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If 200!/10^n is an integer, what is the largest possible value of n?
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28 Apr 2016, 02:36
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58% (00:58) correct 42% (01:40) wrong based on 269 sessions
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If 200!/10^n is an integer, what is the largest possible value of n?
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28 Apr 2016, 04:44
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?
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28 Apr 2016, 09:29
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?
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04 May 2016, 16:15
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?
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25 Jul 2016, 05:16
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?
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28 Jul 2016, 10:44
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) 49I 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?
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28 Jul 2016, 10:53
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?
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28 Jul 2016, 11:00
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?
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