Question

solution plzz...

(5^625)/7
[find Remainder]
​

Answer 1

$\large \textbf{\underline{\underline{Fermat's Theorem}}}\\ \\ \begin{flushleft}\normalsize \textsf{If \ a \ is any positive integer and \ p \ is any prime number such that \ (a,\ p)=1, \ then,}\end{flushleft} \\ \\ \begin{center}\Large a^{p-1}\equiv 1\pmod{p}\end{center}$

Since  $(5,\ 7)=1,$

$\begin{aligned}&5^6\equiv1\pmod{7}\end{aligned}$

Now let's divide 625 by 6.

625 divided by 6 gives quotient 104 and remainder 1.

Hence,

$\begin{aligned}&5^6\equiv1\pmod{7}\\ \\ \Longrightarrow\ \ &\left(5^6\right)^{104}\equiv 1^{104}\pmod{7}\\ \\ \Longrightarrow\ \ &5^{624}\equiv 1\pmod{7}\\ \\ \Longrightarrow\ \ &5^{624}\times 5\equiv 1\times 5\pmod{7}\\ \\ \Longrightarrow\ \ &5^{625}\equiv \textbf{5}\pmod{7}\end{aligned}$

Thus, the remainder is 5.