Challenging problem about inductive functions Let $A_1 = 0$ and $A_2 = 1$. For $n > 2$, the number $A_n$ is defined
by concatenating the decimal expansions of $A_{n−1}$ and $A_{n−2}$ from left to right. For example $A_3 = A_2A_1 = 10$, $A_4 = A_3A_2 = 101$, $A_5 = A_4A_3 = 10110$, and so forth. Determine all $n$ such that $11$ divides $A_n$
Attempt : All I see is this
$A_1 = 0, A_2 = 1$
$A_n = A_{n-1}*10^{fib(n-2)} + A_{n-2}$
Should I go from here?
UPDATE: Function expanded
$A_n = A_{n-3}(10^{fib(n-4)} + 10^{fib(n-4) + fib(n-3) + fib(n-2)} + 10^{fib(n-2)}) + A_{n-4}(1+10^{fib(n-3) + fib(n-2)})$
However I feel like this is wrong because that would entail that at the very end the left operand in the expression is cancelled out because $A_1  = 0$ leaving the expression as purely a multiple of 10.
No I think it is better to see the ending points as $N_3$ and $N_4$

Let  $a_n:=[A_n]\in\mathbb Z_{11}$ so that $11$ divides $A_n$ iff $a_n=[0]$.
Then:$$a_{n+2}=10^{f_{n}}a_{n+1}+a_{n}=\left(-1\right)^{f_{n}}a_{n+1}+a_{n}$$
where $10$ actually stands for $[10]\in\mathbb Z_{11}$ and $-1$ for $[-1]=[10]\in\mathbb Z_{11}$.
It is not difficult to verify that $(-1)^{f_n}=1$ iff $n$ is a multiple of $3$ and $(-1)^{f_n}=-1$ otherwise.
Then:
$a_{3}=-a_{2}+a_{1}$
$a_{4}=-a_{3}+a_{2}$
$a_{5}=a_{4}+a_{3}=a_{2}$
$a_{6}=-a_{5}+a_{4}=-a_{2}+a_{4}=-a_{3}$
$a_{7}=-a_{6}+a_{5}=a_{3}+a_{2}=a_{1}$
$a_{8}=a_{7}+a_{6}=a_{5}=a_{2}$
$a_{9}=-a_{8}+a_{7}=-a_{2}+a_{1}=a_{3}$
$a_{10}=-a_{9}+a_{6}=-a_{3}+a_{2}=a_{4}$
$\cdots$
This reveals that $a_{n+6}=a_{n}$ so only $6$ values have to be checked.
I leave that to you.
