Question

$2^{3n}-7n-1$ is divisible by

• A. $64$
• B. $36$
• C. $49$
• D. $25$

Answer 1

Answer: (C)

Let $P\left(n\right)=2^{3n}-7^n-1$
$\therefore P\left(1\right)=0,P\left(2\right)=49$
$P\left(1\right)$ and $P\left(2\right)$ are divisible by $49$.
Let $P\left(k\right)\equiv 2^{3k}-7k-1=49\lambda ...\left(i\right)$
$\therefore P\left(k+1\right)=23^{3k+3}-7\left(k+1\right)-1$
$=8\left(49\lambda +7k+1\right)-7k-8$ [using $\left(i\right)$]
$=49\left(8\lambda \right)+49k=49\left(8\lambda +k\right)$
Hence, by the principle of mathematical induction
$2^{3n}-7n-1$ is divisible by $49$.