Someone is paying attention

Just looked at the ITA website.
On the box rule fact sheet the F18 & F16 are named..
http://www.tornado.org/info/boxrule_factsheet.asp

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

I <3 Tornadoes! only if they were a tad more affordable.

Quite a good read, even if not all the historical details are the same as I have read earlier. Calling the M20 a modern B class is stretching things a bit in my opinion but it is more or less a philosophical question. Good to see the ITA take initative!
The F-16 definately is on the radar in the catsailing community. Next thing is to become a well known class in the sailing community at large.

To do this we need to post on other sailing forums like SA and www.yachtsandyachting.com

But if we do post in the outside world, we need to be clear, consise, helpfull and accurate. No point going out and pissing everyone off.

I'm finding that F16 is named more often in the broader catamaran sailing scene, especially publications.

Indeed it looks like we are becoming a generally accepted class. Now we only have to make sure we also become an established class the world over. We are halve way there.

The work is not completed yet, but if we keep our noses clean and stick at it then there is no reason why we will not succeed at that as well.

Onwards ! Everybody.

Wouter

This is great to see.

From our prospective, there was a significant amount of internet chatter in the past about the F16. After the Alter Cup last year though, we began to really see the F16 start to become recognized as a real option. There are now in excess of 50 boats in the US, and even in areas where there are no F16s, the F16 class is now regularly listed with the F18, 20's etc as an option for a boat style/group/class. Apparently this is happening inernationaly as well.

No class or style of boat will fit everyone, and the F16 is definitely not an exception. This class of boat is definitely my favorite of the many designs I have sailed and raced in my lifetime. The feedback we get from virtually everyone who gets on one continues to be extremely positive. As such, to some extent the class sells itself becuase of the boat.

While we have not reached critical mass, with 4 builders (and more at least watching it), and boats reasonably spread across the globe, if we can continue the press and promotion of events I feel the class has a great future.

Regards,

Matt

Not too sure what to make of this comment -

---
A-Catamaran, F-18, B Class, etc. builders are not licensed
Anyone can build a boat... and change it without any notice
In the Olympics, €100000 custom boats would be inevitable
---

Are they contrasting the Tornado with less restrictive rules of the other classes (though I'm not sure that's true)? What are they suggesting would lead to €100000 custom boats?

Mark.

There are some more statements in that ITA section that may be considered strenious.

I wouldn't know how to spend €100.000 on a compliant F18 or F16 and make it show up in the performance beyond say a boat made for €25.000.

Alot of people think that money is the only influence factor for performance (more = faster) and that something that is custom made is automatically better. This is quite often based on misunderstanding the situation. One can only improve a product by having it custom made when the builder knows exactly how the product behaves with respect to even very small changes. This may be true for clothing, but this is certainly not true for sailboat design. Sailboat design still is to a large extend a black art, mixed with large amounts of personal preference and emperical knowlegde. It is impracticle, if not impossible, to achieve significant performance differences by custom designing and custom building a racing boat like the F18.

Look at how much money and time is spend on sail development and how little the steps of improvement really are ?

Wouter

Spending GBP100.000 should be no problem. Just invest in enough computer modelling and CFD analyzes. It dont have to make the boat faster, those writing the checks just need to believe it makes the boat faster.

To build boats for sale, you need to be a licensed builder. Otherwise, amateurs are (still) allowed to build one boat a year for personal use. I think they are arguing that "open" rules are bad for cost in the games while the tornado rules are "closed enough" to give everybody a fair chance?

What ? You can't spend 100.000 on exotic tornado snuffers and sail development if you tried ?

There is no reason why you can spend the entire fortune of Bill Gates on a single beach catamaran, but I really don't believe that that will lead to any significant (= measureable) performance gain over say reasonably well designed a 25.000 boat when limiting yourself to say a ruleset like the F18's.




Too many people believe that this is a magic thing. That one only has to put a computer to work on it and that machine will figure out for you what is best. I do alot of computer simulation on noise disturbed data (= real life data) and it simply is not that simple or even that accurate. Alot of people also believe that they are skilled designers when they know how to operate a finite element software package but in reality they are simply not. A very large portion of the designing is to be found in intepreting the computer generated results and quite often you'll see numerical glitches. It take understanding of the algorithms used to make sound intepretations on accuracy and dependability of the generated data. Simply generating a plot with some colourings in it is not accuracy or dependability. Even plots with margins of error are often better at producing a false sense of accuracy and dependability.

An to make things worse. The behaviour of a hull when sailed over a disturbed watersurface is too complex too model accurately in all its details. That means that stochastic analysis (random variables) must be applied to even be able to "run the numbers" on a computer. Stochastic analysis is what you do to predict the outcomes in a Casino and we all know that we can only assign probability levels to those outcomes and thus never be sure. In my experience, probability analysis is something that is widely misunderstood even by well educated engineers. And I must admit it is a very tricky subject that requires you utmost focus to not overlook details that can change the whole model or your results.

To give an example :

Proof which situation has the highest probability of being truthful or else proof that all situations are just as likely the happen :

-1- it is easiest to throw at least the number 6 ones when throwing the dice 6 times
-2- it is easiest to throw at least the number 6 twice when throwing the dice 12 times
-3- it is easiest to throw at least the number 6 three times when throwing the dice 18 times


And then the best part of stochastic processes it that you need to be able to produce a component far more accurately then the difference that was designed into the boat from a prior design. Else the natural variance in production overpowers your designed in difference. Then all that work is just pointless as you will have to produce 10's of boats and test them all to see which boat is actually reflecting what you designed. Much like how the top Laser sailors go through scores of boats and mast to find the best combination. That is OD design for you !

As a matter of fact just producing a [censored] load of models and find the best one by testing them all may actually be the most inexpensive option as you won't need highly accurate (expensive) production tooling and you can still sell the lesser products to part time racers and recreational sailor, thus getting a pretty sizeable return on investment.

Indeed, one can easily spend 100.000 on a golden doornob with platinum and diamond decorations, but in the end of the day it is just a doornob whose function is implemented just as well by a 1 dollar plastic handle we all buy at wallmart.

Wouter

I did not say I tought it made sense <img src="http://www.catsailor.com/forums/images/graemlins/grin.gif" alt="" /> We both remember the brit T alu-mast development project which was used as leverage to get the carbon mast ballot trough. But let us not tear apart what the ITA is trying to do.

I am not quite sure why the sailing community is so fixated on having limits and extensive restrictions to the equipment. This is definitely not an Olympic contraint. Look at all the money and design that goes into the Bobsled event. To carry the argument even further, I think we should make all the runners wear the same shoes, that would equal things up as well.

100K is easy to spend on a boat if you really want. Unless the event supplies the equipment and swaps out every race, there will be variation that is boat and not sailor related. Part of sailing though is fitting the sailor to the boat; Weight, sailing slyle, etc, etc.

The real question is how does this fit into this thread.... Sorry.

Matt

situation -1- Probability is : 1 - (5/6)^6 = 0,67
situation -2- Probability is : 1 - (5/6)^12 - 12*(1/6)*(5/6)^11 = 0,62
situation -3- Probability is : 1 - (5/6)^18 - 18*(1/6)*(5/6)^17 - 18!/2!*(1/6)^2*(5/6)^16 = 0,60


I had to use the symbol "!" to denote the mathematical operation of faculty, which means :

18! = 18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1

As such the expression 18!/2! denotes

(18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1) / (2*1) = 18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3


Now some people will say, hey look the difference between situation -1- and -3- is quite large as it is comparing 0.67 to 0.60, we could have just found the answer by just simulating it (actually throwing the dice and counting the number of times the statements are satisfied !)

In reality the difference is really not that large and actually simulating the results would require 1000's throws to have the variance of the averaged experimental results be smaller then the probabilty difference of 0.67 - 0.60 = 0.07. It requires a good number more experiments to have a sifficiently high accuracy level assigned to your conclusion to make the conclusions dependable that one situation is more likely to happen then another WHILE ASSUMING LABORATORY LIKE CONDITIONS (tightly controlled) during the whole time you are simulating.

All that for a simple problem as stated above.

Now try the same statistical simulating strick in an environment where the conditions can change each 10 minutes and where the overal problem is a 1000 times more complex as we have in sail boat racing. Now try to discover which change to the design has actuall improved the performance of the boat. And now we know why the aging Hobie Tiger is still just as fast in capable hands as the nacra Infusion when both are using similar suits of sails.


Back to the original topic now.

Wouter

I understand what the ITA is trying to do.. Place itself at the fore-front of cat sailing. Making its case for the T as the cat sailing world's olympic representative. So I'm not going to knock them in their efforts.

Personally I agree costs stated by the ITA and reasoning behind the dropping of the T are furfies. But this isn't a game of reality its a game of perception. At least they are trying to get cats back in.

Also, I don't believe there is anything new in the past 2 decades that have been introduced into cat/skiff building.. Nomex/carbon was state of the art back in 1985 and still is today. Sure they could go to boron/carbon but the benefit comes at a cost. Not in money but in brittleness.. To be honest the T isn't the lightest boat around so why bother? Nomex/carbon is a total waste in the steamroller weight classes. But that is only my opinion *smiles* Hell even the A class can be built down to weight without going to nomex/carbon..


Just thought it was cool to see the ITA acknowledge our little class!!

We call it "factorial" in English.

My favorite combinatorial problem: How many ways are there to put N identical balls in M distinct buckets? It's cool problem because you won't get it through brute force, but when you look at it the right way, you can just write down the answer, with only a basic background in combinatorics.

Another bit of brain candy for math lovers: How many zeros are on the end of "100!" ? This requires only very basic (pre algebra) math, and insight.

--Glenn

My first guesstimate is 21...

the theory I've used has worked for 10! and 20!, but can't check higher numbers with my calculator.

Thanks.

Despite having done very large portions of my education in English I still fail to remember some of the correct names. Even if you become a fluent speaker, you will never become a native speaker.

Wouter

Well done! Want to explain why?

--Glenn

I'll go for 24 zeros.

Each of the numbers ending in a zero create a zero on the end so that's 9 zero's
the 100 adds 2 more (so total is 11)
BUT 50 adds 2 zeros (2x50) so that is a total of 12 (we already countered one of them)

Each number that ends in 5 also creates zero's (so that's 10 zeros)

BUT 25 and 75 create 2 zeros (2x25 and 6x75) so that's 2 more

Total 24 zeros....



Don't understand why it's 21.......

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

The number 9 can not be factored in either 2's or 5's and it is not a prime number, it does however remain in the leftover series when calculating 100! and therefor your statement "-reduce all the numbers down to there prime numbers" is incorrect. Also problematic is factorizing a number like 10 to 5's; the result is 2*5 and 2 is most definately not a prime number either.

2*5 = 10 where neither 2 or 5 is a factor of 10. This is a direct counter example to your statement :"only multiplying by 10 or a factor of 10 will produce a trailing zero". The problem here is your use of the word "only"; if you had left that one out then the statement would have been truthful.


Again, like others, you have discribed a method of calculating the number of trailing zero's without proving that this method is actually producing the correct number of trailing zero's. As such you have ONLY proven the minimal amount of trailing zero's to be 24.

This is a small but very important difference in mathematics.

An example; I can show that 20 can be divided by 2 but this doesn't show that it can ONLY be devided by 2.

Interesting stuff right ? It actually factors in with alot of discussions of boat design we have on these forums. There too people "proof" stuff using simple but not "logically thorough" means, making their believes sometimes unfouded or even wrong.

One example of course being that shorter hulls are always slower and as such the F16's can never be as fast as F18's. They could "proof" this by showing that the shorter (and lighter) P16 is significantly slower then the Prindle 18 while the general layout of both designs is almost identical. This statement is in itself truthful but its extrapolation to the conclusion that therefor the smaller (and lighter) F16's must be significant slower then the very similarly designed F18's is simply wrong.

Or Bill Roberts favourite gem ; That where monohull top speeds are determined by Max Speed = 1.54 *sqrt(hull length), multihull top speeds are determined by Max Speed = 4.5 * sqrt(hull length)

You can proof the existance and validity of the first relation but not of the second even though one can show that both produce relative accurate max speeds for a range of boats (but not for the whole range of possible designs).

Showing that some results/statements are correct does not equate to proving that only they are correct or that they are always correct.

Interestingly enough the reverse is however true. A single counter example is enough to completely devalidate any given statement and as such equates as being a fully enclosed (counter) proof.

Wouter

We are talking on an internet forum, not designing airplanes, rockets, medicine and nuclear or chemical reactors.

I'll stick with "simple is best".

Pff, lots of noise about nothing. For 100! just use your computer:

It is not that simple, is it ?

Most computers can not store integer numbers larger then

(2^64 -1) = (18446744073709551616-1) = 18446744073709551615

as they are limited by 64 bits processors and memory slots.

The number 100! is substantially large then that.

So nearly all numerical software switches to floating point number presentation which shortens the number part of the storage space to at least 56 bit as the exponential needs to be stored as well.

This limit the total integer part of the floating point number to :

(2^56 -1) = (72057594037927936-1) = 72057594037927935

Which in turn only allowes floating point numbers to be accurate only to their 17th digit.

This is much much much smaller then the number 100! you give (about 8 times more digits) :

110! = 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518
286253697920827223758251185210916864000000000000000000000000


So the comment "just use your computer" is misleading at best.

You have to have access to a supercomputer (very large computer bus sizes) or special analytical mathematical software packages to even be able to produce the above number in all its digits.

I dare say that the vast majority of the people out there don not have access to either or even understand the need for either in this situation.

Wouter

I'm not a mathematician but one of the best figures must be

Isn't that the formula: B4I4Q RU/18? <img src="http://www.catsailor.com/forums/images/graemlins/grin.gif" alt="" />

Most definitely and I wouldn't mind seeing her 'box rule' going back to a nautical theme <img src="http://www.catsailor.com/forums/images/graemlins/grin.gif" alt="" />

42. Ask Douglas.

Umm, is this a TEST ?

There are 24 factors of 5 in 100!: one for each multiple of 5 in [1,100] and one more for each multiple of 5*5.

There are >50 factors of 2 in 100!: one for each multiple of 2, one for each multiple of 2*2, one for each multiple of 2*2*2, etc.

So, there are 24 factors of 10 in 100! since 10=2*5.

So there are 24 zeros on the end of 100!, since each zero on the end implies a factor of 10.

--Glenn

P.S.: I got it wrong in my head when I implied it was 21.
P.P.S.: Anyone want the shortcut answer to "N balls in M bins?"

No, this is fun!

--Glenn

Haa Haa, now I did do that one at school before I went to Uni.

But I've forgotten how, I'll let someone else discuss it with Wouter!

Wouter, really got something else to do this time. Besides the one thing I hated about probability theory was the balls and containers examples. For some reason that never set right with me. But I'm sure the formula n!/(k!*(n-k)!) is to found used somewhere in that problem.

Wouter

If this is fun then I'm happy being sad (I prefer playing with words rather than figures) <img src="http://www.catsailor.com/forums/images/graemlins/grin.gif" alt="" />

The Python interpreter I used automatically switch to big integer when it becomes too big to fit in a 64 bit int. That's why I picked it, and that why that result is correct.

3.1415926535897932384626433832795028841971693993751058209749

445923078164062862089986280348253421170679821480865132823066

470938446095505822317253594081284811174502841027019385211055

596446229489549303819644288109756659334461284756482337867831

652712019091456485669234603486104543266482133936072602491412

737245870066063155881748815209209628292540917153643678925903

600113305305488204665213841469519415116094330572703657595919

530921861173819326117931051185480744623799627495673518857527

248912279381830119491298336733624406566430860213949463952247

371907021798609437027705392171762931767523846748184676694051

320005681271452635608277857713427577896091736371787214684409

012249534301465495853710507922796892589235420199561121290219

608640344181598136297747713099605187072113499999983729780499

510597317328160963185950244594553469083026425223082533446850

352619311881710100031378387528865875332083814206171776691473

035982534904287554687311595628638823537875937519577818577805

32171226806613001927876611195909216420198938


GOOGLE !!!!!

don't think this kinda "fun " will get folks flocking to join your class (this forum is the worlds window to F16 after all ) , perhaps you're trying to attrack a niche market

good luck whatever .

Well, with Wouter punting, and no other expressed interest, I'll tell the answer to "how many ways are there to put N balls in M bins" to kill this thread:

--

Any set of balls in bins is uniquely represented by a string like "o|ooo|o|||o" where 'o' represents a ball and '|' represents a wall between bins. There are M-1 walls between bins and N balls, so these strings can be generated by starting with N+M-1 walls and choosing N to be balls.

So the answer is "N+M-1 choose N."

--

I first encountered this problem in a graduate level Information Theory course at Caltech, where it stumped a lot of people. It also stumped my research group at Columbia for a couple weeks before I noticed we were working on the same problem. It looks easy because "buckets" are used as sources of "balls" in basic probability problems, but the usual approaches to solving combinatorial problems problem fail on this one.

The problem looks easy, but is very hard, yet the answer is *obvious* basic combinatorics once you look at the problem the right way. That's why it's my favorite combinatorics problem.

--Glenn

OK, you guys answer this one, dead lock, my next boat is F16.

How does a nail, laying flat on the roadway, when driven over by ME, end up sticking straight through the tread of my tire, causing it to go flat? <img src="http://www.catsailor.com/forums/images/graemlins/crazy.gif" alt="" />

the front tire runs over the nail. with the load running over the head and causing the head to flick the point up. a milli-second later the upright but falling nail is in just the correct position to get the back wheel.. after one revolution the back tie is nicely "nailed"..

That means that python switched to a whole different suit of algoritms and stores the result no longer as a number but as a string, as a word processor does with text.

Basically, it stops using the numerical unit in the microprocessor and handles calculations by (its own) software algorithms. = SLOOOOOOOOW, although with todays microprocessors, everything goes fast.

Certainly makes Python a very interesting piece of software.

Wouter

Typical...there's never a good Idiot Savant around when you need one <img src="http://www.catsailor.com/forums/images/graemlins/mad.gif" alt="" />

Will an idiot do?

If you make something idiotproof along comes another idiot

Gato
Stealing the thread slightly but there again it was way over my head anyway. Your F16 is really coming together looking at the website. Have you started painting yet? have you weighed the Hulls?? Did you get the centreboard out OK??? Are you traveling to Mumbles in August???? Have you seen the Sun recently!!!!! <img src="http://www.catsailor.com/forums/images/graemlins/grin.gif" alt="" />
Mark

Or a better and improved idiot <img src="http://www.catsailor.com/forums/images/graemlins/grin.gif" alt="" />
After working with users of IT systems for over 10 years, I am convinced that there is no such thing as idiotproof. No matter how you build things, somebody will do something not anticipated and ruin your day. Training is the only, true, way..

How IS the boat coming together Gato? On hold for the winter or soon two (or is it three) sailing Blades in Suomi?

My cat is doing like the bears, sleeping in the winter.
Still waiting for the mast for the Sydväst Blade, and I am not happy about that.
Mumbles... I have planned to sail around Corsica in June.

Winter is the time to build and prepare for the next season. I usually loathe november and december. The two darkest, longest months in winter with no snow and nothing much to do. This winter we have been working on the Blade project and it has been the shortest winter ever so far <img src="http://www.catsailor.com/forums/images/graemlins/smile.gif" alt="" /> We'll have to come up with a new project for next winter!

What is the problem with having the mast delivered? And what boat are you going to sail around Corsica?

Well, everything from cat parts has been taking a h... of a long time to get here.
Crsica will most likely be done with a Pixie 14, it is possible to put it on top of the car.
If I make a mast that can be taken apart for my Blade I guess that it would be possible to take on top of the car.

Depending on the car, you can put pretty much anything up there. This road-show pulled up at the scandinavian Tornado champs last year..
Owner was pulled over by the police, but they had to let him go as it turned out that it was road legal, much to their surprise. Legal roof loads depends on the car and quite possibly national regulations <img src="http://www.catsailor.com/forums/images/graemlins/smile.gif" alt="" />



Good luck with the Corsica project! Please test whichever boat you take down there thoroughly before leaving. Nothing is worse than having boat problems forcing you to abandon the project. It would be very to interesting to follow the project..