Estonian Mathematical Olympiad

Let $n$ satisfy the condition and $n^{\prime }$ be the number obtained by the replacement of digits. Denote by $x$ the number formed by the digits to the left of the replaced digit and denote by $y$ the number formed by the digits to the right of the replaced digit. Let $k$ be the number of digits in $y$. Then $n=x\cdot 10^{k+1}+4\cdot 10^k+y$ and $n^{\prime }=x\cdot 10^{k+2}+22\cdot 10^k+y$. As $n^{\prime }$ is divisible by $n$, then so must also be $n^{\prime }-n$ and $10n-n^{\prime }$, which yields $$x\cdot 10^{k+1}+4\cdot 10^k+y\mid 9x\cdot 10^{k+1}+18\cdot 10^k,(2)$$ $$x\cdot 10^{k+1}+4\cdot 10^k+y\mid 18\cdot 10^k+9y.(3)$$ We will consider the following cases.

If $x=0$, then by (2), $t(4\cdot 10^k+y)=18\cdot 10^k$ for some integer $t$. As $5(4\cdot 10^k+y)\ge 5\cdot 4\cdot 10^k>18\cdot 10^k>3\cdot 5\cdot 10^k>3(4\cdot 10^k+y)$, the only option is $t=4$. The equation $4(4\cdot 10^k+y)=18\cdot 10^k$ yields $y=\frac{2\cdot 10^k}{4}=5\cdot 10^{k-1}<10^k$. Altogether $n=45\cdot 10^{k-1}$.

If $x=1$, then by (2), $t(14\cdot 10^k+y)=108\cdot 10^k$ for some integer $t$. As $8(14\cdot 10^k+y)\ge 8\cdot 14\cdot 10^k>108\cdot 10^k>6\cdot 15\cdot 10^k>6(14\cdot 10^k+y)$, the only option is $t=7$, but the equation $7(14\cdot 10^k+y)=108\cdot 10^k$ has no integer solutions, because $7\nmid 108\cdot 10^k$.

If $x=2$, then by (2), $t(24\cdot 10^k+y)=198\cdot 10^k$ for some integer $t$. As $9(24\cdot 10^k+y)\ge 9\cdot 24\cdot 10^k>198\cdot 10^k>7\cdot 25\cdot 10^k>7(24\cdot 10^k+y)$, the only option is $t=8$. The equation $8(24\cdot 10^k+y)=198\cdot 10^k$ yields $y=\frac{6\cdot 10^k}{8}=75\cdot 10^{k-2}<10^k$. Altogether $n=2475\cdot 10^{k-2}$.

If $x\ge 3$, then by (3) we have $x\cdot 10^{k+1}+4\cdot 10^k+y\le 18\cdot 10^k+9y$, which yields $x\cdot 10^{k+1}\le 14\cdot 10^k+8y$. On the other hand $x\cdot 10^{k+1}\ge 30\cdot 10^k>22\cdot 10^k>14\cdot 10^k+8y$. The equations contradict each other, so no such $n$ can exist.