A trailing zero is created when 2 and 5 are multiplied. So, to count the number of trailing zeroes, we should count how many 2s and how many 5s are there in x!*y!.AAPL wrote:How many trailing zeroes does x!*y! contain, where x and y are integers and y=x+1?
(1) 9<x<14
(2) 10<y<15
The OA is D.
I don't understand this DS question. Please, can any expert help me with it? Thanks.
Since 2 < 5, thus, we should only count the number of 5s in x!*y! as the number of 2s must be greater than or equal to the number of 5s.
(1) 9 < x < 14
Case 1: Say, x = 10 (minimum value), thus, y = 11 (minimum value). Thus, x!*y! = 10!*11!. We see that there are two 5s in 10! and two 5s in 11!. Thus, the total number of 5s = 4. Since the number of 2s must be greater than the number of 5s, we need not care for the number of 2s. The number of trailing 0s would be 4.
Case 2: Say, x = 13 (maximum value), thus, y = 14 (maximum value). Thus, x!*y! = 13!*14!. We see that there are two 5s in 13! and two 5s in 14!. Thus, the total number of 5s = 4. The number of trailing 0s would be 4. Sufficient.
(2) 10 < y < 15
=> 9 < x < 14.
It is a copy of Statement 1. Sufficient.
The correct answer: D
Hope this helps!
-Jay
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