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Question

For all $n \in N, 2 \cdot 4^{2 n + 1} + 3^{3 n + 1}$ is divisible by

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Solution

For $n = 1,$ we have
$2 \cdot 4^{2 n + 1} + 3^{3 n + 1} = 2 \times 4^{3} + 3^{4} = 209$ which is divisible by 11
For $n = 2,$ we have
$2 \cdot 4^{2 n + 1} + 3^{3 n + 1} = 2 \times 4^{5} + 3^{7} = 4235$ which is divisible by 11
Similarly, it can be checked that ${2.4}^{2 n + 1} + 3^{3 n - 1}$ is divisible by 11 for other values of $n$

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