# How many 0's will be there in 13!? a)2 b)3 c)4 d)5

#### Solution Requireddhonireddy on 17-Jul-2017 23:22

How many 0's will be there in 13!?

a)2
b)3
c)4
d)5
on 28-Jul-2017 13:29

ans:c)4 explanation 13!=13*12*11*10*9*8*7*6*5*4*3*2*1=6227020800
on 28-Jul-2017 14:51

5 pow(1) : 13 Ã· 5 = 2.6, so I get 2 factors of 5 5 pow(2) : 13 Ã· 25 = 0.52, which is less than 1, so I stop here. Then 13! has 2+0=2 trailing zeroes.
on 28-Jul-2017 14:53

5 pow(1) : 13 / 5 = 2.6, so I get 2 factors of 5 5 pow(2) : 13 / 25 = 0.52, which is less than 1, so I stop here. Then 13! has 2+0=2 trailing zeroes. so the answer is a)2
on 29-Jul-2017 10:35

Answer is 4 for dhonireddy question 5! is 120 in 13! 10 is included so 1 more 0 is included we know 7*8*9 =504
on 31-Jul-2017 18:19

ans is 4 because 13!=6,227,020,800 hence it is four
on 5-Aug-2017 15:44

13!=13*12*11*10*9*8*7*6*5*4*3*2*1=6227020800 answer is 4
on 6-Aug-2017 01:05

4
on 9-Aug-2017 15:03

13!=13*12*11*10*9*8*7*6*5*4*3*2*1=6227020800 therefore 4 zero's in 13! option c)4 is the answer
on 13-Aug-2017 15:29

answer is c)4 explanation:13!=13*12*11*10*9*8*7*6*5*4*3*2*1=6,227,020,800
on 15-Aug-2017 22:12

Trailing zeroes : 5*2 , 10 = 2 trailing zeroes. 7*8*9 = 504, and one more zero is included so, ANS: 4
on 17-Aug-2017 10:23

4. 13!= 1*2*3*4*5*6*7*8*9*10*11*12*13=6227020800