XVIII-th Macedonian mathematical Olympiad

A number $B$ having $n-1$ digits '1' and one digit '7' in decimal representation is of the form $B=A_n+6\cdot 10^k$ where $A_n$ is a number having $n$ digits '1', and $0\le k<n$. Notice that if $3|n$ then the sum of the digits of $B$ is $3n+6$. Notice that $$A_1=1,A_2=4,A_3=6,A_4=5,A_5=2,A_6=0(\mathrm{mod}7)$$ $$10^0=1;10^1=3;10^2=2;10^3=6;10^4=4;10^5=5;10^6=1(\mathrm{mod}7)$$ Let $n>6$. Then $n=6t+r$ where $t\ge 0$, $1\le r\le 6$. Then $$A_n=A_{6t+r}=A_{6t}+A_r\cdot 10^{6t}\equiv A_{6t}+A_r\equiv A_r(\mathrm{mod}7)$$ If $6|n$ then $3|n$, so therefore we saw that $B$ is not prime in any case. Suppose $6$ does not divide $n$. Then we saw that $A_n\equiv A_r(\mathrm{mod}7)$ and additionally $A_r$ is not congruent to $0$ modulo $7$. Put $A_r\equiv t(\mathrm{mod}7)$. For this $t$, from the above it follows that we can choose $k$, $0\le k\le 5$, such that $6\cdot 10^k=-10^k\equiv -t(\mathrm{mod}7)$. So this $B$ is divisible by $7$. For $n=6$ we saw that the numbers are composite. We need to check all cases $n\le 5$. For $n=5$ $$A_5+6\cdot 10^2=2-2=0(\mathrm{mod}7)$$ For $n=4$, $1711=29\cdot 59$. In the case $n=3$ we already saw that all numbers are composite. For $n=2$ and $n=1$ all numbers are prime.