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

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

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28 Apr 2016, 03:36
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61% (00:39) correct 39% (01:18) wrong based on 224 sessions

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

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28 Apr 2016, 05:44
1
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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If 200!/10^n is an integer, what is the largest possible value of n?  [#permalink]

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28 Apr 2016, 10:29
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1
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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Re: If 200!/10^n is an integer, what is the largest possible value of n?  [#permalink]

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04 May 2016, 17:15
1
2
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]

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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!
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Re: If 200!/10^n is an integer, what is the largest possible value of n?  [#permalink]

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28 Jul 2016, 11:44
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]

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28 Jul 2016, 11:53
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]

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28 Jul 2016, 12:00
1
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]

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25 Nov 2017, 13:10
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