Call this integer is $ n$ Then $ n=10A+b$ where $ b\in\{0,1,..,9\}$ If we delete the final digit then we has number $ A$ Ipply that $ A|10A+B\Leftrightarrow A|b$ (1) Case 1: $ b=0$ Then (1) true for all $ A>0$ Imply that solution is $ n=10A$ where $ A\in N$ Case 2 $ b$ is different from 0. Then $ A|b$ For $ b=1$ we have $ n\in\{11\}$ For $ b=2$ we have $ n\in\{12,22\}$ For $ b=3$ we have $ n\in \{13,33\}$ For $ b=4$ we have $ n\in \{14,24,44\}$ For $ b=5$ we have $ n\in\{15,55\}$ For $ b=6$ we have $ n\in\{16,26,36,66\}$ For $ b=7$ we have $ n\in\{17,77\}$ For $ b=8$ we have $ n\in\{18,28,48,88\}$ For $ b=9$ we have $ n\in {\19,39,99\}$ An easy problem.