Intresting Factorial!

For students of class 11-12 (age 16+)
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samiul_samin
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Intresting Factorial!

Unread post by samiul_samin » Thu Jan 10, 2019 11:03 am

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

Toky
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Re: Intresting Factorial!

Unread post by Toky » 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 $25$.

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samiul_samin
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Re: Intresting Factorial!

Unread post by samiul_samin » Wed Jan 16, 2019 9:55 pm

I have got another answer by using floor function.


$\lfloor \dfrac{5^2}{5}\rfloor+\lfloor \dfrac{5^1}{5}\rfloor=6$
So the lowest number is $25!$.We have $6$ zero at the end of $25!$

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