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# Let P be the product of the first 10 positive integers. If

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Let P be the product of the first 10 positive integers. If [#permalink]  08 Jul 2003, 19:48
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
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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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Hi [#permalink]  09 Jul 2003, 05:58
What are trailing zeros
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Expert's post
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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Re: Hi [#permalink]  14 Jul 2003, 00:39
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.

Got it?
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
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Short Cut, [#permalink]  14 Jul 2003, 10:07
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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Re: Short Cut, [#permalink]  15 Jul 2003, 04:11
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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I have no idea [#permalink]  16 Jul 2003, 14:06
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, 08:52
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, 08:52
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