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

Question

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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Harley1980 · Accepted Answer

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

bb · Answer

Slight Subject modification for readability. Thank you.

Lucky2783 · Answer

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?

Harley1980 · Answer

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

bgpower · Answer

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!

BrushMyQuant · Answer

Please tag number properties!
Thank you

KarishmaB · Answer

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.

EricImasogie · Answer

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 ?

KarishmaB · Answer

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.

RJ7X0DefiningMyX · Answer

Got the right answer, but took me 03:09. Okay, had I not been sitting on my bed and tried to solve this with a little more focus and speed, the best I would've probably got to is 02:30.

Any quicker approach rather than finding out the number of 2s and 5s?

ccooley   VeritasKarishma

Shivangisharmaa · Answer

But N is not a fraction.. shouldn't the question be how many zeroes are after the 1st non-zero integer, since N is an integer?
Am I going wrong? kindly help

findingmyself · Answer

I don't think so, Took me close to 5 minutes to even understand the solution. Keep aside the answers. You are doing great!!!

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards