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16 Mar 2020, 22:12
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
62% (02:08) correct
38%
(01:46)
wrong
based on 134
sessions
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Which of the following cannot be the number of trailing zeros in n! Where n is a natural number ?
A. 24
B. 32
C. 43
D. 29
E. 62
Posted from my mobile device
A. 24
B. 32
C. 43
D. 29
E. 62
Posted from my mobile device
17 Mar 2020, 04:16
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Raksat
Number of trailing zeroes will depend on number of 5.
REASON :- We will always have more number of 2s as compared to number of 5s, and the trailing 0s would depend on number of 2s and 5s.
Now every multiple of 5 will increase a trailing 0, so 5, 10, 15, 20, 30, 35...will result in ONE extra 0, while 25, 50, 75 will result in TWO extra 0s and 125, 250... will result in THREE extra 0s..
So the zeroes increased will follow pattern 1,1,1,1,2(when n is 25),1,1,1,1,2(when n is 50),1,1,1,1,2(when n is 75),1,1,1,1,2(when n is 100),1,1,1,1,3(when n is 125),1,1,1,1,2(when n is 150)...
so the number of 5s that is NOT possible when 5 increases by 2..
for example 20! gives 4*5s, but 25 gives 6*5s, so 5 number of 5s is NOT possible, meaning there cannot be 5 trailing 0s.
Thus smallest number of trailing 0s that is not possible is 5 and next would be 6 added to it (1,1,1,1,2=6), so 5+6=11, then 17, 23, 29
We have 29 in the choices..
D
Note where we have 125, there will two consecutive numbers that will not be possible .....
17 Mar 2020, 04:59
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Raksat
The sequence of trailing zeros in n! where n is a natural number = {0,0,0,0,1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4,6,6,6,6,6,7,7,7,7,7,8,8,8,8,8,9,9,9,9,9,10,10,10,10,10,12,12,12,12,12,.... }
The number of trailing 0s in n! = {0,1,2,3,4,6,7,8,9,10,12,13,14,15,16,18,...}
The number of missing trailing 0s in n! = {5,11,17,23,29,....}
IMO D
