Question

For all $n\in N$, $3^{2n+2}-8n-9$ is divisible by

• A. $8$
• B. $16$
• C. $32$
• D. $64$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $8$
B $16$
C $32$
D $64$

$3^{2n+2}$
$=9^{n+1}$
$=(1+8{)}^{n+1}$
$=1+8{}^{n+1}{C}_1+64k$ where K is a positive integer.
$=1+8(n+1)+64k$
$=64k+8n+9$
Hence
$3^{2n+2}-8n-9$
$=64k+8n+9-8n-9$
$=64k$
Hence the above expression is divisible by 2,4,8,16,32,64.