what will be the remainder when $43$ divides $32002^{4200}$??
what I did is: $32002\equiv10 \pmod{43}$, how to proceed further?
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See How do I compute $a^b,\bmod c$ by hand? – Bart Michels Apr 14 '15 at 08:28
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Hint: Further to what you observed, $10^{42} \equiv 1 \pmod {43}$ (why?)
Macavity
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1For the reason you figured out above: 32002 and 10 are equivalent (mod 43), which means anywhere you might add, multiply, or use as the base (but not the power) of exponentiation, you can use 10 instead of 32002. ...but in reality, $anything^{42}\equiv 1\pmod {43}$. (well, anything as long as it's coprime to 43, so not 0; 43 is prime though so that's it) – Dan Uznanski Nov 11 '14 at 13:51
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You already mentioned $32002 \equiv 10 \pmod {43}$. Now use $a \equiv b \pmod p \implies a^n \equiv b^n \pmod p$. For a proof, notice $(a-b) | (a^n-b^n)$ – Macavity Nov 11 '14 at 13:52
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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.
Tacet
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