Constant Divisors of 7^(6n) − 6^(6n) For n a positive integer, the expression 7^(6n) − 6^(6n) is divisible by which of the following?

Introduction / Context:
Expressions of the form a^m − b^m have well-known divisibility properties. If d divides m, then a^m − b^m is divisible by a^d − b^d. This problem asks you to recognize constant divisors that work for every positive integer n when the exponent is 6n.

Given Data / Assumptions:

• Expression: 7^(6n) − 6^(6n) with n ≥ 1.
• We are checking divisibility by 13, 127, and 559.
• 6 divides 6n, so 7^(6n) − 6^(6n) is divisible by 7^6 − 6^6.

Concept / Approach:
Since 6 | 6n, a standard result guarantees 7^(6n) − 6^(6n) is divisible by 7^6 − 6^6. Therefore, any factor of 7^6 − 6^6 will also divide 7^(6n) − 6^(6n) for all n. Compute 7^6 − 6^6 and factor (or test) it against the candidate divisors.

Step-by-Step Solution:
Compute 7^6 = 117,649 and 6^6 = 46,656.Difference: 117,649 − 46,656 = 70,993.Check factors: 70,993 ÷ 13 = 5,461 (integer) → divisible by 13.Further factor: 70,993 = 127 × 559 (since 127×500 = 63,500; 127×59 = 7,493; sum = 70,993).Thus 127 and 559 also divide 7^6 − 6^6, and therefore divide 7^(6n) − 6^(6n) for all n.

Verification / Alternative check:
Note also that 7 ≡ −6 (mod 13), so for any even exponent 6n, 7^(6n) − 6^(6n) ≡ (−6)^(6n) − 6^(6n) = 0 (mod 13). The factorization of 70,993 confirms divisibility by 127 and 559 as well.

Why Other Options Are Wrong:
“None of these” is wrong because all three listed numbers divide the expression for every n ≥ 1. Selecting only one of 13, 127, or 559 would be incomplete; hence “All of these” is the only correct choice.

Common Pitfalls:
Attempting to test only n = 1 and missing the general argument; assuming divisibility holds for one modulus but not recognizing all listed moduli are factors of 7^6 − 6^6.

Final Answer:
All of these