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?
 A: 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 $n$.
A: For any natural $n$ we obtain:
$$\begin{aligned}5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}&=2^{n+1}\left(10\cdot5^{2n}+9\cdot6^n\right) \\
&= 2^{n+1}\left(10\cdot(6+19)^{n}+9\cdot6^n\right) \\ 
& \equiv2^{n+1}\left(10\cdot6^n+9\cdot6^n\right) \pmod{19} \\ &\equiv 0 \pmod{19}.\end{aligned}$$
A: Having gotten to $20\cdot2^n\cdot25^n+18\cdot3^n\cdot4^n$, note that $20\equiv1$ mod $19$ while $18\equiv-1$ and $25\equiv6$. It follows that
$$20\cdot2^n\cdot25^n+18\cdot3^n\cdot4^n\equiv2^n\cdot6^n-3^n\cdot4^n\equiv12^n-12^n\equiv0\mod 19$$
Remark: This approach assumes you have modular arithmetic in your toolbox. (Someone added a tag indicating you do, but it may not be the case.)
A: 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
