Lost my first try at typing down the proof due to the forum time-out feature that destroys your message. *@!#$@!*&@#



So here goes again.

- first step :

Each number ending on the digit 0 adds a 0 to the total multiplication. We have 10 such numbers, where the number 100 adds two zero's.

10, 20, 30, 40, 50 , 60, 70, 80, 90 and, 100 => 11 trailing zero's added.

For the remainder of this proof we'll continue with new series of 89 numbers were the above numbers have been removed.



- Second step

Note how the trailing digit of any product of two numbers is totally determined by the trailing digit of each individual number. Example 17*23 = 391 => ends on 1 as 7*3 = 21 and also ends on 1.

By virtue of this result we know that a number ending on the digit 5 will result in a number ending on the digit 0 when it is multiplied by an even number. Example : 25*6 = 120 as 5*6 = 30

As a result we have 10 more numbers that in combination with any even number will multiply into another number with one or more trailing zeros. These numbers are :

5, 15, 25, 35, 45, 55, 65, 75, 85 and 95

If we pick our even numbers carefully then we can garantee that the resulting numbers only have a single trailing zero (and not 2 or more) and that the second last digit is out of the range 1, 2, 3, 4, 6, 7, 8 and 9. This is a handy feature as shall be shown later.

Lets say we multiply these with 4, 6, 2, 8, 12, 14, 16, 18, 22 and 24 (10 and 20 are excluded already as per above) and we find the products :

20, 90, 50, 280, 540, 770, 1040, 1350, 1870 and 2280

Note how all the results have a forelast digit that is part of the set 1, 2, 3, 4, 6, 7, 8 and 9 where the results 50 and 1350 are the only exceptions. We need to remember these numbers for later.

These 10 numbers add 10 trailing zero's to any multiplication they are part off. The may even add more depending with what other numbers the are multiplied with. The only way to be sure is to remove the 10 zero's and keep the following truncated numbers in our remaining multiplication set.

2, 9, 5, 28, 54, 77, 104, 135, 187 and 228.

We take the numbers 5 and 135 from this set and multiply them again with 26 and 28 from the full dataset resulting in :

130 and 3780

Both these numbers have a second digit that now too is of the set 1, 2, 3, 4, 6, 7, 8 and 9.

Again we remove the trailing zero's and add this to our total arriving now at 11+12 = 23 trailing zero's.

Of course we also keep these two truncated numbers 13 and 378 in our remaining dataset from which where we have now also removed the numbers 2, 4, 6, 8, 12, 14, 16, 18, 22, 24, 26 and 28 of course.

thus our subset now looks like
2, 9, 13, 28, 54, 77, 104, 378, 187 and 228. + all other numbers not removed so far.



- Third step

We are now left with a dataseries where all the numbers that end on 5 or 0 have been removed. So all remaining numbers have a trailing digit that is enclosed in the set

1, 2, 3, 4, 6, 7, 8 and 9

Mutual multiplication by numbers ending on a digit from this set is said to be "mathematically closed".

This means that any multiplication between such numbers results in a new number that is also ending on a digit out of this set (and therefor not ending on the digit 5 or 0). Of course this behaviour is recurring.

The overal result is that the multiplication of remaining series of numbers will result in a number that will also end on 1, 2, 3, 4, 6, 7, 8 or 9 and therefor never end with either a 5 or a 0. Therefor this series will never add another trailing zero to the multiplication expressed by 100! irrespectibally of how the numbers are multiplied to eachother. We can therefor scrap his whole series of remaining numbers from our notepad and just focus on the numbers that we had taken out earlier.



- Forth step

That leaves us with the following data sub result :

10, 20, 30, 40, 50 , 60, 70, 80, 90 and, 100 resulting from the numbers ending on 0

+ 12 more times the number 10 resulting from the numbers ending on 5.


This can be rewritten into 23 times the number 10 and the new dataset 1, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Of course the number 5 can be multiplied by 32 out of the old discarded set to produce 160, which breaks down into 10*16 leaving us with :

24 times the number 10 and the dataset 1, 1, 2, 3, 4, 6, 7, 8, 9, 16.

All the numbers of the last data set have a trailing digit from the set 1, 2, 3, 4, 6, 7, 8 and 9. and can in turn never, under multiplication, produce a number with a trailing digit of 5 or 0 again.

And thus we have proven that the number of trailing digits is exactly 24. No less and no more.

That is assuming I have made no error somewhere.

I will have to check my own reasoning in a couple of days to spot oversights and errors. Right now I will most likely only read over them.

Maybe there is a simpler proof, but if there is then I need to let things calm down a bit before I can spot that one. Personally I don't think that the proof is of made of "very basic math" the steps themselves may not be particulary complex but composing a fully "mathematically closed" proof is certainly alot more complex and beyond the level of novices. But that assumes that the above proof is the definate proof, There may always be a very elegant pathway that we (I) haven't discovered yet.

Wouter

Yes, but how many buns make 5....?

I really have to get working on my report now, but I couldn't resist to write a small excel program to numerically work out the number of trailing zero's. Of course excel can't handle sufficiently large numbers to describe the number 100! accurately. But you can proof the number of trailing zero's by multiplying each neightbouring pair of numbers and factoring out any multiples of 10 before executing the next series of neighbouring multiplications.

You stop when all (remaining) neighbouring pairs both have a trailing digit out of the series 1,2,3,4,6,7,8 or 9 where the left over series can never again produce another trailing zero (another 10 that can be factored out). This happens after defactorizing the 2nd series.

Totalling the number of factored out 10's results in 24, ergo the number 100! has 24 trailing zero's.

See the table attached : (POS = number is now Part Of Set with trailing digit 1, 2, 3, 4, 6, 7, 8 or 9)

Or just use my method.

Your method may have produced the (correct) number of 24 trailing zero's but it is not mathematically sound. Basically, you followed a pathway that is not "thoroughly logical" and that is most certainly not "closed". The first meaning that you have not taken the full scope of the problem into account. Example ; who is to say that you have covered all eventualities in your "proof", maybe you missed a special number or special combination ? The second means you have failed to proof that there can AT MOST be 24 trailing zero's.


I'll provide a few more details.

You wrote :




In this method you use the number 2 twice; ones in multiplying 2*50 and once in multiplying 2*25. That is obviously incorrect.

You also say : Each number that ends in 5 also creates zero's (so that's 10 zeros), without showing how this is the case. For example, a number that ends in 5 ONLY creates a trailing zero when multiplied by an even number. Your statement still allows it to create trailing zero's when multiplied with any given number.

You forget to show that such a multiplication can produce more then only 1 trailing zero. Example by using the next available even number = 4 we find : 25 * 4 = 100 and that is adding two zero's to the total instead of just 1. So how do you know that there aren't more then 24 trailing zero's ?

Then you leave out the proof that none of the other possible combinations can produce trailing zero's or can produce numbers that in combination with yet another number can produce zero's.

Even with some leniency regarding your phrasing of the proof (using the number 2 twice), you can at most proof that there are At LEAST 24 trailing zero's. Of course that was never the hard part of the proof. The hard part is to proof that there are AT MOST 24 trailing zero's as then you'll have to find a structure in the remaining series of numbers that prevents this series from ever creating more trailing zero's. Only by combining these two parts can the proof be completed.

Wouter

It was logically thought thru.

I looked at the problem. Understood what was causing the zero's to be produced, calcuated how to work out which numbers produced the zeros and then described the method in simple terms.

Simple is best IMO.

Simon,

We are talking mathematics here. Some words have slightly different meaning there then they do in the broader worlds.

Your approach was indeed logical, but it wasn't "thorougly logical" as in exhaustive.

Your method overlooks part of the possibilities/eventualities and basically implicetly assumes that they can not create more zero's. Assuming something (either explicetly or implicitely) does not equal to proof.



My point was whether you also understood why the other numbers could not produce zero's. That part was completely left out in coverage. Without it the proof is incomplete and not a proof at all. That is the fun about mathematics and the part that makes some proofs so darn difficult.

Point in case: proving Fermat's principle (look it up); we have know that it is true for about 250 years now, but no-one has yet been able to find a "closed" proof for it. Some mathematician thought he had just recently after working on it for some 15 years. Turns out a small part of his very elegant proof was not "thorough", preventing the proof from being "fully closed" and collapsing his whole proof.





Only when you have the option of choosing between two or more pathways that are equal in "thoroughness" and "closedness". A simple pathway that doesn't contain both or either one is absolutely worthless, irrespectibally whether it is simple or not.

Sorry,

I did remove the "lucky" part however as I do not think that was a fair statement

Wouter

Wouter,

OK I did a little digging around the web to try and find out how to show more cleanly what I did. I actually searched for "number of zeros in 100 factorial" via google.

Maybe I had done this calc in the past as part of my studies into Maths as part of my A levels and then deciding weather or not to take maths into a degree (I did not, I studied Computing, Management and software design).

I did essentially what is described here.

This is also worth a read

"interesting" to see what happens with 1,000,000! <img src="http://www.catsailor.com/forums/images/graemlins/grin.gif" alt="" />

More here

and here

Stop IT! My head hurts! <img src="http://www.catsailor.com/forums/images/graemlins/shocked.gif" alt="" />

I'd rather pack decubitus ulcers. <img src="http://www.catsailor.com/forums/images/graemlins/tongue.gif" alt="" />

ewww... just make sure you don't send them to the ER for them to be evaluated....

<img src="http://www.catsailor.com/forums/images/graemlins/laugh.gif" alt="" />

You did in broad lines.

Still, even Paul Kellet is a little bit messy in his mathematics.

At one point he states :"we need to figure the highest powers of 2 and 5 dividing 100! and take the lesser of the two exponents." But he then omits determining the highest power of 2 dividing 100! and omits showing that its power is larger then the power of 5 dividing 100!.

Also it still needs to be proven that the numbers that can not be factorized into 2's and'or 5's can never produce another number that can be factorized along either number after multiplication. Paul Kellet omits that part of the proof as well. Basically he describes a method for calculating the minimum number of trailing zero's, but does not proof that this number is also the maximum number of trailing zero's.



The idea behind his proof is actually very elegant however, although it is beyond novice level. His explanation is certainly beyond novice level mathematics. It can be made full and be worded much simpler, so that is complete and even novices can understand it.

Basically he is saying that the total number of trailing zero's in the number 100! is equal to the number of times you can devide that number by the number 10 without having to write down decimals. As the number 10 can be factorized into a product of the numbers 2 and 5 as 2*5 = 10, you can know the number of "even" (= no resulting decimals) divisions by 10 by factorizing all numbers between 1 to 100 along the numbers 2 and 5. You then group as many of the factored out 2's and 5's together into pair (2*5) as you can (forming 10's) and count these combo's => the number of combo's produces the minimum number of trailing 0's.

a small example for the other readers out here ;

the result of 55*54*53*52*51*50 has 3 trailing zero's. Proof :

55 factors in 11*5
54 factors in 27*2
53 can not be factored in multiple of 2's or 5's
52 factors in 26*2
51 can not be factored in multiple of 2's or 5's
50 factors in 5*5*2

All the leftover parts that can not be further factorized along either the numbers 2 or 5 are members of the "number space" that satisfies teh condition "the number is an integer with its last digit being 1,2,3,4,6,7,8 or 9'. This space is closed under multiplication, meaning that no multiplication using numbers from this number space can result in a number that does not belong to this number space as well. As a consequence, the result (multiplication of leftovers) can never produce any trailing zero's as that digit is not part of the set 1,2,3,4,6,7,8 and 9. This is turn proofs that the method described above not only gives the minimum amount of trailing zero's but also the maximum amount of trailing zero's

Combining these results we have found 3 times a factor 5 and three times a factor 2 allowing us to form 3 combo's of 2*5=10 ergo the resulting multuplication has 3 trailing zero's


P.S. I used the identifier "number space" here for clearity, but I think the correct name for such a mathematic body is a "ring"

Wouter

and of course I too omitted part of the whole proof. That proofs my point at how easy it is to not close up a proof !

Everything was covered in my last post except for the numbers 2's (or 5's) that can not be taken up into a combination (2*5=10). Can this excess in number 2's (or 5's) result in trailing zero's ?

No, as all the numbers that can be fully factorized into a multiplication of 2's is a closed number space (ring) as well. Multiplication using numbers that are members of this number space (ring) all have a last digit that belongs to the set 2,4,6 and 8. Therefor there exist no multiplication of any given number of 2's that can produce a number that ends with the digit 0. Therefor this number can never be evenly divided by 10. As the set 2,4,6 and 8 is a subset of the set 2,3,4,6,7,8 and 9 the numbers space (ring) for powers of 2's is wholely enclosed in the numbers space (ring) for the leftovers as identified in my former posting. As a result, the multiplication of the leftovers from factorizing and the excess of 2's can never produce a number that ends with either a 5 or 0 as the last digit. Therefor this series of leftover numbers can never produce additional trailing zero's.

The proof when an excess of 5's is had goes along the line of reasoning.

So ! Now the proof is mathematically complete and closed.


Can you tell that I do this kind of stuff more often !


Ehh, is someone still paying attention ?! <img src="http://www.catsailor.com/forums/images/graemlins/grin.gif" alt="" />

Wouter

Wouter, put down your calculator and GO TO BED! <img src="http://www.catsailor.com/forums/images/graemlins/grin.gif" alt="" />

I'm going to need 2 more beers just to flush that out.

WHAT TF is going on here?

W- shouldn't you be doing something more important? dunno what.. but still

Actually I was doing these proofs as a break from the work that I need to get done this night. I have to deliver a report dealing with system identification tomorrow.

One of the plots ;




Notice how both signals, input and ouput, are both very noise like. The output has a very strong noise like disturbance superimposed over it, just to make things extra interesing !

I have to use various mathematical tools based on probability theory to identify the mechanical system that relates the two signals. Actually similar methods are used to have mobile phones and satelite communication work. These signals are strongly disturbed by outside interference as well.

A collegue once described these methods as standing on one side of the hall where a rock concert is held trying to hear what your girlfriend is yelling at you from the far away other side of the hall.

I have done that and now I'm completing my report.

Wouter

Wouter, I understand the elegance of a formal proof. I also operate in the real world now, where time is money, and Simple is usually best.

Now what about Pi to 100! places....

24 is correct for trailing zeoes of 100!...

The logic required for n!...
-only multiplying by 10 or a factor of 10 will produce a trailing zero

-multiplying 5 by an even number will produce a factor of 10

-there will be more even numbers than multiples of 5, so 5's are the limiting number in the next step.

-reduce all the numbers down to there prime numbers

-counting all the times 5 occurs will give you the number of trailing zeroes.

Actually I was told many years back that the number 100! denotes more elements then there are particles in the universe. So I think that I'll pass on :




With respect to :




There are things in the real world like designing airplanes, rockets, medicine and nuclear or chemical reactors were getting things absolutely right is paramount and trumps any "time is money" or "simple is best" considerations.

Or at least that should be the case.

Make an error in some computer scripting and your PC crashes, make an error in some control system stabilizing a nuclear reactor or a robot arm and people loose lives and body parts. In case of designing nuclear reactors you can not even dependent on prototyping to show the errors. The meltdown or blowing up is such a severe accident that a designer can never take the risk of it happening.

I'm active in the field of such systems.

Wouter