parmenides51 wrote: Let $s(a)$ denote the sum of digits of a given positive integer a. The sequence $a_1, a_2,..., a_n, ...$ of positive integers is such that $a_{n+1} = a_n+s(a_n)$ for each positive integer $n$. Find the greatest possible n for which it is possible to have $a_n = 2008$. Any solution?

$a_n = 2008 \Rightarrow a_{n-1} = 1985, 2003$ There does not exist $x\in \mathbb{N}$ for which $x+s(x)=1985$. $a_{n-1} = 2003 \Rightarrow a_{n-2} = 1978$ $a_{n-2} = 1978 \Rightarrow a_{n-2} = 1961$ $a_{n-3} = 1961 \Rightarrow a_{n-4} = 1939$ $a_{n-4} = 1939 \Rightarrow a_{n-5} = 1919$ There does not exist $x\in \mathbb{N}$ for which $x+s(x)=1919$. So the answer is $\boxed{n=6}$

This is $not$ a nice problem... how is this a $JBMO$ question?