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

We want to determine the remainder when $$(11)^{1011} + (1011)^{11}$$ is divided by 9. First, observe that $$11 \equiv 2 \pmod{9}$$ so that $$11^{1011} \equiv 2^{1011} \pmod{9}$$. To compute this, note that the powers of 2 modulo 9 cycle with period 6:

$$2^1 \equiv 2, \quad 2^2 \equiv 4, \quad 2^3 \equiv 8, \quad 2^4 \equiv 7, \quad 2^5 \equiv 5, \quad 2^6 \equiv 1 \pmod{9}$$

Since $$1011 = 6 \times 168 + 3$$, it follows that $$2^{1011} \equiv 2^3 \equiv 8 \pmod{9}$$.

Next, we reduce $$1011^{11}$$ modulo 9. Writing $$1011 = 112 \times 9 + 3$$ shows that $$1011 \equiv 3 \pmod{9}$$, hence $$1011^{11} \equiv 3^{11} \pmod{9}$$. Because $$3^2 = 9 \equiv 0 \pmod{9}$$, one factor of $$3^2$$ makes the product a multiple of 9, so $$3^{11} = 3^2 \times 3^9 = 9 \times 3^9 \equiv 0 \pmod{9}$$.

Adding these results gives

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

The correct answer is Option B: $$8$$.