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If 200!/10^n is an integer, what is the largest possible value of n?

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Bunuel
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If 200!/10^n is an integer, what is the largest possible value of n? [#permalink] New post   28 Apr 2016, 03:36
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If 200!/10^n is an integer, what is the largest possible value of n?

A. 40
B. 42
C. 44
D. 48
E. 49
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Abhishek009
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If 200!/10^n is an integer, what is the largest possible value of n? [#permalink] New post   28 Apr 2016, 10:29
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Bunuel wrote:
If 200!/10^n is an integer, what is the largest possible value of n?

A. 40
B. 42
C. 44
D. 48
E. 49


The question actually asks the highest power of 10 which divides 200! ( For a number to be an integer - Without any remainder all the trailing zeroe's must be divided by the denominator)

10 = 2 x 5


200 factorial will have 49 as -

200/5 = 40
40/5 = 8
8/5 =1

So answer will be (E) 49
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raarun
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If 200!/10^n is an integer, what is the largest possible value of n? [#permalink] New post   28 Apr 2016, 05:44
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We have to basically count the number of 5s withing 200
200/5 =40
200/5^2=8
200/5^3=1

We know for sure that there will be atleast 49 even numbers(multiples of 2)within 200

So there are 49 10s in 200!

The ans is E
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Re: If 200!/10^n is an integer, what is the largest possible value of n? [#permalink] New post   04 May 2016, 17:15
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The question is essentially asking "what is the number of trailing zeros in 200!".

Number of trailing zeros in n! is calculated as n/5 + n /\(5^{2}\) + n/\(5^{3}\)+...+n/\(5^{k}\) where \(n\leq{5^k}\)

So, number of trailing zeros in 200! = 200/5 + 200/25 + 200/125 = 40 + 8 + 1 = 49.
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Re: If 200!/10^n is an integer, what is the largest possible value of n? [#permalink] New post   25 Jul 2016, 06:16
Bunuel wrote:
If 200!/10^n is an integer, what is the largest possible value of n?

A. 40
B. 42
C. 44
D. 48
E. 49


\(10^n\) can be written as \((2*5)^n\) = \(2^n*5^n\)

SO we have to figure out how many 2 and 5 are in 200! and then we can remove the common number of 2 and 5 (thus removing 10) from the expression

2 is present 197 times in 200!
5 is present 49 times in 200!

Since 2*5 = 10 , so taking 49 as common we can remove \((2*5)^{49}\) or \(10^{49}\) from 200!
Answer is E
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Re: If 200!/10^n is an integer, what is the largest possible value of n? [#permalink] New post   28 Jul 2016, 11:44
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Abhishek009 wrote:
Bunuel wrote:
If 200!/10^n is an integer, what is the largest possible value of n?

A. 40
B. 42
C. 44
D. 48
E. 49


The question actually asks the highest power of 10 which divides 200! ( For a number to be an integer - Without any remainder all the trailing zeroe's must be divided by the denominator)

10 = 2 x 5


200 factorial will have 49 as -

200/5 = 40
40/5 = 8
8/5 =1

So answer will be (E) 49


I didnt understand what you mean here.
Here is what i did =>
Number of two's are adequate , we need the number of 5's
5,5,5,5,5^2,5,5,5,5,5^2,5,5,5,5,5^2,5,5,5,5,5^2,5,5,5,5,5^3,5,5,5,5,5^2,5,5,5,5,5^2,5,5,5,5,5^2 => 49
Smash that E
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If 200!/10^n is an integer, what is the largest possible value of n? [#permalink] New post   28 Jul 2016, 11:53
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stonecold wrote:

I didnt understand what you mean here.
Here is what i did =>
Number of two's are adequate , we need the number of 5's
5,5,5,5,5^2,5,5,5,5,5^2,5,5,5,5,5^2,5,5,5,5,5^2,5,5,5,5,5^3,5,5,5,5,5^2,5,5,5,5,5^2,5,5,5,5,5^2 => 49
Smash that E


Seems like unnecessary counting and chances of skipping and missing a digit are high.
Also time factor is crucial.
Why don't you apply an easier formula to get the exact numbers of 5's in less than three or four steps ?
Frankly not to be harsh .. you method is rudimentary and inundated with numerous opportunities for silly careless mistakes, not to mention the eons it will take for such a counting.
Imagine if the question was \(\frac{3195!}{10^n}\)
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Re: If 200!/10^n is an integer, what is the largest possible value of n? [#permalink] New post   28 Jul 2016, 12:00
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Bunuel wrote:
If 200!/10^n is an integer, what is the largest possible value of n?

A. 40
B. 42
C. 44
D. 48
E. 49


For 200!/10^n to be an integer. 200! must be a multiple of 10^n or (2*5)^n

Since 5 appears less than 2 does in 200!, 5 is the deciding factor in value of n

200/5= 40
200/5^2= 9

5 appears 40+9= 49 times, which is the maximum value of n

E is the answer
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Re: If 200!/10^n is an integer, what is the largest possible value of n? [#permalink] New post   13 Jul 2020, 23:22
Bunuel wrote:
If 200!/10^n is an integer, what is the largest possible value of n?

A. 40
B. 42
C. 44
D. 48
E. 49


Hi Bunuel,

I understand the concept.

Just wanted to ask if you can provide links of different problems with the same concept?
I want to know how one can frame different questions from this same concept.

Thanks.
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Bunuel
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Re: If 200!/10^n is an integer, what is the largest possible value of n? [#permalink] New post   14 Jul 2020, 00:04
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minirana wrote:
Bunuel wrote:
If 200!/10^n is an integer, what is the largest possible value of n?

A. 40
B. 42
C. 44
D. 48
E. 49


Hi Bunuel,

I understand the concept.

Just wanted to ask if you can provide links of different problems with the same concept?
I want to know how one can frame different questions from this same concept.

Thanks.


Check the following topics from our Special Questions Directory:

12. Trailing Zeros
13. Power of a number in a factorial
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