Lets consider the base case of remainder of 1 when divided by 2 and remainder of 2 when divided by 3. The smallest number satisfying this 5, which is LCM(2, 3) - 1.
This idea stems from the fact that the answer can be written in the below form for positive integers \begin{align} X &= 2X_{1} + 1\newline X &= 3X_{2} + 2\newline X &= 10X_{9} + 9\newline \end{align}
which can be rearranged to be written as \begin{align} X + 1 &= 2(X_{1} + 1)\newline X + 1 &= 3(X_{2} + 1)\newline X + 1 &= 10(X_{9} + 1)\newline \end{align}
Meaning, the answer we are looking for is LCM(2, 3, .., 10) - 1