Question

\( (2^{3n} - 1) \) is divisible by

• A. 6
• B. 7
• C. 8
• D. 9

Hint:

Use modular exponentiation and Fermat's Little Theorem to test for divisibility.

Answer 1

The Correct Option is B

We know a standard identity:
\( 2^3 = 8 \Rightarrow 2^3 \equiv 1 \pmod{7} \Rightarrow (2^3)^n \equiv 1^n = 1 \pmod{7} \)
Thus, \( 2^{3n} \equiv 1 \pmod{7} \Rightarrow 2^{3n} - 1 \equiv 0 \pmod{7} \)
Hence, \( 2^{3n} - 1 \) is divisible by 7.