Question

The remainder when 3^21 is divided by 7

Answer 1

According to Fermat's theorem, if 'p' is a prime number and 'a' is any integer where both 'a' and 'p' are coprimes, then,

$a^{p-1}\equiv 1\pmod{p}$

Here, '7' is a prime number, '3' too, and (3, 7) = 1. Hence,

$\begin{aligned}&3^6\equiv 1\pmod{7}\\ \\ \Longrightarrow\ \ &(3^6)^3\equiv 1^3\pmod{7}\\ \\ \Longrightarrow\ \ &3^{18}\equiv 1\pmod{7}\\ \\ \Longrightarrow\ \ &3^{18}\times 3^3\equiv 1\times 3^3\pmod{7}\\ \\ \Longrightarrow\ \ &3^{21}\equiv 27\pmod{7}\\ \\ \Longrightarrow\ \ &3^{21}\equiv \bold{6}\pmod{7}\end{aligned}$

Hence 6 is the remainder.