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

Question

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

Abhishek009 · Accepted Answer

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

raarun · Answer

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

unverifiedvoracity · Answer

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.

LogicGuru1 · Answer

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

stonecold · Answer

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

LogicGuru1 · Answer

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}

Divyadisha · Answer

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

minirana · Answer

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.

Bunuel · Answer

Check the following topics from our  Special Questions Directory :

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

bumpbot · Answer

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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?

Prep Toolkit

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