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terminating Zeros

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Senior Manager
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terminating Zeros [#permalink] New post 17 Oct 2008, 10:05
Please explain the approach for the soln. How many terminating zeroes does 200! have?

(A) 40
(B) 48
(C) 49
(D) 55
(E) 64
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Re: terminating Zeros [#permalink] New post 17 Oct 2008, 10:52
nikhilpoddar wrote:
C


What am I missing below?

Multiplication of every 10 numbers from 1 to 90 will have 2 terminating zeros (e.g. 2*5 and 10, 12*15 and 20, etc.). Thus, total 18 zeros. Same is true for numbers from 101 to 190

Thus 18 + 18 = 36 so far.

Additionally, multiplication of 100, 92 and 95 will have 3 terminating zeros. The same for 192, 195, 200.

Hence, 36 + 6 = 42 in total.

I cannot think of other terminating zeros.
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Re: terminating Zeros [#permalink] New post 17 Oct 2008, 11:31
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scthakur wrote:
nikhilpoddar wrote:
C


What am I missing below?

Multiplication of every 10 numbers from 1 to 90 will have 2 terminating zeros (e.g. 2*5 and 10, 12*15 and 20, etc.). Thus, total 18 zeros. Same is true for numbers from 101 to 190

Thus 18 + 18 = 36 so far.

Additionally, multiplication of 100, 92 and 95 will have 3 terminating zeros. The same for 192, 195, 200.

Hence, 36 + 6 = 42 in total.

I cannot think of other terminating zeros.


we must count the no. of 5's in 200!.
200/5 = 40
200/25 = 8
200/125 = 1
Total = 49
Hence, C
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Re: terminating Zeros [#permalink] New post 17 Oct 2008, 11:37
nikhilpoddar wrote:
scthakur wrote:
nikhilpoddar wrote:
C


What am I missing below?

Multiplication of every 10 numbers from 1 to 90 will have 2 terminating zeros (e.g. 2*5 and 10, 12*15 and 20, etc.). Thus, total 18 zeros. Same is true for numbers from 101 to 190

Thus 18 + 18 = 36 so far.

Additionally, multiplication of 100, 92 and 95 will have 3 terminating zeros. The same for 192, 195, 200.

Hence, 36 + 6 = 42 in total.

I cannot think of other terminating zeros.


we must count the no. of 5's in 200!.
200/5 = 40
200/25 = 8
200/125 = 1
Total = 49
Hence, C


But, are we not counting the same number multiple times here? For example, 50 is counted for 200/5 as well as for 200/25, etc.? Am I missing something here?
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Re: terminating Zeros [#permalink] New post 17 Oct 2008, 13:53
No, we are not double counting
50 = 5 * 5 *2
the first five gets calculated when dividing by 5. (i.e 1)
the second five gets calculated when dividing by 25. (i.e 1)
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Re: terminating Zeros [#permalink] New post 17 Oct 2008, 19:57
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A good rule:

First point:
we can get 0 when we multiply 2 with 5 , mean for one 0 we need one pair of 2 and 5

say if we need 4 zero then we need 4 pair of 2 and 5.. 2^4 5^5 .. no other posibility is there..

second point:
the max. no of 0 = min. occu. of 2 OR 5 ..

2^3 * 5^4 = how many zero.. not 4 but 3 because at max we can make 3 pair only..

2^5 * 5^2 = how many zero.. not 5 but 2 because at max we can make 2 pair only..

2^5 * 5^5 = how many zero.. 5 because at max we can make 5 pair only..

Point 3: its obvious that we have less no. of 5 mulitple in 100 dn 200 so just divide 200 by 5,25,125 ...just sum those number and you will get your answer.

Thanks
Re: terminating Zeros   [#permalink] 17 Oct 2008, 19:57
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