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$(a_1 + a_2 + ... + a_n) \mod k = (a_1 \mod k + a_2 \mod k + a_3 \mod k + ... + a_n \mod k) \mod k$

$\therefore N \mod 8= (1! \mod 8 + 2! \mod 8 + ... + 2020! \mod 8) \mod 8$

$8 = 2*4 \text{ and } n! = n * (n-1)! \text{ and } 0! = 1$

$n! \mod 8 = 0, \forall n \ge 4$, since it'll have a factor of $2,4$

$1! = 1, 2! = 2, 3! = 6$

$\therefore N \mod 8= (1 + 2 + 6 + 0 + 0 + ... + 0) \mod 8 = 1$

Answer :- 1.

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