Intresting Factorial!

 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Intresting Factorial!
For which lowest positive integer $n$, the last six digits of $n!$ is $0$ ?
Re: Intresting Factorial!
The number is $25$. First we have to think how $0$ produces at last position. It must be done by multiplying by $10$. So we need to determine the factorial of lowest number that produces $6$, 10s...
$25! = 1\times2\times3\times4\times... ...\times24\times25$
Here
i. $2\times5 = 10$
ii. $10 = 1\times10$
iii. $12\times15 = 180 = 18\times10$
iv. $20 = 2\times10$
v. $22\times25 = 550 = 55\times10$
vi. $55 \times 4 = 220 = 22 \times 10$ (55 got from $v$.)
This is how we get $6$ , 10s for factorial $25$. So the answer is $25$.
$25! = 1\times2\times3\times4\times... ...\times24\times25$
Here
i. $2\times5 = 10$
ii. $10 = 1\times10$
iii. $12\times15 = 180 = 18\times10$
iv. $20 = 2\times10$
v. $22\times25 = 550 = 55\times10$
vi. $55 \times 4 = 220 = 22 \times 10$ (55 got from $v$.)
This is how we get $6$ , 10s for factorial $25$. So the answer is $25$.

 Posts: 1007
 Joined: Sat Dec 09, 2017 1:32 pm
Re: Intresting Factorial!
I have got another answer by using floor function.