Q93 / 180 · COMEDK 2022 · Mathematics
$\left(2^{3 n}-1\right)$ is divisible by
- A6
- Correct answerB7
- C8
- D9
Solution
The expression is $2^{3n} - 1$.
We can rewrite $2^{3n}$ as $(2^3)^n$.
Since $2^3 = 8$, the expression becomes $8^n - 1$.
Using the algebraic identity $x^n - 1 = (x - 1)(x^{n-1} + x^{n-2} + \dots + 1)$, we substitute $x = 8$.
$8^n - 1 = (8 - 1)(8^{n-1} + 8^{n-2} + \dots + 1) = 7(8^{n-1} + 8^{n-2} + \dots + 1)$.
This shows that the expression is always divisible by 7 for any positive integer $n$.
Answer: 7
We can rewrite $2^{3n}$ as $(2^3)^n$.
Since $2^3 = 8$, the expression becomes $8^n - 1$.
Using the algebraic identity $x^n - 1 = (x - 1)(x^{n-1} + x^{n-2} + \dots + 1)$, we substitute $x = 8$.
$8^n - 1 = (8 - 1)(8^{n-1} + 8^{n-2} + \dots + 1) = 7(8^{n-1} + 8^{n-2} + \dots + 1)$.
This shows that the expression is always divisible by 7 for any positive integer $n$.
Answer: 7
Hence, the correct answer is option: 'B'.