Remainder Again?

Approach one: simplification of base to negative $13^{682}\equiv(-1)^{682}\equiv1\mbox{ (mod 7)}$

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Approach two: simplification of base to positive $13^{682}\equiv6^{682}\equiv36^{341}\equiv1^341\equiv1\mbox{ (mod 7)}$

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Approach three: simplification of index by Fermat's Little Theorem $13^{682}\equiv13^{6\times113+4}\equiv13^4\equiv28561\equiv1\mbox{ (mod 7)}$

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Approach four: binary $682=512+128+32+8+2$ $13^1\equiv-1\mbox{ (mod 7)}$ $13^2\equiv1\mbox{ (mod 7)}$ $13^4\equiv1\mbox{ (mod 7)}$ $13^8\equiv1\mbox{ (mod 7)}$ $13^{16}\equiv1\mbox{ (mod 7)}$ $13^{32}\equiv1\mbox{ (mod 7)}$ $13^{64}\equiv1\mbox{ (mod 7)}$ $13^{128}\equiv1\mbox{ (mod 7)}$ $13^{256}\equiv1\mbox{ (mod 7)}$ $13^{512}\equiv1\mbox{ (mod 7)}$ $13^{682}=13^{512+128+32+8+2}=13^{512}\times13^{128}\times13^{32}\times13^8\times13^2\equiv1\times1\times1\times1\times1=1\mbox{ (mod 7)}$

Faster method: $13 \equiv -1 \pmod{7} \Rightarrow 13^{682} \equiv (-1)^{682} \equiv 1 \pmod{7}$

What we are looking here is $13^{682}\equiv x \pmod{7}$ First approach: Euler's totient function. We know that $\varphi(7)=6$ Thus, we can reduce the exponent by subtracting $6k$ where k is an integer such that $682>6k$ so, $13^{682-113*6}\equiv x \pmod{7}$ $13^{4}\equiv x \pmod{7}$ $28561\equiv1\pmod{7}$