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I have this question:

What is the remainder of $$\sum_{x=1}^{312} x \times x!$$ (or just simply) $$(1! \times 1) + (2! \times 2) + (3! \times 3) + \dots + (312! \times312)$$ Divided by $2016$?

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closed as off-topic by Crostul, Carl Mummert, user223391, projectilemotion, Shailesh Feb 6 '17 at 0:15

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  • 1
    $\begingroup$ What contest is this from? Is it still on-going? $\endgroup$ – Joel Reyes Noche Feb 5 '17 at 13:41
  • $\begingroup$ What contest is this from? $\endgroup$ – Joel Reyes Noche Feb 12 '17 at 13:59
3
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HINT:

$$n\cdot n!=(n+1-1)\cdot n!=(n+1)!-(n)!$$

Do you know Telescoping Series?

$$\sum_{r=n}^m r\cdot r!= (m+1)!-n!$$

Here $n=1, m=312$

Now $2016=7\cdot3^2\cdot2^5$

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