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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.

_________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

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?

_________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

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 .

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 .

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.

_________________

Brainless

gmatclubot

Re: I have no idea
[#permalink]
25 Jul 2003, 08:52