The remainder when $11^{1011} + 1011^{11}$ is divided by $9$ is

We know that:

$11^{1011} \equiv 2^{1011} \pmod{9}$

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Let's examine the pattern of powers of $2$ modulo $9$:

$2^1 \equiv 2 \pmod{9}$

$2^2 \equiv 4 \pmod{9}$

$2^3 \equiv 8 \pmod{9}$

$2^4 \equiv 16 \equiv 7 \pmod{9}$

$2^5 \equiv 14 \equiv 5 \pmod{9}$

$2^6 \equiv 10 \equiv 1 \pmod{9}$

We see the pattern repeats every $6$ powers, so we need to find $1011 \mod 6$:

$1011 = 6 \times 168 + 3$

Therefore, $2^{1011} \equiv 2^3 \equiv 8 \pmod{9}$

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For $1011^{11}$, we first determine $1011 \mod 9$:

Using the digit sum rule: $1 + 0 + 1 + 1 = 3$

So $1011 \equiv 3 \pmod{9}$, which means $1011^{11} \equiv 3^{11} \pmod{9}$

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Examining powers of $3$ modulo $9$:

$3^1 \equiv 3 \pmod{9}$

$3^2 \equiv 9 \equiv 0 \pmod{9}$

Since $3^2 \equiv 0 \pmod{9}$, any higher power will also be $0$ modulo $9$.

Therefore, $3^{11} \equiv 0 \pmod{9}$

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Finally, $11^{1011} + 1011^{11} \equiv 8 + 0 \equiv 8 \pmod{9}$

The remainder is $\boxed{8}$