Divisibility of a recursive sequence.

$a_{n} = (n+1)*a_{n-1} - n*a_{n-2}$ -> $a_{n} - a_{n-1} = n*(a_{n-1} - a_{n-2})$

Set $V_{n} = a_{n} - a_{n-1}$ , after that we have $V_{n} = n*V_{n-1} =n*(n-1)*(n-2)...(n-(n-3))*V_{2} = n!$ (as $V_{2}=2=2*1$ )

So $a_{n} = n! + a_{n-1} = \sum_{k=1}^n (k!)$ for the first 10 terms, only $a_{4} , a_{8}, a_{10}$ are divisible by 11; and we also have $\sum_{k=1}^{10} (k!)$ is divisible by 11 that means for every $n \geq 11$ we have $a_{n} = 11*t + \sum_{k=11}^n (k!)$ (where $11*t = \sum_{k=1}^{10} (k!)$ ). And since all $n! (n \geq 11)$ are divisible by 11, all $a_{n}$ where $n \geq 11$ are divisible 11

---> We have 993 terms which are divisible by 11

$\displaystyle \begin{array}{c}\\ a_1\equiv 1 \pmod {11}\\ a_2\equiv 3 \pmod {11}\\ a_3\equiv 4 \times 3-3\times 1 \equiv 9 \pmod {11}\\ a_4 \equiv 5\times 9-4\times 3 \equiv 33 \equiv 0 \pmod {11}\\ a_5 \equiv -5\times 9\equiv -45 \equiv 10 \pmod {11}\\ a_6 \equiv 7\times 10 \equiv 4 \pmod {11}\\ a_7 \equiv 8\times 4-7\times 10 \equiv -38 \equiv 6 \pmod {11}\\ a_8 \equiv 9\times 6-8\times 4\equiv 22 \equiv 0 \pmod {11}\\ a_9 \equiv -9 \times 6 \equiv 1 \pmod{11}\\ a_{10} \equiv 11 \times 1 \equiv 0 \pmod{11}\\ a_{11} \equiv -11\times 1\equiv 0 \pmod {11}\\ \end{array}$

The subsequent terms turn out to be divisible by 11 because both $a_{n-2}, a_{n-1} \equiv 0\pmod {11}$ . Hence, the terms $a_4$ , $a_8$ and the terms from $a_{10}$ to $a_{1000}$ are all divisible by 11 giving us a total of $\boxed{993}$ terms.