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# N! is ending with x zeroes. (N + 3)! is ending with (x + 2) zeroes. Ho

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DS Forum Moderator
Joined: 19 Oct 2018
Posts: 1826
Location: India
N! is ending with x zeroes. (N + 3)! is ending with (x + 2) zeroes. Ho  [#permalink]

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24 May 2019, 17:57
5
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Difficulty:

55% (hard)

Question Stats:

68% (02:32) correct 32% (03:24) wrong based on 22 sessions

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N! is ending with x zeroes. (N + 3)! is ending with (x + 2) zeroes. How many such N's will exist, given N is positive integer less than 50?

A. 5
B. 6
C. 7
D. 8
E. 9
SVP
Joined: 20 Jul 2017
Posts: 1506
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: N! is ending with x zeroes. (N + 3)! is ending with (x + 2) zeroes. Ho  [#permalink]

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24 May 2019, 19:08
1
nick1816 wrote:
N! is ending with x zeroes. (N + 3)! is ending with (x + 2) zeroes. How many such N's will exist, given N is positive integer less than 50?

A. 5
B. 6
C. 7
D. 8
E. 9

We have to understand that to increase 2 number of 0's, (n+3) should have either '25' or '50' {25x4 = 100, 50x4 = 200) and 'n' SHOULD NOT have any of '25' and '50'. Only then there would be an addition of 2 0's

So, possible values of n are 22, 23, 24, 47, 48 and 49.

IMO B.
ISB School Moderator
Joined: 08 Dec 2013
Posts: 776
Location: India
Concentration: Nonprofit, Sustainability
Schools: ISB
GMAT 1: 630 Q47 V30
WE: Operations (Non-Profit and Government)
Re: N! is ending with x zeroes. (N + 3)! is ending with (x + 2) zeroes. Ho  [#permalink]

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24 May 2019, 19:20
3
nick1816 wrote:
N! is ending with x zeroes. (N + 3)! is ending with (x + 2) zeroes. How many such N's will exist, given N is positive integer less than 50?

A. 5
B. 6
C. 7
D. 8
E. 9

I've used two concepts-

1. In any n! the number of zeroes is primarily contributed by 5s in the product. (we have more 2s in comparison to 5)
2. Box Function- to calculate the number of zeroes in a factorial. Trailing Zeroes: https://gmatclub.com/forum/how-many-tra ... l#p2190991

Interesting facts- Number of 5s gets bumped at 25, 50 etc.

Possible combinations: (N, N+3); N<50
22, 25
23, 26
24, 27

47, 50
48, 51
49, 52

So, 6 Ns are possible. B
CEO
Joined: 03 Jun 2019
Posts: 2889
Location: India
GMAT 1: 690 Q50 V34
WE: Engineering (Transportation)
N! is ending with x zeroes. (N + 3)! is ending with (x + 2) zeroes. Ho  [#permalink]

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28 Mar 2020, 04:16
1
nick1816 wrote:
N! is ending with x zeroes. (N + 3)! is ending with (x + 2) zeroes. How many such N's will exist, given N is positive integer less than 50?

A. 5
B. 6
C. 7
D. 8
E. 9

N+3 = 5^2k = 25k form
N+3 = {25,50}
N = {22,23,24,47,48,49}

IMO B
_________________
Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com
N! is ending with x zeroes. (N + 3)! is ending with (x + 2) zeroes. Ho   [#permalink] 28 Mar 2020, 04:16