BTGmoderatorDC wrote: ↑Tue May 05, 2020 1:51 am
If 60! is written out as an integer, with how many consecutive 0’s will that integer end?
A. 6
B. 12
C. 14
D. 42
E. 56
OA
C
Source: Manhattan Prep
A trailing 0 is formed when 2 is multiplied to 5. Thus, we must count the number of 2s and that of 5s in 60!. Lesser of the count of 2s and 5s will determine the no. of trailing 0s.
Since 5 > 2, there would be fewer 5s than 2s in 60!. So, we should count no. of 5s in 60!.
We know that 60! = 1.2.3.4.5...60.
To get the no. of 5s, let's read the table of 5 till 60.
It's 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60.
Each of the 12 numbers 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 have one 5 except 25 and 50; they have two 5s each.
So, there are 12 + 1 + 1 = 14 fives.
So, there would be 14 consecutive 0s in 60!.
The correct answer:
C
Hope this helps!
-Jay
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