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?

Okay, so you have shown the expression to be equal to $20\cdot 2^n\cdot 25^n+18\cdot3^n\cdot4^n=20\cdot 50^n+18\cdot12^n$ and as @mwt as shown in comments , write $20=19+1$ and $18=19-1$ to get the expression equal to $19(50^n+12^n)+50^n-12^n$. Now we know that $a^n-b^n$ is divisible by $a-b$ for any natural number $n$. If you don't know that, you can prove it by factorizing $a^n-b^n$.
So $50^n-12^n$ is divisble by $38$ and so $19$ divides the whole expression for any natural number $n$.
