Is it possible to (uniquely) reverse modulo operation by solving multiple equations with the same original integer?

I've tried searching but I haven't been able to find an answer. There are similar questions about reversing modulo operation here on stackexchange, but I haven't found a question which is applicable to my problem.

All answers I've found on reversing modulo says that you cannot uniquely determine the original answer. (At least if I understand them right.)

But let's say we know that:

$x \!\mod 10 = 13\$ and $\ x \!\mod 13 = 2$.

Can we with this method uniquely determine $x$ if we have $n$ amount of these equations? I'm guessing that if $n \rightarrow \infty$ we can do it, but would this be the only case?

I hope I'm making sense, thanks!

• Chinese Remainder Theorem – Matthew Towers Mar 18 '16 at 22:19
• Well, wouldn't we only be able to determine a solution to modulo $\prod_n m$? – fleablood Mar 18 '16 at 22:20
• If you add the product of all modulos to $x$, then that is another solution. – Henricus V. Mar 18 '16 at 22:23

If you have the $n$ equations: $$x\bmod m_1=m_2$$ You won't be able to determine $x$ better than $$\text{lcm}(m_1,m_2,\ldots,m_n)$$ ($\text{lcm}$ meaning "least common multiple").