<s>[INST] Compute $5^{10,000}\equiv \mod 52$ I'm unsure how to solve questions like these, we are given an example below but I'm still confused about how they got from $5^{100\times10+0}$ to $5^{0}$.
ie) Compute $5^{1000} \bmod 77$
$$
5^{1000}=5^{(166⋅6+4)}=5^4=25^2=4^2=16=2 \pmod 7\\
5^{1000}=5^{(100⋅10+0)}=5^0=1 \pmod{11}\\
5^{1000}=2⋅2⋅11+1⋅8⋅7=44+56=100=23 \pmod{77}
$$
 [/INST] One uses the effective Chinese remainder theorem:  
Let $7u+11v=1$ a Bézout's relation between $7$ and $11$. Then $$(x\equiv a\mod 7  \;\text{ and }\; x\equiv b\mod 11)\iff (x\equiv 7bu+11av\mod 7\times11).$$
One ingredient is Lil' Fermat: as $7$ and $11$ are prime, $x^6\equiv1\mod 7$ for all $x\not \equiv 0\mod 7$ and  $x^{10}\equiv1\mod 11$ for all $x\not \equiv 0\mod 11$, hence $5^{10,000}\equiv 5^{10,000\bmod 6}\mod 7$ and $5^{10,000}\equiv 5^{10,000\bmod 10}\mod 11$.
This is the general method. However, in the present case, it is faster to observe that $$5^4=625\equiv 1\mod 52, \;\text{ hence }\; 5^{10,000}=(5^4)^{2500}\equiv 1^{2500}\mod 52.$$
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