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Re: terminating Zeros [#permalink]
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.

Re: terminating Zeros [#permalink]
17 Oct 2008, 11:31

1

This post received 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

Re: terminating Zeros [#permalink]
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?

Re: terminating Zeros [#permalink]
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)

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

1

This post received KUDOS

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

gmatclubot

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

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