# Solve the following linea congruence: 17x congruence 3(mod 210)

Solve the following linea congruence: 17x congruence 3(mod 210)
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Step 1
Given:
$17x\equiv 3\left(\text{mod}210\right)$
Therefore,
$x={17}^{-1}\cdot 3\left(\text{mod}210\right)$
Find ${17}^{-1}\left(\text{mod}210\right)$
$210=17×12+6$
$17=6×2+5$
$6=5×1+1$
Trace the steps backward.
$1=6-5$
$=6-\left(17-6×2\right)$
$=3×6-17$
$=3\left(210-17×12\right)-17$
$=3×210-36×17-17$
$=3×210-37×17$
Thus, ${17}^{-1}\left(\text{mod}210\right)=-37\left(\text{mod}210\right)=173\left(\text{mod}210\right)$
Step 2
Hence,
$x\equiv {17}^{-1}\cdot 3\left(\text{mod}210\right)$
$\equiv 173\cdot 3\left(\text{mod}210\right)$
$\equiv 519\left(\text{mod}210\right)$
$\equiv 99\left(\text{mod}210\right)$
Thus, $x\equiv 99\left(\text{mod}210\right)$
Therefore, $17×99=1683\equiv 3\left(\text{mod}210\right)$
Step 3
Result:
Thus, $x\equiv 99\left(\text{mod}210\right)$
Therefore, $17×99=1683\equiv 3\left(\text{mod}210\right)$