number theory problem

[tushar7]

Find the remainder when $1!+2!+3!....+99!$ is divided by $30$.

[AntiviruShahriar]

$6!$ থেকে বড় শংখ্যার factorial গুলো $30$ দ্বারা বিভাজ্য।
অর্থাৎ, $1!+2!+3!+4!+5!$ $\equiv$ $3$ (mod 30)
ans: $3$.......

[viivviiave]

wait
i don't understand. both 5! and 6! are also divisible by 30

[tushar7]

wait
i don't understand. both 5! and 6! are also divisible by 30

i did not get what you meant ....but explaing a liitle bit

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

[AntiviruShahriar]

wait
i don't understand. both 5! and 6! are also divisible by 30

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]

wait
i don't understand. both 5! and 6! are also divisible by 30

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.

yup!!!

[Dipan]

I can't understand the question...if the given series is divided by 30 how can we get remainder????/

[tushar7]

from $5!$ to $99!$ is divisible bt 30 .

[leonardo shawon]

Then 1! 2! 3! 4! ?? That mean 288?

[HandaramTheGreat]

Then 1! 2! 3! 4! ?? That mean 288?

you've multiplied them... was the question like that?