, 2011

Let $a_n = a_{n,1}a_{n,2}\dots a_{n,k_n}$ be the decimal representation of $a_n$. Note that if $k_n$ is even, then $a_n \equiv_{11} a_{n,1} - a_{n,2} + \dots - a_{n,k_n}$ and $b_n \equiv_{11} a_{n,k} - a_{n,k-1} + \dots - a_{n,1} \equiv -a_n$. Hence $a_{n+1} = a_n + b_n \equiv_{11} 0$, so $a_{n+1}$ is divisible by $11$. Also note that if $a_n$ is divisible by $11$, then $b_n$ is also divisible by $11$, and hence $a_{n+1}$ is divisible by $11$. In order to prove that $a_7$ cannot be prime it suffices to prove that one of $a_1, a_2, \dots, a_6$ has an even number of digits.

For a contradiction, suppose that all six numbers have an odd number of digits. Note that $a_{n+1}$ can have at most one more digit than $a_n$, so in fact $a_1, \dots, a_6$ all have the same number of digits.

Let $a_{1,1} = a$ and $a_{1,k_1} = d$. Then the first digit of $a_2$ is either $a+d$ or $a+d+1$, and since $a_2$ has the same number of digits as $a_1$, $a+d < 10$. Hence the units digit of $a_2$ equals $a+d$.

Thus the first digit of $a_3$ is at least $2(a+d) < 10$, and the final digit of $a_3$ is at least $2(a+d)$. Continuing in this way, we see that the first digit of $a_4$ is at least $4(a+d)$, the first digit of $a_5$ is at least $8(a+d)$ and the first digit of $a_6$ is at least $16(a+b)$. However, $16(a+d)$ is certainly not less than $10$, which implies that $a_6$ must have more digits than $a_1$, a contradiction. This shows that $a_7$ must be divisible by $11$, and hence cannot be prime.