# What is the remainder when 1!+2!+3!+4!+5!+…+50! is divided by 5!?

What is the remainder when 1!+2!+3!+4!+5!+.......+50! is divided by 5!

My Approach

$1$+$2$+$6$+$24$+$5$!/$5$!+$6 . 5$!/$5$!+$7$ .$6$ . $5$!/$5$!....so on

$33$+$1$+$6$+$42$+......

I am not getting the correct answer as the solution is getting complex.

Can anyone guide me how to approach the problem?

• You asked almost the same question here...Using the same tools leads to a very similar solution... – user37238 Nov 16 '15 at 9:34

Hint: Terms of $5!$ onwards are divisble by $5!$, so you only need the remainer of $1! +2!+3!+4!$.
As Lee said $5!$ onwards all are divisible by $5!$ so we need remainder of($1!+2!+3!+4!$) 33 so remainder is 33.
• @ArchisWelankar: Would you care to repeat that comment in plain English? FYI $5!=120$. – Marc van Leeuwen Nov 16 '15 at 11:13
• Cant you write $33=6.5+3$ thats what i want to say let the sum be x .so $x+30+3$ whats the remainder??. Hope now you are clear on that. – Archis Welankar Nov 16 '15 at 11:19