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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?

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    $\begingroup$ You asked almost the same question here...Using the same tools leads to a very similar solution... $\endgroup$ – user37238 Nov 16 '15 at 9:34
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Hint: Terms of $5!$ onwards are divisble by $5!$, so you only need the remainer of $1! +2!+3!+4!$.

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  • $\begingroup$ Okay I haven't thought of remainder After 5! all remainder will be 0. $\endgroup$ – justin takro Nov 16 '15 at 9:44
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As Lee said $5!$ onwards all are divisible by $5!$ so we need remainder of($1!+2!+3!+4!$) 33 so remainder is 33.

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    $\begingroup$ Isn't the remainder when 33 is divided by 5! 33 itself? $\endgroup$ – R_D Nov 16 '15 at 10:11
  • $\begingroup$ The 5! Is over all the summatjon sk we can split 33 as 30+3 $\endgroup$ – Archis Welankar Nov 16 '15 at 10:27
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    $\begingroup$ @ArchisWelankar: Would you care to repeat that comment in plain English? FYI $5!=120$. $\endgroup$ – Marc van Leeuwen Nov 16 '15 at 11:13
  • $\begingroup$ 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. $\endgroup$ – Archis Welankar Nov 16 '15 at 11:19
  • $\begingroup$ @ArchisWelanker. Yes, 3 is the remainder when dividing by 5, but the question is what's the remainder when dividing by 120. So 33. $\endgroup$ – RemcoGerlich Nov 16 '15 at 11:30

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