If n is a positive integer, then $2\cdot 4^{2n+1}+3^{3n+1}$is divisible by :

$\text{ Let }P\left(n\right)=2\cdot 4^{2n+1}+3^{3n+1}$

Then $P\left(1\right)=2\cdot 4^3+3^4=209$which is divisible by 1 1 but not divisible by 2,7 or 27 .
Further, let $\mathrm{P}\left(\mathrm{k}\right)={2.4}^{2\mathrm{k}+1}+3^{3\mathrm{k}+1}$is divisible by 11, that is,

$2\cdot 4^{2\mathrm{k}+1}+3^{3\mathrm{k}+1}=11\mathrm{q}$ for some integer q. Now

$\mathrm{P}(\mathrm{k}+1)=2\cdot 4^{2\mathrm{k}+3}+3^{3\mathrm{k}+4}$

$=2\cdot 4^{2k+1}\cdot 4^2+3^{3k+1}\cdot 3^3$

$=16\cdot 2\cdot 4^{2\mathrm{k}+1}+27\cdot 3^{3\mathrm{k}+1}$

$=16\cdot 2\cdot 4^{2\mathrm{k}+1}+(16+11)\cdot 3^{3\mathrm{k}+1}$

$=16\left[2\cdot 4^{2\mathrm{k}+1}+3^{3\mathrm{k}+1}\right]+11\cdot 3^{3\mathrm{k}+1}$

$=16\cdot 11\mathrm{q}+11\cdot 3^{3\mathrm{k}+1}$

$=11\left(16\mathrm{q}+3^{3\mathrm{k}+1}\right)=11\mathrm{m}$

where $m=16q+3^{3k+1}$is another integer.

$\therefore \mathrm{P}(\mathrm{k}+1)$ is divisible by 1 1.