Notice $ n=2^a5^b $ for some non-negatives $ a,b $ since $ n=a_1+a_2+... $ Suppose $ a_k $ is the last element in the sequence $ \{a_n\} $ such that $ a_k \neq 0 $ Now, there are three cases: (1) $ k=a=b $: Then we get $ 10 \le n=a_1+a_2+...=1 $, contradition. (2) $ k=a>b $: Then $ 2^k \le n=a_1+a_2+... \le 9k \rightarrow k<6 \Rightarrow k=3, n=8 $ (3) $ k=b>a $ Then $ 5^k \le n=a_1+a_2+... \le 9k \rightarrow k<2 \Rightarrow $ no such solution.