1) Number with one digit is clearly solution. 2) Let $N=10a+b$ with $1\le a\le 9$ and $1\le b\le 9$ The problem give us that $10a+b|10b+a$ Case 1 : $10b+a=2(10a+b)$ $\iff$ $8b=19a$ but $19$ can't divide the left hand side. Case 2 : $10b+a=3(10a+b)$ $\iff$ $7b=29a$ but $29$ can't divide the left hand side. Case 3 : $10b+a=4(10a+b)$ $\iff$ $2b=13a$ but $13$ can't divide the left hand side. Case 4 : $10b+a=k(10a+b)$ with $k\ge5$. In this case if $a\ge 2$ $10b+a\ge 5*10*2=100$ but $100$ doesn't have 2 digits. So $a=1$ $10b+1=10k+kb$ $\iff$ $b(10-k)=10k-1$ So $10-k|10k-1$ so $10-k\in\{1.3.9\}$ we compute that there are no solutions. 3) Let $N=100a+10b+c$ with $1\le a,c\le 9$ and $0\le b\le 9$ The problem give us that $100a+10b+c|100c+10b+a$ Case 1 : $100c+10b+a=2(100a+10b+c)$ $\iff$ $98c=199a+10b$ so $c>2a$ and $c\ge3a$ so $a\in\{1.2.3\}$ but $a$ is even so $a=2$ $98c=398+10b$ we find that $c\equiv 1\pmod5$ so $c=6$ but that implies that $b=19$ this is clearly impossible. Case 2 : $100c+10b+a=3(100a+10b+c)$ $\iff$ $97c=299a+20b$ so $c>3a$ and $c\ge4a$ so $a\in\{1.2\}$ but if we look $\pmod4$ we find that $a\equiv 3c\pmod4$ so if $a=2$ $c\equiv 2\pmod4$ but $c\ge 8$ and there are no digits which respect this conditions and $a=1$ we find that $c\equiv3\pmod4$ but we have $c\ge4$ so $c=7$ but that give us $b=19$ impossible Case 3 : $100c+10b+a=k(100a+10b+c)$ with $k\ge4$ $\iff$ $(100-k)c=(10k-10)b+(100k-1)a$ so $c>ka$ so $10>c\ge a (k+1)\ge5a$ so $a=1$ so $100c+10b+1=k(100+10b+c)$ so if we look the number of hundred we find that $k=c$ or $k+1=c$ but $k=c$ give us $10b(c-1)+c^2-1=0$ but it's clear that $10b(c-1)+c^2-1>0$. If $k=c-1$ we have $20b=10bc+c^2-c$ but $k\ge4$ so $10bc\ge40b\ge20b$ and $c^2>c$ so it is impossible.