If $n$ is a positive integer, then $2.4^{2 n+1}+3^{3 n+1}$ is divisible by:

Let $P(n)=2 \cdot 4^{2 n+1}+3^{3 n+1}$
Then $P(1)=2.4^3+3^4=209$, which is divisible by 11 but not divisible by 2,7 or 27 .
Further, let $P(k)=2.4^{2 k+1}+3^{3 k+1}$ is divisible by 11 , that is,
$2.4^{2 k+1}+3^{3 k+1}=11 q$ for some integer q. Now
$\begin{aligned}<br />\mathrm{P}(\mathrm{k}+1) & =2 \cdot 4^{2 \mathrm{k}+3}+3^{3 \mathrm{k}+4} \\<br />& =2 \cdot 4^{2 \mathrm{k}+1} \cdot 4^2+3^{3 \mathrm{k}+1} \cdot 3^3 \\<br />& =16 \cdot 2 \cdot 4^{2 \mathrm{k}+1}+27 \cdot 3^{3 \mathrm{k}+1} \\<br />& =16 \cdot 2 \cdot 4^{2 \mathrm{k}+1}+(16+11) \cdot 3^{3 \mathrm{k}+1} \\<br />& =16\left[2 \cdot 4^{2 \mathrm{k}+1}+3^{3 \mathrm{k}+1}\right]+11 \cdot 3^{3 \mathrm{k}+1} \\<br />& =16 \cdot 11 \mathrm{q}+11 \cdot 3^{3 \mathrm{k}+1} \\<br />& =11\left(16 \mathrm{q}+3^{3 \mathrm{k}+1}\right)=11 \mathrm{~m}<br />\end{aligned}$
where $m=16 q+3^{3 k+1}$ is another integer.
$\therefore \mathrm{P}(\mathrm{k}+1)$ is divisible by 11 .