real answer for any n:
Spoiler:
Subtract the sum of the digits of the base 5 representation of n, then divide by 4.

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Re: How many zeros ????

Postby munchman » Thu Jul 02, 2009 11:53 am UTC

Sorry to post so late, but consider the following:
Spoiler:
There are no zeroes at the end of "100!", simple!
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