
Solution: Let $m$ be an integer number such that $m=b_{0}+b_{1} \cdot 5+$ $\cdots+b_{n} \cdot 5^{n}$. Denote $f(m)=2 b_{0}+2 b_{1} \cdot 10+\cdots+2 b_{n} \cdot 10^{n}$. It is clear that $\{f(m) \mid m \in \mathbb{N}\}$ is the set of integer numbers with only even digits in their decimal representation. Since $f\left(m_{1}\right)10^{n}$. Thus $2^{n}<30$ and therefore $n \leq 4$. If $n=4$, we get from (1) that $b_{0}+4 b_{1}+10 b_{2}=50 b_{3}+1250 b_{4} \geq 1250$, which is imposible. In the same way it is easy to show that $n \geq 3$, i.e., $n=3$. In this case $b_{0}+4 b_{1}+10 b_{2}=50 b_{3}$. Obviously $b_{3}=1$ and $b_{0}=b_{1}$, because $b_{0}-b_{1}$ is divisible by 5 . As a result we have the equation $b_{0}+2 b_{2}=10$, and its solutions are $b_{0}=2, b_{2}=4$ and $b_{0}=4, b_{2}=3$. Therefore all integer numbers $m$ with the required property are $m=2+2 \cdot 5+4 \cdot 5^{2}+5^{3}=$ 237 and $m=4+4 \cdot 5+3 \cdot 5^{2}+5^{3}=224$.



Problem 11.3. Let $\left\{a_{m}\right\}_{m=1}^{\infty}$ be a sequence of integer numbers such that their decimal representations consist of even digits ( $a_{1}=2$, $\left.a_{2}=4, a_{3}=6, \ldots\right)$. Find all integer numbers $m$ such that $a_{m}=12 m$.
