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terminating Zeros

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Senior Manager
Joined: 18 Jun 2007
Posts: 285

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17 Oct 2008, 10:05
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Please explain the approach for the soln. How many terminating zeroes does 200! have?

(A) 40
(B) 48
(C) 49
(D) 55
(E) 64
SVP
Joined: 17 Jun 2008
Posts: 1529

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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.
Manager
Joined: 14 Jan 2006
Posts: 95
Schools: HKUST

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17 Oct 2008, 11:31
1
KUDOS
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
SVP
Joined: 17 Jun 2008
Posts: 1529

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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?
Manager
Joined: 14 Jan 2006
Posts: 95
Schools: HKUST

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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)
Manager
Joined: 09 Oct 2008
Posts: 93

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17 Oct 2008, 19:57
1
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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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terminating Zeros

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