number theory problem
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
number theory problem
Find the remainder when $1!+2!+3!....+99!$ is divided by $30$.

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Re: number theory problem
wait
i don't understand. both 5! and 6! are also divisible by 30
i don't understand. both 5! and 6! are also divisible by 30
Re: number theory problem
i did not get what you meant ....but explaing a liitle bitviivviiave wrote:wait
i don't understand. both 5! and 6! are also divisible by 30
you can clearly see $6!=1.2.3.4.5.6$ so its clearly divisible by 30 and from 6! any 'integer factorial' would be divisible by 30 . so we just need to think about $1!$ to $5!$ and i you are right that $5!$ is divisible by 30

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Re: number theory problem
yup i did'nt think about $5!$......it was my mistake but as $a$ $\equiv$ $a+n$ $(mod n)$, $5!$ can't change the ans.viivviiave wrote:wait
i don't understand. both 5! and 6! are also divisible by 30

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Re: number theory problem
yup!!!AntiviruShahriar wrote:yup i did'nt think about $5!$......it was my mistake but as $a$ $\equiv$ $a+n$ $(mod n)$, $5!$ can't change the ans.viivviiave wrote:wait
i don't understand. both 5! and 6! are also divisible by 30
Re: number theory problem
I can't understand the question...if the given series is divided by 30 how can we get remainder????/
Re: number theory problem
from $5!$ to $99!$ is divisible bt 30 .
 leonardo shawon
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Re: number theory problem
Then 1! 2! 3! 4! ?? That mean 288?
Ibtehaz Shawon
BRAC University.
long way to go .....
BRAC University.
long way to go .....

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Re: number theory problem
you've multiplied them... was the question like that?leonardo shawon wrote:Then 1! 2! 3! 4! ?? That mean 288?