Call this number is $ m = {a_1..a_n}_{10}$ where $ a_i\in\{1,...,9\}$ The new number is $ m' = {a_2..a_na_1}_{10}$ Consider the equation $ m' = km$ $ \Leftrightarrow k(10^{n - 1}a_1 + T) = 10T + a_1$ where $ T = {a_2...a_n}_{10}$ We prove that this equation has no solution on the case $ k\in{5,6,8\}}$ Easy to check if $ k > 5$ then $ a_1 = 1$ Case 1 $ k = 5$ Imply that $ 5.10^{n - 1}a_1 + 5T = 10T + a_1$ Imply that $ 5|a_1$ so $ a_2 > 9$ (tradition!) So it has no solution on the case $ k = 5$ Case 2 $ k = 6$ $ 6.10^{n - 1}a_1 + 6T = 10T + a_1$ $ \Leftrightarrow a_1(6.10^{n - 1} + 1) = 4T$ Imply that $ 2|a_1$ (tradition!) Case 3$ k = 8$ $ (8.10^{n - 1} - 1)a_1 = 2T$ imply that $ 2|a_1$ so it has no solution. result If $ k\in\{5,6,8\}$ then no exist m satisfy the condition.