$2^{3 n}-7 n-1$ is divisible by

Let $f(n) = 2^{3n} - 7n - 1 = 8^n - 7n - 1$.

For $n = 1$, $f(1) = 8^1 - 7(1) - 1 = 8 - 8 = 0$, which is divisible by any integer.

For $n = 2$, $f(2) = 8^2 - 7(2) - 1 = 64 - 14 - 1 = 49$.

For $n = 3$, $f(3) = 8^3 - 7(3) - 1 = 512 - 21 - 1 = 490$.

Since $f(2) = 49$ and $f(3) = 490 = 10 \times 49$, the expression $2^{3n} - 7n - 1$ is divisible by 49 for all $n \ge 1$.

Using the binomial expansion, $8^n = (1 + 7)^n = 1 + ^{n}C_{1}(7) + ^{n}C_{2}(7^2) + ^{n}C_{3}(7^3) + \dots + ^{n}C_{n}(7^n)$.

Substituting this into the expression: $f(n) = (1 + 7n + ^{n}C_{2}(49) + ^{n}C_{3}(343) + \dots) - 7n - 1 = ^{n}C_{2}(49) + ^{n}C_{3}(7^3) + \dots$.

Every term in the expansion is a multiple of 49, therefore $f(n)$ is divisible by 49.

Answer: 49