Question

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

• A. 2
• B. 7
• C. 11
• D. 27

Answer 1

Answer: (C)

Let $P(n)=2.4^{2n+1}+3^{3n+1}$
Then $P(1)=2.4^3+3^4=209,$ which is divisible by 11 but not divisible by 2, 7 or 27.
Further, let $P(k)=2.4^{2k+1}+3^{3k+1}$ is divisible by 11, i.e., $2.4^{2k+1}+3^{3k+1}=11q$ for some integer $q$.
Now $P(k+1)=2.4^{2k+3}+3^{3k+4}$
$=2.4^{2k+1}.4^2+3^{3k+1}.3^3$
$=\mathrm{16.2.}{4}^{2k+1}+27.3^{3k+1}$
$=\mathrm{16.2.}{2}^{2k+1}+\left(16+11\right).3^{3k+1}$
$=16\left[2.4^{2k+1}+3^{3k+1}\right]+11.3^{3k+1}$
$=16.11q+11.3^{3k+1}$
$=11\left(16q+3^{3k+1}\right)=11m$
where $m=16q+3^{3k+1}$ is another integer.
$\therefore p(k+1)$ is divisible by 11.
$\therefore P(n)=2.4^{2n+1}+3^{3n+1}$ is divisible by 11
for all $n\in N$.