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IF N=5^5*..., then how many zeroes exist between decimal and first non

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IF N=5^5*..., then how many zeroes exist between decimal and first non  [#permalink]

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New post 02 Apr 2015, 08:25
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IF N=5^5*10^10*15^15*20^20*25^25*30^30*35^35*40^40*45^45*50^50 , then how many zeroes exist between decimal and first non zero digit of N ?

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Re: IF N=5^5*..., then how many zeroes exist between decimal and first non  [#permalink]

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New post 02 Apr 2015, 08:33
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CrazyIvan wrote:
IF N=5^5*10^10*15^15*20^20*25^25*30^30*35^35*40^40*45^45*50^50 , then how many zeroes exist between decimal and first non zero digit of N ?

150
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For calculating 0 at the end we should calculate quantity of "2" and "5" in the final number.

in all numbers which end at 5 we don't have any "2" but have a lot of "5".
So in final number we will have more "5" than "2" and we need to calculate quantity of "2" in each number
10 - one "2" in 10 exponent = 10
20 - two "2" in 20 exponent = 40
30 - one "2" in 30 exponent = 30
40 - three "2" in 40 exponent = 120
50 - one "2" in 50 exponent = 50

10 + 40 + 30 + 120 + 50 = 250
Answer is C
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Re: IF N=5^5*..., then how many zeroes exist between decimal and first non  [#permalink]

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New post 02 Apr 2015, 08:36
Slight Subject modification for readability. Thank you.
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Re: IF N=5^5*..., then how many zeroes exist between decimal and first non  [#permalink]

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New post 02 Apr 2015, 11:27
CrazyIvan wrote:
IF N=5^5*10^10*15^15*20^20*25^25*30^30*35^35*40^40*45^45*50^50 , then how many zeroes exist between decimal and first non zero digit of N ?

150
200
250
300
350



I have no different method than what has been provided by Harley1980.
thats the most obvious and efficient way IMO.

does anyone have urls for such questions?
Dont even know which category these questions fall into.
chetan2u any idea?
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Re: IF N=5^5*..., then how many zeroes exist between decimal and first non  [#permalink]

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New post 02 Apr 2015, 11:32
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Lucky2783 wrote:
CrazyIvan wrote:
IF N=5^5*10^10*15^15*20^20*25^25*30^30*35^35*40^40*45^45*50^50 , then how many zeroes exist between decimal and first non zero digit of N ?

150
200
250
300
350



I have no different method than what has been provided by Harley1980.
thats the most obvious and efficient way IMO.

does anyone have urls for such questions?
Dont even know which category these questions fall into.
chetan2u any idea?


I think this is Number properties
Here is some examples:
how-many-zeros-does-100-end-with-100599.html
how-many-zeros-are-the-end-of-142479.html
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Re: IF N=5^5*..., then how many zeroes exist between decimal and first non  [#permalink]

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New post 04 Aug 2015, 06:15
Awesome question!

Can this somehow be solved by applying @Bunuel's approach in dividing by 5 and it's exponents (25, 125, etc.)?

Thank you!
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Re: IF N=5^5*..., then how many zeroes exist between decimal and first non  [#permalink]

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New post 05 Aug 2015, 20:08
Please tag number properties!
Thank you
CrazyIvan wrote:
IF N=5^5*10^10*15^15*20^20*25^25*30^30*35^35*40^40*45^45*50^50 , then how many zeroes exist between decimal and first non zero digit of N ?

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Re: IF N=5^5*..., then how many zeroes exist between decimal and first non  [#permalink]

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New post 05 Aug 2015, 21:28
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bgpower wrote:
Awesome question!

Can this somehow be solved by applying @Bunuel's approach in dividing by 5 and it's exponents (25, 125, etc.)?

Thank you!


No. If N were

N = 5*10*15*20*25*... and so on, then you could have put all the 5s together and you would have been left with 1*2*3*4*5*6*7...
Here, to find the number of 5s, you could have used Bunuel's approach of dividing by 5 and then 25 etc. Note that that approach works for factorials only.

OR if N were

N = 5^5 * 10^5 * 15^5 * 20^5 * 25^5 ... and so on, then you could have also used that method by raising the whole thing to a common power of 5 and then proceeding as done above.
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Re: IF N=5^5*..., then how many zeroes exist between decimal and first non  [#permalink]

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New post 31 Oct 2015, 23:11
How does the approach ,shown by Harley1980, take into account the n! counting method of n/5 + n/5^2 + n/5^3......n/5^k ?

I cant see how this counting method can be used in this question ?
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Re: IF N=5^5*..., then how many zeroes exist between decimal and first non  [#permalink]

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New post 02 Nov 2015, 20:37
EricImasogie wrote:
How does the approach ,shown by Harley1980, take into account the n! counting method of n/5 + n/5^2 + n/5^3......n/5^k ?

I cant see how this counting method can be used in this question ?


It doesn't since we cannot use that method here. Every term has a different exponent so you have to account for the exponents too. In that method, the exponent is always 1.
10! = 1*2*3*4*5*6*7*8*9*10

Here, instead we have
N=5^5*10^10*15^15*20^20*25^25*30^30*35^35*40^40*45^45*50^50
There are 2s only in alternate terms
10^10 * 20^20 * 30^30 * 40^40 * 50^50

So you just consider independent terms and that's all.
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Re: IF N=5^5*..., then how many zeroes exist between decimal and first non  [#permalink]

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Re: IF N=5^5*..., then how many zeroes exist between decimal and first non &nbs [#permalink] 27 Jul 2018, 18:34
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