Find number of zeroes at the right end in n!

Question

Find the number of zeroes at the right end in n! given n is a positive integer.

Statement 1: n is divisible by 8
Statement 2: n is divisible by 25

tanmaykothari2710 · Answer

No of Zeros = Maximum power of 5 in any factorial.

Statement 1 : N is divisible by 8 .
N can hold : 8,16,24,32.....
Multiple answer exist , So A is not Sufficient .

Statement 2 : N is divisible by 25 .
N can hold : 25,50,75......
Multiple answer exist , So B is not Sufficient

Combining both A & B 
N can hold : 200, 400 ....
Multiple answer exist , So both are not Sufficient.

Hence Correct answer : E

Nups1324 · Answer

Number of zero depends on the number of 2s and 5s right? I'm assuming you didn't mention about 2 because it's implicit. Just confirming. Thank you.

Posted from my mobile device

CrackverbalGMAT · Answer

A zero is obtained by the multiplication of a 2 and a 5.

A method to find the number of 0's at the end of n! is to divide n by 5. (We need to find only for 5 as number of 2's will always be greater than number of 5's for any n.)

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 8

n! = 8!, 16!, 24! etc

8! =  \frac{8}{5} . The Quotient = 1. Therefore 8! has 1 zero to its right.

16! =  \frac{16}{5} . The Quotient = 3. Therefore 16! has 3 zeroes to its right.

Since we get 2 different values, Statement 1 is Insufficient.

Answer Option possible is B, C or E

Statement 1 :  n is divisible by 25

n! = 25!, 50!, 75! etc

25! =  \frac{25}{5} . The Quotient = 5

Since the Quotient is  \geq 5 , we divide the Quotient again =  \frac{5}{5}  = 1

Therefore 25! has 5 + 1 = 6 zeroes to its right.

50! =  \frac{50}{5} . The Quotient = 10.

Since the Quotient is  \geq 5 , we divide the Quotient again =  \frac{10}{5}  = 2

Therefore 50! has 10 + 2 = 12 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 8 and 25 = LCM(8,25) = 200

n! = 200!, 400!, 600! etc

200! =  \frac{200}{5} . The Quotient = 40

Since the Quotient is  \geq 5 , we divide the Quotient again =  \frac{40}{5}  = 8

Since the Quotient is  \geq 5 , we divide the Quotient again =  \frac{8}{5}  = 1

Therefore 200! has 40 + 8 + 1 = 49 zeroes to its right.

400! =  \frac{400}{5} . The Quotient = 80

Since the Quotient is  \geq 5 , we divide the Quotient again =  \frac{80}{5}  = 16

Since the Quotient is  \geq 5 , we divide the Quotient again =  \frac{16}{5}  = 3

Therefore 400! has 80 + 16 + 3 = 99 zeroes to its right.

Again we get 2 different values. Both statements combined are insufficient.

Option E

Arun Kumar

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Find number of zeroes at the right end in n!

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