Question

Given $a \equiv 1 \pmod{7}$, $b \equiv 2 \pmod{7}$, and $c \equiv 6 \pmod{7}$, what is the remainder when $a^{81} b^{91} c^{27}$ is divided by 7?

Answer 1

5 (mod 7)

Explanation:

Answer 2

`\begin{cases}a\equiv1\pmod7\\b\equiv2\pmod7\\c\equiv6\pmod7\end{cases}`


`a^(81)\equiv1^(81)\equiv1\pmod7`


`b^(91)\equiv2^(91)\pmod7`


`2^(91)\equiv(2^3)^(30)*2^1\equiv8^(30)*2\pmod7`

`8\equiv1\pmod7`

`2\equiv2\pmod7`

`\implies2^(91)\equiv1^(30)*2\equiv2\pmod7`


`c^(27)\equiv6^(27)\pmod7`


`6^(27)\equiv(-1)^(27)\equiv-1\equiv6\pmod7`


`\implies a^(81)b^(91)c^(27)\equiv1*2*6\equiv12\equiv5\pmod7`