Find the remainder when 67⁹⁹ is divided by 7.

Explanation:-

$$\eqalign{ & {\text{Remainder}}\,{\text{of}}\,\frac{{{{67}^{99}}}}{7} = = R \Rightarrow \frac{{{{\left( {63 + 4} \right)}^{99}}}}{7} \cr & [{\text{63}}\,{\text{is}}\,{\text{divisible}}\,{\text{by}}\,{\text{7}}\,{\text{for}}\,{\text{any}}\,{\text{power,}}\,{\text{so}}\,{\text{required}}\,{\text{remainder}}\,{\text{will}}\,{\text{depend}}\,{\text{on}}\,{\text{the}}\,{\text{poer}}\,{\text{of}}\,{\text{4}}] \cr & {\text{Required}}\,{\text{remainder}}: \cr & \frac{{{4^{99}}}}{7} = = R = = \frac{{{4^{\left( {96 + 3} \right)}}}}{7} \cr & \frac{{{4^3}}}{7} \Rightarrow \frac{{64}}{7} \Rightarrow \frac{{\left( {63 + 1} \right)}}{7} = = R \Rightarrow 1 \cr & \cr & {\text{Note}}: \cr & \cr & \frac{4}{7}\,{\text{remainder}} = 4 \cr & \frac{{\left( {4 \times 4} \right)}}{7} = \frac{{16}}{7}\,{\text{remainder}} = 2 \cr & \frac{{\left( {4 \times 4 \times 4} \right)}}{7} = \frac{{64}}{7} = 1 \cr & \frac{{\left( {4 \times 4 \times 4 \times 4} \right)}}{7} = \frac{{256}}{7}\,{\text{remainder}} = 4 \cr & \frac{{\left( {4 \times 4 \times 4 \times 4 \times 4} \right)}}{7} = 2 \cr & \cr}$$
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. A^x/B.