# x &#x2261;<!-- ≡ --> 11 <mspace width="1em" /> ( mod <mspace width="0.333em" /> 84

$x\equiv 11\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}84\right)$
$x\equiv 23\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}36\right)$
I have the bulk of the work done for this;
$x=11+84j$
$x=23+36k$
$⇒11+84j\equiv 23\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}36\right)$
$⇒84j\equiv 12\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}36\right)$
$⇒12j\equiv 12\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}36\right)$
$⇒j\equiv 1\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}36\right)$
$⇒j=1+36n$
Thus this system is true for any $x$ of the form
$x=11+84\left(1+36n\right)=95+3024n$
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Harold Cantrell
Correct is:

Therefore as claimed.

Ayanna Trujillo
$x\equiv 11\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}84\right)$, so $x=84k+11$, and $x\equiv 23\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}36\right)$, so $x=36n+23$. So $84k+11=36n+23$, and $84k-36n=12$. So $7k-3n=1$. So $3n=7k-1$, and $n=\frac{7k-1}{3}=2k+\frac{k-1}{3}$. Thus $3|k-1$, and $k=3t+1$, and $n=2\left(3t+1\right)+t=7t+2$. So $x=36n+23=36\left(7t+2\right)+23=252t+95$ with $t\in \mathbb{Z}$