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24 May 2019, 18:57
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
51% (02:12) correct
49%
(02:50)
wrong
based on 93
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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
A. 5
B. 6
C. 7
D. 8
E. 9
24 May 2019, 20:08
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nick1816
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
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.
24 May 2019, 20:20
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nick1816
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
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
