If $$(27)^{999}$$ is divided by 7, then the remainder is:

We have to find the remainder obtained when the very large power $$27^{999}$$ is divided by $$7$$. In other words, we must evaluate $$27^{999}\pmod 7.$$ Working entirely in modular arithmetic keeps the numbers small and the calculations manageable.

First, reduce the base modulo $$7$$:

$$27 = 7\times 3 + 6 \; \Longrightarrow \; 27 \equiv 6 \pmod 7.$$

So the original expression can be rewritten as

$$27^{999} \equiv 6^{999}\pmod 7.$$

Now, instead of tackling the exponent $$999$$ directly, let us examine the powers of $$6$$ modulo $$7$$ to see whether they follow a simple repeating pattern.

Compute the first few powers:

$$6^1 \equiv 6 \pmod 7,$$

$$6^2 = 36 \equiv 36 - 35 = 1 \pmod 7,$$

$$6^3 = 6^2\cdot 6 \equiv 1\cdot 6 = 6 \pmod 7,$$

$$6^4 = (6^2)^2 \equiv 1^2 = 1 \pmod 7.$$

From these results we observe the clear two-term cycle

$$6,\,1,\,6,\,1,\,\dots$$

In particular,

$$\bigl(6^1\bigr)\equiv 6,\qquad \bigl(6^2\bigr)\equiv 1,$$

and therefore

$$6^{\text{odd exponent}}\equiv 6,\qquad 6^{\text{even exponent}}\equiv 1 \pmod 7.$$

The exponent we need is $$999$$, which is odd. Hence

$$6^{999}\equiv 6 \pmod 7.$$

But $$6$$ is already less than $$7,$$ so it is itself the remainder.

Therefore, when $$27^{999}$$ is divided by $$7,$$ the remainder is $$6$$.

Hence, the correct answer is Option C.