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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!

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

Search for "Chinese remainder theorem" for more information.

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