 # The problem is asking to find c given that- c ≡ 13a(mod19) and the variable a ≡ 11 (mod19). Parker Bird 2022-07-17 Answered
The problem is asking to find c given that-
$c\equiv 13a\left(\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}19\right)$ and the variable $a\equiv 11\left(\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}19\right)$.
I've tried to solve this using algebra and using the theorems but I can't seem to work it out.
Any suggestions?
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Step 1
Well, there's no problem: $c\equiv 13×11=143\equiv 10\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}19\right).$
Step 2
Thus $c=10+19k,\phantom{\rule{.5em}{0ex}}k\in \mathbf{Z}$. The only solution between 0 and 18 is for $k=0$, i.e. $c=10$

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