Find the largest positive integer $n$ such that $2^n$ divides $3^{4096} −1$. I dont know how to start with this questions.It is an olympiad sum.

HINT:
Observe that  the highest power of $2$ that divides $3^2-1$ is $3$
Now $3^{2n}=(1+8)^n\equiv1\pmod8\implies3^{2n}+1\equiv2\pmod8\equiv2\pmod4$
So, the highest power of $2$ that divides $3^{2n}+1$ is $1$ for integer $n\ge0$
So, the highest power of $2$ that divides $3^{(2^2)}-1=(3^2-1)(3^2+1)$ will be $3+1$
So, the highest power of $2$ that divides $3^{(2^3)}-1=(3^2-1)(3^2+1)(3^4+1)$ will be $3+1+1$
Observe that $4096=2^{12}$
