prove that $f_n = 37111111...111$ is never prime

Question

Let $$f_n = 37111111...111$$ with n 1's. Prove that $$f_n$$ will never be prime for $$n\ge1.$$

I tried to look $$f_n$$ in mod(p), assuming $$f_n$$ is prime, for the sake of contradiction. I also tried to apply Wilson and Fermat's small theorem.

I'm sure there must be a simple factorization which I'm overseeing.

Answer 1

Hint: the modular bases you are looking for are $3, 7, 13$ and $37$ for different values of $n$. One of these divides every member of the sequence.

Answer 2

Hint $\ $ Examining prime factorizations of the first $\,\color{#c00}6\,$ elements of $\,f_N = 334\cdot 10^{N}\!-1$ reveals

$\, 10^{\large\color{#c00}6}\!-1 = 3^3\cdot 7\cdot 11\cdot 13\cdot 37,\,$ and $\,f_N$ is divisible by one of those prime powers $\,p^i$ for all $\,N <\color{#c00} 6,\,$ therefore $\,f_N\,$ is divisible by one of those $\,q = p^i\,$ for all $\,N,\,$ so $\,f_N/9\,$ is composite for all $\,N\ge 0.\,$

Proof $\ $ Dividing $\,N\,$ by $\,6\,$ yields $\,N = K+6J\,$ for $\,K < 6.\ $ By hypothesis some $\,q\mid f_K$

${\rm mod}\ q\!:\ \color{#c00}{10^{\,6}}\equiv\color{#c00} 1\,\Rightarrow\, \color{#0a0}{10^N} \equiv 10^{K+6J}\equiv 10^K(\color{#c00}{10^6})^J\equiv 10^K\color{#c00}1^J\equiv \color{#0a0}{10^K}\ $ by $ $ Congruence Rules

So $\,q\mid f_K\Rightarrow\, 0\equiv f_K\! \equiv 334\cdot \color{#0a0}{10^K}\!-1\equiv 334\cdot \color{#0a0}{10^N}\!-1 \equiv f_N\Rightarrow\, q\mid f_N,\,$ properly by $\,q < f_0 \le f_N$

Remark $\ $ This is a typical application of covering congruences. See also this question, and see this question for a polynomial analog, and a link to a paper of Schinzel.