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If the greatest divisor of $30 \cdot 5^{2 n}+4 \cdot 2^{3 n}$ is $p, \forall n \in N$ and that of $2^{2 n+1}-6 n-2$ is $q, \forall n \in N$, then $p+q=$
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Verified Answer
The correct answer is:
52
We have, $30 \cdot 5^{2 n}+4 \cdot 2^{3 n}$
Put
$\begin{aligned}
n & =1 \\
30 \times 25+4 \times 8 & =750+32=782 \\
782 & =2 \times 23 \times 17
\end{aligned}$
Put $x=2$
$30 \times 625+4 \times 64=18750+256=19006$
In Eqs. (i) and (ii), we get
Greatest divisor $=2 \times 17=34$
$\therefore \quad p=34$
Now in $2^{2 n+1}-6 n-2$
Put $n=1$
$8-6-2=0$
Put
$n=2$
$32-12-2=18$
Put $n=3$
$128-18-2=108$
$\because$ Greatest divisor $=18$
$\begin{aligned}
\therefore \quad q & =18 \\
\therefore p+q=34+18 & =52
\end{aligned}$
Put
$\begin{aligned}
n & =1 \\
30 \times 25+4 \times 8 & =750+32=782 \\
782 & =2 \times 23 \times 17
\end{aligned}$
Put $x=2$
$30 \times 625+4 \times 64=18750+256=19006$
In Eqs. (i) and (ii), we get
Greatest divisor $=2 \times 17=34$
$\therefore \quad p=34$
Now in $2^{2 n+1}-6 n-2$
Put $n=1$
$8-6-2=0$
Put
$n=2$
$32-12-2=18$
Put $n=3$
$128-18-2=108$
$\because$ Greatest divisor $=18$
$\begin{aligned}
\therefore \quad q & =18 \\
\therefore p+q=34+18 & =52
\end{aligned}$
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