Find the residue

A simple way to find the answer is substituting $n = 1$ ; we get $3^6 - 26 = 703 = 4\cdot 169 + \boxed{27}$ .

But we must prove that it works for all positive integers $n$ .

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$3^{3n+3} = (3^3)^{n+1} = (1 + 2\cdot 13)^{n+1}.$ Use the binomial expansion: $\dots = \binom{n+1}{0} 1^{n+1} + \binom{n+1}{1} 1^n\cdot (2\cdot 13) + \binom{n+1}{2} 1^{n-1}\cdot (2\cdot 13)^2 + \cdots + \binom{n+1}{n+1} (2\cdot 13)^{n+1}.$ From the second term onward, each term is a multiple of $13^2 = 169$ . Therefore, modulo 169, we have $\dots \equiv 1 + (n+1)\cdot 2\cdot 13 = 1 + 26(n+1) = 27 + 26n.$ Finally, $3^{3n+3} - 26n \equiv 27 + 26n - 26 n = \boxed{27}.$