$\large 3^{6n}- 2^{6n}$

Is always divisible by which number(largest in options)?

nbelongs to natural numbers

None
2
35
3
34
5
36
6

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Since $3^{6n} - 2^{6n} \; = \; (3^6)^n - (2^6)^n \; = \; 729^n - 64^n \; = \; (729 - 64)\sum_{j=0}^{n-1}729^{n-j} \times 64^j$ is always, in fact, divisible by $729-64 = 665 = 5\times7\times19$ , it is certainly always divisible by $\boxed{35}$ , but $665$ is the largest possible common divisor (it divides $3^{6n} - 2^{6n}$ for all $n$ , and is equal to $3^{6n} - 2^{6n}$ when $n=1$ ).