<s>[INST] What should $n$ be equal to, so that $5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}$ is completely divisible by $19$? What should $n$ be equal to, so that the number:
$$5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}$$
is completely divisible by 19? I broke it into this:
$$20\cdot 2^{n}\cdot 25^{n}+18\cdot 3^{n}\cdot 4^{n}$$ But what should i do next?
 [/INST] For any natural $n$ we obtain:
$$\begin{aligned}5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}&=2^{n+1}\left(10\cdot5^{2n}+9\cdot6^n\right) \\
&= 2^{n+1}\left(10\cdot(6+19)^{n}+9\cdot6^n\right) \\ 
& \equiv2^{n+1}\left(10\cdot6^n+9\cdot6^n\right) \pmod{19} \\ &\equiv 0 \pmod{19}.\end{aligned}$$
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