Saudi Arabian Mathematical Competitions

Let $n \geq 2$ be such an integer. Because $n=a_{1}+a_{2}+\cdots$, there exists a positive integer $k>0$ such that $a_{k} \neq 0$ and $a_{i}=0$ for all integers $i>k$.

Because $\frac{10^{k}}{n}=\overline{a_{1}a_{2}\cdots a_{k}}$ is an integer and $\frac{10^{k-1}}{n}=\overline{a_{1}a_{2}\cdots a_{k-1},a_{k}}$ is not an integer, there exist two non-negative integers $a, b \geq 0$ such that $n=2^{a} \cdot 5^{b}$, with $\max\{a, b\}=k$. We have three cases:

a. If $k=a=b$, then $n=10^{k} \geq 10$ and $a_{1}+a_{2}+\cdots+a_{k}=1 \neq n$, contradiction.

b. If $k=b>a$, then
$$5^{k} \leq 2^{a} \cdot 5^{k}=n=a_{1}+a_{2}+\cdots+a_{k} \leq 9k.$$
For $k=2$, we have $5^{2}=25>18=9 \times 2$. If $k \geq 2$ and $5^{k}>9k$, then $5^{k+1}>5 \times 9k=9(k+4k)>9(k+1)$. This implies that $k=1$ and therefore $a=0$, that is $n=5$. But $\frac{1}{n}=0.2$ and $n \neq 2$, contradiction.

c. If $k=a>b$, then
$$2^{k} \leq 2^{k} \cdot 5^{b}=n=a_{1}+a_{2}+\cdots+a_{k} \leq 9k.$$
For $k=6$, we have $2^{6}=64>54=9 \times 6$. If $k \geq 6$ and $2^{k}>9k$, then $2^{k+1}=2^{k}+2^{k}>9k+2^{6}>9(k+1)$. This implies that $k \leq 5$.

Now, assume that $b \geq 1$. We have
$$10 \times 2^{k-1} \leq n=a_{1}+a_{2}+\cdots+a_{k} \leq 9k.$$
For $k=1$, this is not satisfied. Assume that $10 \times 2^{k-1}>9k$. We have $10 \times 2^{k}>9k+10 \times 2^{k-1} \geq 9k+10>9(k+1)$, contradiction.

Hence, $b=0$ and $n=2^{k}$.

By checking for $k=1,2,3,4,5$ we find that only $n=2^{3}$ has this property.

Hence $n=8$.