Remainder

$a = 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! + 10! + 11! \\ a = 1 + (1 \times 2) + 6 + \color{#3D99F6}{(12 \times 2) + (12 \times 2 \times 5) + ... (12 \times 2 \times 5 \times 6 \times \ldots 11)} \\ a = \color{#3D99F6}{12 \times ( 2 + (2 \times 5) + \ldots + (2 \times 5 \times \ldots 11))} + (1 + 2 + 6) \\ \color{#D61F06}{\therefore \text{ When a divided by `4' leaves the remainder as (1+2+6) and follows Euclid's division lemma as (1+2+6) < 12}} \\ \implies \text{Remainder = } 1 + 2 + 6 = 9$

Because every factorial 4! and above has factors 4 and 3, they are all divisible by 12, thus one needs to consider only 1!, 2!, and 3!, which added together gives 9. Since this is not 12 or greater, this is the answer.