Question

What values of $n$ satisfy the following statements (given n is a natural number)?
$2^{3n}-1$ is divisible by $7$

• A. Even values of $n$
• B. Odd values of $n$
• C. All values of $n$
• D. Cannot say

Answer 1

The correct option is C All values of $n$

As we see $2^{3n}$-1=$8^n-1$

can be written as ${(7+1)}^n$ $=$ ${(}^n{C}_0{.7}^n{+}^n{C}_1{.7}^{n-1}........{.}^n{C}_{n-1}.7{+}^n{C}_n)$=${(}^n{C}_0{.7}^n{+}^n{C}_1{.7}^{n-1}........{.}^n{C}_{n-1}.7+1)-1$

So, $8^n-1$ $=$ ${(}^n{C}_0{.7}^n{+}^n{C}_1{.7}^{n-1}........{.}^n{C}_{n-1}.7)$

Taking $7$ common, we get

$7{(}^n{C}_0{.7}^{n-1}{+}^n{C}_1{.7}^{n-2}........{.}^n{C}_{n-1})$

So, it is of the form $7k$, hence divisible by $7$.