XVI Junior Macedonian Mathematical Olympiad

We use the equality: $\stackrel{k}{\overbrace{\mathrm{11...1}}}=\frac{10^k-1}{9}$. We get $$\frac{\stackrel{2n}{\overbrace{\mathrm{11...1}}}}{11}=\frac{10^{2n}-1}{11\cdot 9}=\frac{(10^n-1)(10^n+1)}{11\cdot 9}.$$ For $n=1$, $\stackrel{2n}{\overbrace{\mathrm{11...1}}}/11=\frac{11}{11}=1$ which is not a prime number. For $n=2$, $$\frac{\stackrel{2n}{\overbrace{\mathrm{11...1}}}}{11}=\frac{(10^2-1)(10^2+1)}{11\cdot 9}=\frac{99\cdot 101}{99}=101.$$ We will show that for $n>2$, the number $\stackrel{2n}{\overbrace{\mathrm{11...1}}}/11$ is composite. We consider two cases: $n$ being even and $n$ being odd.

Case 1. Let $n$ be even. Then $10^{2k}$ gives residue 1 when divided by 9 and 11. Since $\text{gcd}(9,11)=1$ we get that $10^n-1$ is divisible by 99 and the number $$\stackrel{2n}{\overbrace{\mathrm{11...11}}}=\frac{(10^n-1)(10^n+1)}{11\cdot 9}=\frac{(10^n-1)}{99}(10^n+1)$$ is a product of two natural numbers bigger than 1.

Case 2. Let $n$ be odd. Then $10^{2k+1}$ gives residue -1 when divided by 11. The number $10^{2k+1}+1$ is divisible by 11, and $10^{2k+1}-1$ is divisible by 9. The number $$\stackrel{2n}{\overbrace{\mathrm{11...11}}}=\frac{(10^n-1)(10^n+1)}{11\cdot 9}=\frac{(10^{2k+1}-1)(10^{2k+1}+1)}{9}$$ is a product of two natural numbers bigger than 1 and hence it is composite.