<s>[INST] Prove that $6^n=6^{n+5} (mod 100)$ How can we prove that $6^n=6^{n+5} (\text{mod}~ 100)$? I tried by writing $6^{n+5}=7776 \cdot 6^n = 76 \cdot 6^n (\text{mod}~ 100)$ but this approach does not lead to the above result.
 [/INST] $$6^5-1=5(6^4+6^3+6^2+6+1)$$ and since
$$6^4+6^3+6^2+6+1\equiv(1+1+1+1+1)(\mod5)=5,$$ we obtain $6^6-1$ is divisible by $25$.
Also, $6^n$ is divisible by $4$ for all $n\geq2.$
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