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?

You can prove by induction that $20\cdot2^n\cdot25^n+18\cdot3^n\cdot4^n$
$=20\cdot50^{n}+18\cdot12^{n}$ is divisible by $19$.
It's obviously true for $n=0$, since $20+18=38=2\cdot19$.
Now assume $19$ divides $20\cdot50^{n}+18\cdot12^{n}$.
Then $20\cdot50^{n+1}+18\cdot12^{n+1}=50\cdot20\cdot50^{n+1}+12\cdot18\cdot12^{n+1}$
$=12\cdot(20\cdot50^n+18\cdot12^{n})+38\cdot20\cdot50^n$ is divisible by $19$.
QED
