Find the remainder when 6799 is divided by 7.

Answer: Option C

$$\begin{array}{rl}& \text{Remainder of}\frac{{67}^{99}}{7}\\ & \text{or, }R=\frac{{\left(63+4\right)}^{99}}{7}\end{array}$$
63 is divisible by 7 for any power, so required remainder will depend on the power of 4
Required remainder
$$\begin{array}{rl}& \frac{4^{99}}{7}==R==\frac{4^{\left(96+3\right)}}{7}\\ & \frac{4^3}{7}\Rightarrow \frac{64}{7}\Rightarrow \frac{\left(63+1\right)}{7}==R\Rightarrow 1\\ & \\ & \text{Note}:\\ & \frac{4}{7}\text{remainder}=4\\ & \frac{\left(4\times 4\right)}{7}=\frac{16}{7}\text{remainder}=2\\ & \frac{\left(4\times 4\times 4\right)}{7}=\frac{64}{7}=1\\ & \frac{\left(4\times 4\times 4\times 4\right)}{7}=\frac{256}{7}\text{remainder}=4\\ & \frac{\left(4\times 4\times 4\times 4\times 4\right)}{7}=2\end{array}$$
If we check for more power we will find that the remainder start repeating themselves as 4, 2, 1, 4, 2, 1 and so on. So when we get A number having greater power and to be divided by the other number B, we will break power in (4n + x) and the final remainder will depend on x i.e. $$\frac{{\text{A}}^{\text{x}}}{\text{B}}$$