Always divisible — For any positive integer n, the number 6^(2n) - 1 is always divisible by which of the following?

Introduction / Context:
Expressions of the form a^(2n) - 1 often have predictable divisibility features because they factor as (a^n - 1)(a^n + 1). This question focuses on 6^(2n) - 1 and asks whether 5, 7, both, or neither always divide it. Modular arithmetic provides a fast, general proof.

Given Data / Assumptions:

• n is any positive integer.
• Expression: 6^(2n) - 1.
• We test divisibility by 5 and by 7.

Concept / Approach:
Reduce the base 6 modulo 5 and modulo 7. For mod 5, note that 6 ≡ 1 (mod 5), so powers of 6 simplify drastically. For mod 7, note that 6 ≡ -1 (mod 7), and even exponents of -1 give 1. These two quick observations resolve the question without heavy computation.

Step-by-Step Solution:
Modulo 5: 6 ≡ 1 (mod 5) → 6^(2n) ≡ 1^(2n) ≡ 1 (mod 5) → 6^(2n) - 1 ≡ 0 (mod 5).Modulo 7: 6 ≡ -1 (mod 7) → 6^(2n) ≡ (-1)^(2n) ≡ 1 (mod 7) → 6^(2n) - 1 ≡ 0 (mod 7).Therefore, the expression is divisible by both 5 and 7 for every positive integer n.

Verification / Alternative check:
Use factorization: 6^(2n) - 1 = (6^n - 1)(6^n + 1). Since consecutive even-odd neighbors differ by 2, at least one factor is divisible by 3, but here the mod 5 and mod 7 arguments already confirm both divisibilities independently.

Why Other Options Are Wrong:

• 11: Not always a divisor for general n.
• 5 or 7 alone: Both actually divide for all n, so singling out only one is incomplete.
• Neither: Contradicts the modular results.

Common Pitfalls:
Overcomplicating with large exponents; forgetting that 6 ≡ 1 (mod 5) and 6 ≡ -1 (mod 7); misapplying exponent rules to modular arithmetic.

Final Answer:
Both (B) and (C)