Any ideas ? Thanks

yuyang wrote: Any ideas ? Thanks if $a_i$ is of even no. Of digits for any $i\in N$ then $11|a_{n}$ for all $n\geq i+1$ If $a_i$ is of odd no. Of digits then $2a_i\equiv a_{i+1} \mod 11$ Also if $a_1$ is of odd no. Of digits see $a$ is of course not true because try proving you will get $k<7$ such that $a_k$ is of even digits or $11|a_k$. then for $b$ try getting some $k<5$ such that $a_k$ is of even digits...

For b), take $a_1=10031$. For a), the answer is no. Observe that once $a_i$ has an even number of digits, $a_{i+1}$ will be divisible by $11$. It can be checked that none of the single-digit possibilities for $a_1$ lead to $11$. Thus, we need to start out with an odd number of digits for $a_1$ (say $2k+1$ digits) and not obtain a number with $2k+2$ digits until the very last step (so $a_6<10^{2k+1}$). This is because $a_{n+1}$ cannot have two digits more than $a_n$. Let $p_i,q_i$ be the first and last digits of $a_i$. Until $a_7$, we can never carry the one's digit, because then we'd also gain a digit in the front. However, it's easy to see that $q_2$ is at least $1$ and from there on, it doubles (at least) in each step, which means that $q_5\geq 8$, so $a_6$ has $2k+2$ digits and $11\mid a_7$, a contradiction.