remainder when $43$ divides $32002^{4200}$ what will be the remainder when $43$ divides $32002^{4200}$??
what I did is:
$32002\equiv10 \pmod{43}$,
how to proceed further?
 A: Hint: Further to what you observed, $10^{42} \equiv 1 \pmod {43}$ (why?)
A: From Euler's theorem (here Fermat's little theorem is sufficient, however...)
$$\left(\gcd(a,b) = 1 \right)\Longrightarrow a^{\varphi(b)} \equiv 1 \pmod{b}$$
Where $\varphi$ is totient function. Here $\gcd(43, 3202) = 1 \wedge \varphi(43) = 42$ (43 is prime, so all number $\lbrace1,2,...,42\rbrace$ are coprime). So,
$$\begin{align*}\begin{split}
\gcd(43,3202)= 1~~ \Longrightarrow~~&3202^{\varphi(43)} &\equiv 1 \pmod{43} \Leftrightarrow \\
&3202^{42} & \equiv 1 \pmod{43} \end{split}\end{align*}$$
Assume that $3202^{4200} = 3202^{42\cdot100} = \left(3202^{42}\right)^{100}  = \left(3202^{\varphi(43)}\right)^{100} \cdot 3202^{0}$ and by simple transformations, obtain:
$$\begin{align*}\begin{split}
3202^{4200} &\equiv \left(3202^{\varphi(43)}\right)^{100} &\pmod{43} \Longleftrightarrow \\
3202^{4200} &\equiv \left(1\right)^{100} &\pmod{43}\Longleftrightarrow \\
3202^{4200} &\equiv 1 & \pmod{43}\end{split}\end{align*}$$
Thus, the number you are looking for is 1.
