If the first two digits are $a$ and $b$ (with $a\neq 0$) then the six terms will be $a$, $b$, $a+b$, $a+2b$, $2a+3b$ and $3a+5b$, so we must have $3a+5b \leq 9$.
If $b=0$ then $a$ can be $1$, $2$ or $3$. If $b=1$ then $a$ can only be $1$, and $b$ cannot be greater than $2$.
Hence there are four possibilities, namely $101123$, $202246$, $303369$ and $112358$.
This problem is taken from the UKMT Mathematical Challenges.