# Question on application of Chinese Remainder Theorem x <mspace width="thickmathspace" />

Question on application of Chinese Remainder Theorem
$x\phantom{\rule{thickmathspace}{0ex}}\equiv \phantom{\rule{thickmathspace}{0ex}}3\phantom{\rule{thickmathspace}{0ex}}\left(\text{mod}\phantom{\rule{thickmathspace}{0ex}}30\right)$
$x\phantom{\rule{thickmathspace}{0ex}}\equiv \phantom{\rule{thickmathspace}{0ex}}5\phantom{\rule{thickmathspace}{0ex}}\left(\text{mod}\phantom{\rule{thickmathspace}{0ex}}56\right)$
I have a system of modular equation that I want to solve. However, I thought that this system has no solution because the modulos are not coprime. Further, attempting to solve using chinese remainder theorem:
$x\phantom{\rule{thickmathspace}{0ex}}\equiv \phantom{\rule{thickmathspace}{0ex}}56p\phantom{\rule{thickmathspace}{0ex}}+\phantom{\rule{thickmathspace}{0ex}}30q$
where $p$ is such that
$56p\phantom{\rule{thickmathspace}{0ex}}\equiv \phantom{\rule{thickmathspace}{0ex}}3\phantom{\rule{thickmathspace}{0ex}}\left(\text{mod}\phantom{\rule{thickmathspace}{0ex}}30\right)\phantom{\rule{thickmathspace}{0ex}}$
and $q$ is such that
$30q\phantom{\rule{thickmathspace}{0ex}}\equiv \phantom{\rule{thickmathspace}{0ex}}5\phantom{\rule{thickmathspace}{0ex}}\left(\text{mod}\phantom{\rule{thickmathspace}{0ex}}56\right)\phantom{\rule{thickmathspace}{0ex}}$
However, again the modular inverses of these do not exist.
Yet, one solution to this system of modular inequalities is 1293. How come the Chinese remainder theorem gives that no solution exists?
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Govorei9b
Better if you write it as
$x\equiv 3\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}2\right),$
$x\equiv 3\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}15\right),$
$x\equiv 5\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}8\right),$
$x\equiv 5\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}7\right).$
The ones with 2 and 8 are consistent, as the one with 2 is just asking for an odd number, so the system is equivalent to
$x\equiv 3\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}15\right),$
$x\equiv 5\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}8\right),$
$x\equiv 5\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}7\right),$
where now the moduli are coprime. That is the requirement for the Chinese Remainder Theorem.
So we combine second and third again to get
$x\equiv 3\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}15\right),$
$x\equiv 5\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}56\right).$
For the 56 one, we have
$5,61,117,173,229,285,341,397,453,\dots$
and
$453\equiv 3\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}15\right).$