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Eternal Intern
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Let P be the product of the first 10 positive integers. If [#permalink]
 08 Jul 2003, 20:48
Let P be the product of the first 10 positive integers. If P/ 10 ^ X is an integer, what is the maximum possible value of x?
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For P/(10 ^ X) to be integral, P must have at least X trailing zeroes. Figure out a way to determine how many zeroes P has.
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
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10!/10^X = an integer
so X is the maximal number of tens in 10!
10!=1*2*3*4*5*6*7*8*9*10
so, 10! includes two tens
X=2
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What are trailing zeros
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stolyar wrote: 10!/10^X = an integer
so X is the maximal number of tens in 10!
10!=1*2*3*4*5*6*7*8*9*10
so, 10! includes two tens
X=2
BRAVO! 
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It would seem helpful to know facts like
5! = 120
6! = 720
Just in case you have brain cramp when this comes up as Q#27 and you are starting to get tired of math, you need only do a few calcs...
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Curly05 wrote: What are trailing zeros
100 has 2 trailing zeroes.
21000 has 3 trailing zeroes.
1200000 has 5 trailing zeroes.
One billion (US) has 9 trailing zeroes.
Google has 100 trailing zeroes.
Got it?
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
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My short cut would be, in order to be divisable for 10^x, it means that the number must be able to divisable for 5 and 2 at least. If you look into the following: 10x9x8x7x6x5x4x3x2x1 , you will see that besides 10 and 5, there are not others qualify for 5 factor anymore. So the answer must be 2  .
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minghoo wrote: My short cut would be, in order to be divisable for 10^x, it means that the number must be able to divisable for 5 and 2 at least. If you look into the following: 10x9x8x7x6x5x4x3x2x1 , you will see that besides 10 and 5, there are not others qualify for 5 factor anymore. So the answer must be 2 . what would you do if the P = 200! ?
 ?
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Well, Brainless, give it a show, I think Stolyar's way is the best.
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In stolyar's method also, you will need to count for 200!
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Re: I have no idea [#permalink]
25 Jul 2003, 09:52
Curly05 wrote: Well, Brainless, give it a show, I think Stolyar's way is the best.
We all know it that to get a trailing zero , we need a 5 as one of the factors of the number. So in 10!, we need to just find how many numbers are there which are multiples of 5 . Clearly they are 5 and 10. So in 10!, The maximum possible value of X ( as aked in the question ) is 10/5 = 2.
Now lets take 20!, The maximum possible value of X will be 20/5 = 4.
Now lets make it even bigger , 100!
Now we have 100/5 = 20 numbers which are multiples of 5
100/25 = 4 numbers which are multiples of 25
100/125 = 0 , so we dont have to go further
So there will be 24 trailing zeroes, hence maximum possible value for X would be 24.
Now I guess , you will be able to figure it out the maximum possible value for X in 200! or infact for any factorial , no matter how big it is ..
BTW, I do agree Stolyar is the BEST.
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Re: I have no idea
[#permalink]
25 Jul 2003, 09:52
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