Question

A divisor of \[ 3^{6n}+56n-1,\qquad n\in\mathbb{N} \] is

• A. \(729\)
• B. \(625\)
• C. \(676\)
• D. \(784\) \bigskip

Hint:

For expressions of the form \[ a^{2m}-1, \] always use \[ a^{2m}-1=(a^m-1)(a^m+1). \] Then check divisibility of consecutive even factors or use modular arithmetic to identify constant divisors.

Answer 1

The Correct Option is D

Concept: Express the exponential term modulo the required divisor and use congruence properties to identify the greatest constant divisor.

Step 1: Since \(3^6=729\equiv1\pmod{784}\), it follows that \(3^{6n}\equiv1\pmod{784}\) for every positive integer \(n\).

Step 2: Also, \(56n=7\times8n\) and \(3^{6n}-1\) contributes the remaining factor required for divisibility by \(784\). Hence, \(3^{6n}+56n-1\equiv0\pmod{784}\).

Step 3: Therefore, the greatest constant divisor is \(\boxed{784}\).