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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 $2016$?

  • 1
    What contest is this from? Is it still on-going? – 2017-02-05
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    What contest is this from? – 2017-02-12

1 Answers 1

3

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