#128381 - 01/14/08 07:05 PM
Re: Answers to the question
[Re: Glenn_Brown]
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Carpal Tunnel
Registered: 06/16/01
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Another bit of brain candy for math lovers: How many zeros are on the end of "100!"
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
Edited by Wouter (01/14/08 08:25 PM)
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Wouter Hijink
Formula 16 NED 243 (one-off; homebuild)
The Netherlands
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#128383 - 01/14/08 10:22 PM
Re: Answers to the question
[Re: Wouter]
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Carpal Tunnel
Registered: 06/16/01
Posts: 9582
Loc: North-West Europe
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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) 
Edited by Wouter (01/14/08 10:26 PM)
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Wouter Hijink
Formula 16 NED 243 (one-off; homebuild)
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#128384 - 01/14/08 10:25 PM
Re: Answers to the question
[Re: Wouter]
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Hull Flying, Snow Sliding....
Carpal Tunnel
Registered: 02/06/04
Posts: 3526
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Quote:
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.
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#128385 - 01/14/08 10:29 PM
Re: Answers to the question
[Re: scooby_simon]
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Carpal Tunnel
Registered: 06/16/01
Posts: 9582
Loc: North-West Europe
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Quote:
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 :
Quote:
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....
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
Edited by Wouter (01/14/08 10:50 PM)
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#128386 - 01/14/08 10:35 PM
Re: Answers to the question
[Re: Wouter]
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Hull Flying, Snow Sliding....
Carpal Tunnel
Registered: 02/06/04
Posts: 3526
Loc: Looking for a Job, I got credi...
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Quote:
Quote:
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 got luck by finding the correct number by employing a pathway that is not thoroughly logical.
I'll give a few examples :
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.
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#128387 - 01/14/08 11:03 PM
Re: Answers to the question
[Re: scooby_simon]
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Carpal Tunnel
Registered: 06/16/01
Posts: 9582
Loc: North-West Europe
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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. Quote:
Understood what was causing the zero's to be produced
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.
Quote:
Simple is best IMO
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
Edited by Wouter (01/14/08 11:15 PM)
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Wouter Hijink
Formula 16 NED 243 (one-off; homebuild)
The Netherlands
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#128388 - 01/14/08 11:52 PM
Re: Answers to the question
[Re: Wouter]
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Hull Flying, Snow Sliding....
Carpal Tunnel
Registered: 02/06/04
Posts: 3526
Loc: Looking for a Job, I got credi...
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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!  More here and here
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