### Intresting Factorial!

Posted:

**Thu Jan 10, 2019 11:03 am**For which lowest positive integer $n$, the last six digits of $n!$ is $0$ ?

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Posted: **Thu Jan 10, 2019 11:03 am**

For which lowest positive integer $n$, the last six digits of $n!$ is $0$ ?

Posted: **Tue Jan 15, 2019 12:29 am**

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 $25$.

Posted: **Wed Jan 16, 2019 9:55 pm**

I have got another answer by using floor function.