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If $2\left(4^{2 n+1}\right)+3^{3 n+1}$ is divisible by $k, k>1$ for all $n \in N$, then the value of $k$ is
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Verified Answer
The correct answer is:
11
The given expression
$\begin{aligned} & 2\left(4^{2 n+1}\right)+3^{3 n+1}=P(n) \\ & \because \quad P(n)=8\left(4^{2 n}\right)+3\left(3^{3 n}\right) \\ & \text {For } n=1 \text {, } \\ & \because \quad n \in N \\ & P(n=1)=128+81=209 \\ & \end{aligned}$
is divisible by 19 and 11.
Similarly, $\quad P(n=2)=2048+2187=4235$ is divisible by 11 but not divisible by 19.
So, the given expression $2\left(4^{2 n+1}\right)+3^{3 n+1}$ is divisible by
For all
$k=11$
$n \in N \text {. }$
$\begin{aligned} & 2\left(4^{2 n+1}\right)+3^{3 n+1}=P(n) \\ & \because \quad P(n)=8\left(4^{2 n}\right)+3\left(3^{3 n}\right) \\ & \text {For } n=1 \text {, } \\ & \because \quad n \in N \\ & P(n=1)=128+81=209 \\ & \end{aligned}$
is divisible by 19 and 11.
Similarly, $\quad P(n=2)=2048+2187=4235$ is divisible by 11 but not divisible by 19.
So, the given expression $2\left(4^{2 n+1}\right)+3^{3 n+1}$ is divisible by
For all
$k=11$
$n \in N \text {. }$
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