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24 Aug 2020, 22:28
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
81% (00:47) correct
19%
(00:56)
wrong
based on 69
sessions
History
Date
Time
Result
Not Attempted Yet
Find the number of zeroes at the right end in n! given n is a positive integer.
Statement 1: n is divisible by 4
Statement 2: n is divisible by 5
Statement 1: n is divisible by 4
Statement 2: n is divisible by 5
24 Aug 2020, 23:49
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Statement 1: Given that n is divisible by 4
which implies that n can be 4,8,12....
For the number to have 0 it should either have 5*2 and 10 in the factorial
For n=4, n!=4!=24 (0 zeros)
For n=8, n!=8!= xxx0 (Since 8! will have one 5 and 2 we will have 1 zero)
for n=12, n!=12!= xxxx00 (Since 12! will have 10, 5 and 2 we will have 2 zeros)
Hence Statement 1 is insufficient
Statement 2: Given that n is divisible by 5
which implies that n can be 5,10,15...
For n=5, n!=5!=120 (1 zero)
For n=10, n!=10!= xx00 (Since 10! will have 10, 5 and 2 we will have 2 zeros)
Hence Statement 2 is insufficient
Statement 1 & 2 taken together
n is divisible by both 5 and 4 implying that n is divisible by 20
n=20,40....
For n =20, 20! (Has 20,15,10,5,2 which will give 4 zeros)
For n=40, 40! (has 40,35,30,25,20,15,10,5 and 2 which will lead to 8 zeros)
Hence both are insufficient
OA is E
which implies that n can be 4,8,12....
For the number to have 0 it should either have 5*2 and 10 in the factorial
For n=4, n!=4!=24 (0 zeros)
For n=8, n!=8!= xxx0 (Since 8! will have one 5 and 2 we will have 1 zero)
for n=12, n!=12!= xxxx00 (Since 12! will have 10, 5 and 2 we will have 2 zeros)
Hence Statement 1 is insufficient
Statement 2: Given that n is divisible by 5
which implies that n can be 5,10,15...
For n=5, n!=5!=120 (1 zero)
For n=10, n!=10!= xx00 (Since 10! will have 10, 5 and 2 we will have 2 zeros)
Hence Statement 2 is insufficient
Statement 1 & 2 taken together
n is divisible by both 5 and 4 implying that n is divisible by 20
n=20,40....
For n =20, 20! (Has 20,15,10,5,2 which will give 4 zeros)
For n=40, 40! (has 40,35,30,25,20,15,10,5 and 2 which will lead to 8 zeros)
Hence both are insufficient
OA is E
25 Aug 2020, 07:53
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A method to find the number of 0's at the end of n! is to divide n by 5.
If the quotient obtained is \(\geq 5\), then divide the quotient again by 5.
This process is repeated until the quotient obtained is less than 5. We then add all the quotients, which gives our required answer.
Note: Do not worry about the remainder. Just the quotient obtained.
For eg. number of trailing 0's for 100!
Divide 100 by 5, Q = 20
Divide 20 by 5, we get Q = 4
Therefore the number of trailing 0's = 20 + 4 = 24
In the question above
Statement 1: n is divisible by 4
n! = 4!, 8!, 12! etc
4! = 24 has no zeroes to the end (as there is no 5).
8! = \(\frac{8}{5}\), Quotient = 1. Therefore 8! has 1 zero to its right.
Since we get 2 different values, Statement 1 is Insufficient.
Answer Option possible is B, C or E
Statement 2: n is divisible by 5
n! = 5!, 10!, 15! etc
5! = \(\frac{5}{5}\), Quotient = 1. Therefore 5! has 1 zero to its right.
10! = \(\frac{10}{5}\), Quotient = 2. Therefore 10! has 2 zeroes to its right.
Again we get 2 different values. Statement 2 is Insufficient.
Answer Options possible is C or E
Combining both the statements : n is divisible by 4 and 5 = LCM(4,5) = 20
n! = 20!, 40!, 60! etc
20! = \(\frac{20}{5}\), Quotient = 4. Therefore 20! has 4 zeroes to its right.
40! = \(\frac{40}{5}\), Quotient = 8.
\(\frac{8}{5}\), Quotient = 1.
Therefore 40! has 8 + 1 = 9 zeroes to its right.
Again we get 2 different values. Both statements combined are insufficient.
Option E
Arun Kumar
If the quotient obtained is \(\geq 5\), then divide the quotient again by 5.
This process is repeated until the quotient obtained is less than 5. We then add all the quotients, which gives our required answer.
Note: Do not worry about the remainder. Just the quotient obtained.
For eg. number of trailing 0's for 100!
Divide 100 by 5, Q = 20
Divide 20 by 5, we get Q = 4
Therefore the number of trailing 0's = 20 + 4 = 24
In the question above
Statement 1: n is divisible by 4
n! = 4!, 8!, 12! etc
4! = 24 has no zeroes to the end (as there is no 5).
8! = \(\frac{8}{5}\), Quotient = 1. Therefore 8! has 1 zero to its right.
Since we get 2 different values, Statement 1 is Insufficient.
Answer Option possible is B, C or E
Statement 2: n is divisible by 5
n! = 5!, 10!, 15! etc
5! = \(\frac{5}{5}\), Quotient = 1. Therefore 5! has 1 zero to its right.
10! = \(\frac{10}{5}\), Quotient = 2. Therefore 10! has 2 zeroes to its right.
Again we get 2 different values. Statement 2 is Insufficient.
Answer Options possible is C or E
Combining both the statements : n is divisible by 4 and 5 = LCM(4,5) = 20
n! = 20!, 40!, 60! etc
20! = \(\frac{20}{5}\), Quotient = 4. Therefore 20! has 4 zeroes to its right.
40! = \(\frac{40}{5}\), Quotient = 8.
\(\frac{8}{5}\), Quotient = 1.
Therefore 40! has 8 + 1 = 9 zeroes to its right.
Again we get 2 different values. Both statements combined are insufficient.
Option E
Arun Kumar
