Recurrence Relation

We can rewrite the equation into : $(a_n - a_{n-1} ) = n(a_{n-1} - a_{n-2}) \implies \frac{a_n - a_{n-1}}{a_{n-1} - a_{n-2}} = n$

$\prod_{i=3}^{n} \frac{a_n - a_{n-1}}{a_{n-1} - a_{n-2}} = n(n-1)(n-2) \ldots 3$

Then, by telescoping, it's easy to see that :

$\frac{a_n-a_{n-1}}{a_2 - a_1} = \frac{n!}{2} \implies a_n - a_{n-1} = n!$

Then, from above we obtain that, for $n \ge 11$ , we have : $a_n - a_{n-1} = n! \equiv 0 \pmod{11}$ $a_n = a_{n-1} \pmod{11}$ . So, we just need to check from $a_1$ to $a_{11}$

Now, by recurrence we obtain : $a_1 \equiv 1 \pmod{11}$ $a_2 \equiv 3 \pmod{11}$ $a_3 \equiv a_2 + 3! \equiv 9 \pmod{11}$ $a_4 \equiv a_3 + 4! \equiv 0 \pmod{11}$ $a_5 \equiv a_4 + 5! \equiv -1 \pmod{11}$ $a_6 \equiv a_5 + 6! \equiv 4 \pmod{11}$ $a_7 \equiv a_6 + 7! \equiv 6 \pmod{11}$ $a_8 \equiv a_7 + 8! \equiv 0 \pmod{11}$ $a_9 \equiv a_8 + 9! \equiv 1 \pmod{11}$ $a_{10} \equiv a_9 + 10! \equiv 0 \pmod{11}$ $a_{11} \equiv a_{10} + 11! \equiv 0 \pmod{11}$

Then, from the first one we obtain $a_n \equiv 0 \pmod{11} \forall n \ge 11$ and $n \in \mathbb{N}$

So, from $a_1$ to $a_{1000}$ , the number that multiple 11 are a 4, a 8, and {a n} {n \ge 10}, The total is $(1000 - 10 + 1) + 1 + 1 = \boxed{993}$

To first attempt to a notice a pattern, simply compute the first few terms modulo 11, and the residues of the first few terms are $a_1 \equiv 1, a_2 \equiv 3, a_3 \equiv 9, a_4 \equiv 0, a_5 \equiv 10, a_6 \equiv 4, a_7 \equiv 6, a_8 \equiv 0, a_9 \equiv 1, a_{10} \equiv 0, a_{11} \equiv 0$ . From then on, it is clear that every term afterwards will be a multiple of 11, as if $11|a_{n-1}, a_{n-2}$ , then clearly $11 | (n+1)a_{n-1} - na_{n-2} = a_{n}$ .

So $11|a_n$ for $n=10, 11, \ldots, 1000$ (which results in $991$ values of $n$ ), as well as $n=4, 8$ . So then there are $991+2 = \boxed{993}$ values of $n$ .

$a_n=(n+1)a_{n-1}-na_{n-2}$

So, $a_n-a_{n-1}=n(a_{n-1}-a_{n-2})$

As $a_2-a_1=2, a_n-a_{n-1}=n!$

Compute the first 11 terms. $a_4$ , $a_8$ and $a_{10}$ will be divisible by 11.

$a_{11}=a_{10}+11!$ . As both $a_{10}$ and $11!$ are divisible by 11, $a_{11}$ is divisible by 11.

$a_{12}=a_{11}+12!$ . As both $a_{11}$ and $12!$ are divisible by 11, $a_{12}$ is divisible by 11. Similarly all terms above $a_{10}$ are divisible by 11.

$a_4$ , $a_8$ and all terms from $a_{10}$ to $a_{1000}$ are divisible by 11. Number of terms divisible by 11 $= 2+(1000-10+1) = 993$

At first we simply compute the first few terms modulo $11$ and the residues of the first few terms are ,

$a_1 \equiv 1 \pmod{11}$

$a_2 \equiv 3 \pmod{11}$

$a_3 \equiv 9 \pmod{11}$

$a_4 \equiv 0 \pmod{11}$

$a_5 \equiv -1 \pmod{11}$

$a_6 \equiv 4 \pmod{11}$

$a_7 \equiv 6 \pmod{11}$

$a_8 \equiv 0 \pmod{11}$

$a_9 \equiv 1 \pmod{11}$

$a_{10} \equiv 0 \pmod{11}$

$a_{11} \equiv 0 \pmod{11}$

From then on, it is clear that every term afterwards will be a multiple of $11$ , as if $11| a_{n-1} , a_{n-2}$ then clearly $11| (n+1)a_{n-1} -na_{n-2}$ . That means $11|a_n$ .

So, $11|a_n$ where $n∈S=$ { $4,8,10,11,12.......999,1000$ } and $n(S) =\boxed {993}$

a {n}=(n+1)(a {n-1} - n(a_{n-2})

solving for n, we have n=\frac {a {n}-a {n-1}}{a {n-1}-a {n-2}}

if n=n, we have n=\frac {a {n}-a {n-1}}{a {n-1}-a {n-2}} if n=n-1 we have n-1=\frac {a {n-1}-a {n-2}}{a {n-2}-a {n-3}} if n=n-2 we have n-2=\frac {a {n-2}-a {n-3}}{a {n-3}-a {n-4}} ... if n=n-(n-3)=3 we have 3=\frac {a {3}-a {2}}{a {2}-a {1}}

notice that if we multiply all the equations, we have a telescoping product, thus we have

n(n-1)(n-2)(n-3)...(4)(3)=\frac {a {n}-a {n-1}}{a {2}-a {1}}

rearranging the equation, we have n(n-1)(n-2)(n-3)..(4)(3)(a {2}-a {1}) = a_{n} - a{n-1}

since a {2}=3 and a {1}=1, then a {2}-a {1}=2

this implies that n(n-1)(n-2)(n-3)..(4)(3)(a {2}-a {1}) = a {n} - a{n-1} n(n-1)(n-2)(n-3)..(4)(3)(2) = a {n} - a{n-1} n!=a_{n} - a{n-1}

we have a {n}=n! + a {n-1} since n! \equiv 0 \pmod{11} for n \geq 11 this implies that a {n} \equiv 0 \pmod{11} for n \geq 11 this implies that if a {10} \equiv 0\pmod{11} then a {11} upto a {1000} is divisible by 11, thus we have to check the terms a {1} upto a {10}

a 1 \equiv 1\pmod{11} a 2 \equiv 3\pmod{11} a 3 \equiv 9\pmod{11} a 4 \equiv 0\pmod{11} a 5 \equiv 10\pmod{11} a 6 \equiv 4\pmod{11} a 7 \equiv 6\pmod{11} a 8 \equiv 0\pmod{11} a 9 \equiv 1\pmod{11} a {10} \equiv 0\pmod{11}

notice that a {10} \equiv 0 \pmod{11} then a {11} upto a {1000} are divisible by 11, so we just have to count the terms from a 1 to a 10 that are not divisible by 11 and subtract it from 1000, thus only a 4 , a 8, and a {10} are divisible by 11, thus there are 7 who are not divisible, thus

1000-7 =993 answer: 993

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

First, we need to simplify the equation above into

$a_{n}-a_{n-1}=n(a_{n-1}-a_{n-2})$

By substituting value $n=k,(k-1),(k-2),\cdots3$ we have

$a_{k}-a_{k-1}=k(a_{k-1}-a_{k-2})$

$a_{k-1}-a_{k-2}=(k-1)(a_{k-2}-a_{k-3})$

$a_{k-2}-a_{k-3}=(k-2)(a_{k-3}-a_{k-4})$

$\vdots$

$a-{3}-a_{2}=3(a_{2}-a_{1})$

Product all the equation above which give new equation

$(a_{k}-a_{k-1})(a_{k-1}-a_{k-2})(a_{k-2}-a_{k-3})\cdots(a_{3}-a_{2})=(k(k-1)(k-2)\cdots3)((a_{k-1}-a-{k-2})(a_{k-2}-a_{k-3})(a_{k-3}-a_{k-4})\cdots(a_{2}-a_{1}))$

Simplify the equation above as we get

$a_{k}-a_{k-1}=\frac{k!}{2!}(a_{2}-a_{1})$

Because $(a_{2}-a_{1})=2$ then

$a_{k}-a_{k-1}=k!$

So we may conclude that $a_{n}=(n+1)a_{n-1}-na_{n-2}$ can be expressed as $a_{n}=a_{n-1}+n!$

Since for $n\geq11, a_{n}$ is divisible by $11.$ Then we must check for $n=1,2,3, \cdots10$

$a_{1}\equiv1\pmod{11},a_{2}\equiv3\pmod{11},a_{3}\equiv9\pmod{11},a_{4}\equiv0\pmod{11}$ $a_{5}\equiv-1\pmod{11},a_{6}\equiv4\pmod{11},a_{7}\equiv6\pmod{11},a_{8}\equiv0\pmod{11}$ $a_{9}\equiv1\pmod{11},a_{10}\equiv0\pmod{11}$

Hence, the satisfying value for $n$ are $a_{4},a_{8},a_{10}$ and $a_{11}-a_{1000}$ which is $3+990=993$

From the equation : $a_{n}$ = ( n +1). $a_{n-1}$ - n . $a_{n-2}$

We have $a_{n}$ - $a_{n-1}$ = n . ( $a_{n-1}$ - $a_{n-2}$ )

similarly $a_{n-1}$ - $a_{n-2}$ = ( n -1) . ( $a_{n-2}$ - $a_{n-3}$ )

$a_{n-2}$ - $a_{n-3}$ = ( n -2) . ( $a_{n-3}$ - $a_{n-4}$ )

                 ..................

$a_{3}$ - $a_{2}$ = 3. ( $a_{2}$ - $a_{1}$ )

Because $a_{2}$ - $a_{1}$ = 3 -1 =2 , hence $a_{n}$ - $a_{n-1}$ = n!

and $a_{n}$ = 1! +2! +3! + ... + n!

if 1 =< n =< 10 , only $a_{4}$ , $a_{8}$ , $a_{10}$ are the mutiples of 11

n >= 11, n! is the mutiple of 11 with any n , so from $a_{11}$ to $a_{1000}$ are the mutiples of 11

finally, we have 3 + (1000-11):1+1 = 993

From the given equation, we have: $a_n-a_{n-1}=n(a_{n-1}-a_{n-2})$ .

Therefore, $a_n-a_{n-1}=n(a_{n-1}-a_{n-2})$ $=n(n-1)(a_{n-2}-a_{n-3})=...=\frac{n!}{2!} (a_2-a_1)$ .

Because $a_2-a_1=2!$ , so $a_n=n!+a_{n-1}$ .

Because if $n \geq 11$ , $n! \equiv 0 (mod 11)$ or $a_n \equiv a_{n-1} (mod 11)$ . Therefore, we only need to check from $a_1$ to $a_{10}$ .

By recurrence, we have from $a_1$ to $a_{10}$ , only $a_4,a_8,a_{10}$ are multiples of 11. This also leads to $a_n$ is a multiple of 11 with all $n \geq 11$ .

Therefore, the answer is: (1000-11+1) + 3 = 993

We'll first experiment with few first values and see what happens. We have (This is found out by simply plugging in values for $n$ ):

$a_1 \equiv 1 \pmod{11}$

$a_2 \equiv 3 \pmod{11}$

$a_3 \equiv 9 \pmod{11}$

$a_4 \equiv 0 \pmod{11}$

$a_5 \equiv 10 \pmod{11}$

$a_6 \equiv 4 \pmod{11}$

$a_7 \equiv 6 \pmod{11}$

$a_8 \equiv 0 \pmod{11}$

$a_9 \equiv 1 \pmod{11}$

$a_{10} \equiv 0 \pmod{11}$

$a_{11} \equiv 0 \pmod{1}$

Now it is clear that any $a_n$ with $n \geq 12$ will be divisible by $11$ because for example for $n = 12$ we get $a_{12} \equiv 13 \cdot 0 - 12 \cdot 0 \equiv 0 \pmod{11}$ . So there are only $7$ terms not divisible by $11$ , thus there are $1000 - 7 = \boxed{993}$ terms divisible by $11$ .

In this problem, computing values up to a point seems like it might work.

To make this a little easier, we work in modulo 11 and rewrite our equation: $a_{n}=n(a_{n-1}-a_{n-2})+a_{n-1}$ Now, $a_{1}\equiv 1 \pmod {11}$ $a_{2}\equiv 3 \pmod {11}$ $a_{3}\equiv 9\equiv 3\cdot 2+3 \pmod {11}$ $a_{4}\equiv 0\equiv 4\cdot 6+9 \pmod {11}$ $a_{5}\equiv 10\equiv 5\cdot 2+0 \pmod {11}$ $a_{6}\equiv 4\equiv 6\cdot 10+10 \pmod {11}$ $a_{7}\equiv 6\equiv 7\cdot 5+4 \pmod {11}$ $a_{8}\equiv 0\equiv 8\cdot 2+6 \pmod {11}$ $a_{9}\equiv 1\equiv 9\cdot 5+0 \pmod {11}$ $a_{10}\equiv 0\equiv 10\cdot 1+1 \pmod {11}$ $a_{11}\equiv 0\equiv 11\cdot 10+0 \pmod {11}$ It is easy to see that for $n>11$ $a_{n}\equiv 0 \pmod {11}$ This follows from the fact that $0\cdot 0+0=0$ Now, looking at our table, we have 7 nonzero values modulo 11, so the answer is $1000-7=\boxed{993}$

Lemma: $a_n=a_{n-1}+n!, n\ge 2$

Proof: By induction in $n$ . If $n=2$ , we have $a_2=3=1+2=a_1+2!\\$ Fixed $m\ge 2, m\in \mathbb N$ suppose that law $a_n=a_{n-1}+n!$ is true for $n=m\\$ , that is, $a_m=a_{m-1}+m!\Rightarrow a_m-a_{m-1}=m!\\$

Where we have $a_{m+1}=(m+2)*a_{m}-(m+1)*a_{m-1}=\\$ $(m+1)*(a_{m}-a_{m-1})+a_{m}=(m+1)*m!+a_{m}$ $=(m+1)!+a_m\\$ And by Induction Principle, the lemma is proved.

Now, since that $11|n!, n\ge 11$ , by the Lemma, we have $a_n\equiv a_{10}\pmod{11}, n\ge 11$ . By the Lemma, it's easy to see that:

$\\$ $a_1\equiv 1\pmod{11}$

$a_2\equiv 3\pmod{11}$

$a_3\equiv 3+3!\equiv 9\pmod{11}$

$a_4\equiv 9+4!\equiv 0\pmod{11}$

$a_5\equiv 0+5!\equiv -1\pmod{11}$

$a_6\equiv -1+6!\equiv 4\pmod{11}$

$a_7\equiv 4+7!\equiv 6\pmod{11}$

$a_8\equiv 6+8!\equiv 0\pmod{11}$

$a_9\equiv 0+9!\equiv 1\pmod{11}$

$a_{10}\equiv 1+10!\equiv 0\pmod{11}$

Since we have $a_n\equiv a_{10}\equiv 0\pmod{11}, n\ge 11$ , there are $1000-11+1=990$ values of $n, 11\le n\le 1000$ such that $11|a_n$ and more 3 values $n=4, 8, 10<11$ , totaling $993$ values.

From $a_n=(n+1)a_{n-1}-na_{n-2}$ we have $a_n-a_{n-1}=n(a_{n-1}-a_{n-2}=...=n!$ This gives $a_n=n!+a_{n-1}=....=n!+(n-1)!+...+1$ It's easy to check that $a_n$ is a multiple of 11 <=> n=4,8,10 and $n \geq 11$ Hence the answer is 993

As is standard for most problems involving recurrence relations, we first start out by trying to find patterns from the first few elements.

Using the recurrence stated, we get the first $5$ elements to be $1, 3, 9, 33, 153$ . Observing consecutive differences, we get $3-1 = 2, 9-3 = 6, 33 - 9 = 24, 153 - 33 = 120$ . These differences are $2!, 3!, 4!, 5!$ , respectively. Thus, $a_k = \displaystyle \sum_{n=1}^k n!$ This can be proven easily.

$a_k = (k+1)((k-1)! + (k-2)! + \ldots + 1!) - (k)((k-2)! + \ldots + 1!) =$

$(k-1)! + (k - 2)! + \ldots + 1! + (k)((k - 1)! + (k - 2)! + \ldots + 1! -$

$((k - 2)! + (k - 1)! + \ldots + 1!))) =$

$(k - 1)! + (k - 2)! + ... + 1! + (k)((k - 1)!) = \displaystyle \sum_{n=1}^k n!$

Next, we must look for multiples of 11. From the first 5 terms, it is clear that the fourth term is a multiple of 11. We look to find any more multiples of 11 until we get to $a_{10}$ , as if $a_{10}$ is a multiple of 11 then all of the following terms are as well, but if it is not then none of the following terms are multiples of 11.

> This is due to the fact that for for every term $a_k$ with $k \geq 11$ the difference between each consecutive term is a multiple of 11 because $n!$ is a multiple of 11 when $n \geq 11$ . Thus, for all $a_k$ with $k \geq 11$ $a_k \equiv m \pmod{11}$ , i.e. the residue when considered mod 11 is consistent.

However, the numbers get very large, so it is not feasible to deal with them unless we consider them in a more representative fashion using the following method:

$a_5$ can be expressed as $a_4 + 5!(1)$ . Furthermore, $a_6$ can be expressed as $a_4 + 5!(1 + 6)$ , and $a_7$ as $a_4 + 5!(1 + 6 + 42)$ , and finally $a_8$ as $a_4 + 5!(1 + 6 + 42 + 336) = a_4 + 5!(385)$ . Since $385$ is a multiple of 11 and $a_4$ is a multiple of 11, $a_8$ must then be a multiple of 11 and, using the same logic, since neither 1, 7, nor 49 are multiples of 11 and 11 is prime, $a_5, a_6,$ and $a_7$ are not multiples of 11.

Using the same method with $a_8$ , we come to $a_{10} = 9!(1 + 10) \rightarrow a_{10}$ is a multiple of 11.

Since $a_{10}$ is a multiple of 11, every term $a_k$ with $k \geq 11$ is also divisible by 11. Thus, the number of integer values of n from 1 to 1000 that satisfy the equation is $2 + 991 \Rightarrow 993$

Manipulating the recursive formula given, we get $a_n - a_{n-1} = n(a_{n-1} - a{n-2}).$ Let $b_n = a_{n+1}-a_n$ ; then this becomes $b_n = (n+1)b_{n-1}$ , after shifting the indices up by one. Because $a_1 = 1$ and $a_2 = 3,$ we know that $b_1 = 2.$ Using our formula for $b_n,$ we calculate the next few values of $b_n,$ reduced $\mod{11}$ : $b_1 = 2,$ $b_2 = 3(2) = 6,$ $b_3 = 4(6) \equiv 2,$ $b_4 \equiv 5(2) = 10,$ $b_5 \equiv 6(10) \equiv 5,$ $b_6 \equiv 7(5) \equiv 2,$ $b_7 \equiv 8(2) \equiv 5,$ $b_8 \equiv 9(5) \equiv 1,$ $b_9 \equiv 10(1) = 10,$ $b_{10} \equiv 11(10) \equiv 0.$

Notice that because $b_{10} \equiv 0 \pmod{11},$ every term after it will also be congruent to $0, \mod{11},$ as all we are doing to find these values is multiplying and reducing. Because we defined $b_n$ as the difference between consecutive terms of $a_n,$ we can use these values to find the terms of $a_n \pmod {11}$ : $a_1 = 1,$ $a_2 = 3,$ $a_3 = a_2 + b_2 = 3 + 6 = 9,$ $a_4 = a_3 + b_3 \equiv 9 + 2 \equiv \boxed{0},$ $a_5 \equiv 0 + 10 = 10,$ $a_6 \equiv 10 + 5 \equiv 4,$ $a_7 \equiv 4 + 2 = 6,$ $a_8 \equiv 6+5 \equiv \boxed{0},$ $a_9 \equiv 0 + 1 = 1,$ $a_{10} \equiv 1 + 10 \equiv \boxed{0}.$

Because all of $b_{10}, b_{11}, b_{12}, \ldots$ are zero, so are $a_{10}, a_{11}, a_{12}, \ldots.$ From the list above, we see that $a_4$ and $a_8$ are two additional terms that are multiples of $11,$ so in total there are $2 + (1000 - 10 + 1) = \boxed{993}$ terms in $a_1, a_2, \ldots, a_{1000}$ that are multiples of $11.$

Rewrite the recurrence relation as $a_n - a_{n-1} = n (a_{n-1} - a_{n-2})$ . With the starting value $a_2-a_1 = 2$ , we can show (by induction) that $a_n - a_{n-1} = n!$ . Hence,

$a_n= (a_n - a_{n-1} ) + (a_{n-1} - a_{n-2} ) + \ldots + (a_2 - a_1) + a_1 = n! + (n-1)! + \ldots + 2! + 1 ! .$

We check the first few values $1 \leq n \leq 11$ , and find that $11 \mid a_4, a_8, a_{10}, a_{11}$ . For $n > 11$ , since $a_n = n! + a_{n-1}$ , and $n!$ is a multiple of 11, we see by induction that they are all multiples of $11$ . Hence, only $a_1, a_2, a_3, a_5, a_6, a_7, a_9$ are not multiples of $11$ , so we have to exclude 7 of them. Thus, $\boxed{993}$ numbers are multiples of 11.

Using recursive memoization (aka dynamic programming) in Python:

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11

memo = {1:1, 2:3}
def a(n):
    if n in memo:
        return memo[n]
    memo[n] = (n+1)*a(n-1)-n*a(n-2)
    return memo[n]
count = 0
for i in xrange(1, 1001):
    if a(i) % 11 == 0:
        count += 1
print "Answer:", count

a1 = 1

a2 = 3

a3 = 4(a2) - 3(a1) = 9

a4 = 5(a3) - 4(a2) = 33 = 0 mod 11 (in equation below, I used the mod value since all I care about is if it is divisible by 11)

a5 = 6(a4) - 5(a3) = -45 = -1 mod 11 (I used -1 to keep calc easier)

a6 = 7(a5) - 6(a4) = -7 = 4 mod 11

a7 = 8(a6) - 7(a5) = 39 = 6 mod 11

a8 = 9(a7) - 8(a6) = 22 = 0 mod 11

a9 = 10(a8) - 9(a7) = -54 = 1 mod 11

a10 = 0(a9) - 10(a8) = 0 mod 11 (Note: since it is mod 11, multiplying by 11 is equivalent to multiplying by 0)

a11 = 1(a10) - 0(a9) = 0 mod 11

Now that we have two 0 mod 11 terms in a row, all future terms will now be divisible by 11 since the past two terms will have been divisible by 11.

a4, a8, a10 - a1000 = 1+1+991 terms = 993 terms are divisible by 11

Let Bn= An - A(n-1) So Bn=B(n-1)*n then Bn/B(n-1) = n multiplying all those equations we have Bn/B2=n!/2 but B2=2 then Bn=n! But we have An= Bn+B(n-1)+...+B2+A1=n!+(n-1)!+...+1! Now we observe each k! in Z(11) 1!=1(mod11) 2!=2(mod11) 3!=6(mod11) 4!=2(mod11) 5!=10(mod11) 6!=5(mod11) 7!=2(mod11) 8!=5(mod11) 9!=1(mod11) 10!=10(mod11) So A4, A7 and A10 are divisible by 11. Every k!, k>10, is divisible by 11 so Ai=A10+ 11Ti for some integer Ti, for every i>10. But A10 is divisible by 11 and then every Ai, i>10, is divisible by 11. So the answer is: (1000-11+1)+3=993

We note that the first few values of $a_n$ are 1, 3, 9, 0, 10, 4, 6, 0, 1, 0, 0 mod 11. Now that $a_{10}$ and $a_{11}$ are both 0, we can see that all subsequent values will be zero, so there are only the above seven values which are nonzero.

This gives 993 which are multiples of 11.

Claim: If $a_{n-1},a_{n-2}$ are both multiples of $11$ , then so is $a_n$ . Proof: Assume $a_{n-1},a_{n-2}$ are multiples of $11$ . Then there exist integers $k,l$ such that $a_{n-1}=11k,a_{n-2}=11l$ .

$a_n=(n+1)a_{n-1}+na_{n-2}=(n+1)(11k)+n(11l)=11(nk+k+nl)$ By closure of the integers, $nk+k+nl$ is an integer, so $a_n$ is divisible by 11.

Evaluate the first eleven terms: $(1,3,9,33,153,873,5913,46233,409113,4037913,43954713)$ . Modulo $11$ , the first eleven terms are $1,3,9,0,10,4,6,0,1,0,0$ . We then claim that $a_n$ is a multiple of $11$ for all $n\geq 10$ , making the inductive proof using $n=10,11$ as base cases (proof by evaluation) and noting that the inductive step is the first claim, already proven.

Therefore, for $n\in\{1,...,1000\}$ , the only values of $n$ for which $a_n$ is not a multiple of $11$ must have $n<10$ ; looking at our initial evaluations, we can narrow this down to $n\in\{1,2,3,5,6,7,9\}$ .

There are $7$ integers values of $n$ from $1$ to $1000$ inclusive such that $a_n$ is not a multiple of 11; therefore, there are $1000-7=993$ integer values of $n$ from $1$ to $1000$ inclusive such that $a_n$ is a multiple of 11.

a {n} - a {n-1}=n(a {n-1} - a {n-2}), a {n-1} - a {n-2}=(n-1)(a {n-2} - a {n-3}), a {n-2} - a {n-3}=(n-2)(a {n-3} - a {n-4}), . . a {3} - a {2}=3(a {2} - a {1})=3(3-1)=3*2, a {2} - a {1}=3-1=2; So, we can easily find that, a {n} - a {n-1} = n!, a {n-1} - a {n-2}=(n-1)!, a {n-2} - a {n-3}=(n-2)!, . . a {3} - a {2}=3!, a {2} - a {1}=2!; By adding these (n-1) equations, all the a {i} 's will be canceled out, where i is a positive integer and 2 \leq i \leq (n-1). So, we shall get, a {n} - a {1} = \displaystyle \sum {j=2}^n j! \Rightarrow a {n}= \displaystyle \sum {j=1}^n j! [where a {1}=1, given]. Note that, 11 divides \displaystyle \sum {j=1}^10 j! and for all positive integer k \geq 11; 11 also divides k!. So, when n \geq 10, a {n} is divisible 11. So, for 10 \leq n \leq 1000, there are 991 such a {n}'s.
Now we just need to find, how many a {n}'s are divisible by 11, for 1 \leq n \leq 9. It is very easy to calculate a {n}, where 1 \leq n \leq 9. Then, we shall find that only a {4} and a {8} are divisible by 11. So, the total number of such a_{n} is (991+2) = 993 [answer]

It can be checked that $\begin{aligned} a_3\equiv4a_2-3a_1&\equiv9\pmod{11}\\ a_4\equiv5a_3-4a_2&\equiv0\pmod{11}\\ a_5\equiv6a_4-5a_3&\equiv10\pmod{11}\\ a_6\equiv7a_5-6a_4&\equiv4\pmod{11}\\ a_7\equiv8a_6-7a_5&\equiv6\pmod{11}\\ a_8\equiv9a_7-8a_6&\equiv0\pmod{11}\\ a_9\equiv10a_8-9a_7&\equiv1\pmod{11}\\ a_{10}\equiv11a_9-10a_8&\equiv0\pmod{11}\\ a_{11}\equiv12a_{10}-11a_9&\equiv0\pmod{11} \end{aligned}$ Also, if $a_n\equiv a_{n+1}\equiv0\pmod{11}$ for some integer $n$ then $a_{n+2}\equiv(n+3)a_{n+1}+(n+2)a_n\equiv0\pmod{11}$ Thus if $P_n$ is the statement that $a_n\equiv a_{n+1}\equiv0\pmod{11}$ , then $P_{10}$ holds and $P_n\implies P_{n+1}$ . Therefore $P_n$ holds for all $n\geq10$ . In particular, this means that 11 divides all $a_n$ with $n\geq10$ .