How do I reduce the following and remove the $11$ from LHS, $$ 11x \equiv 4 \ \ \text{mod }50 $$
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1Do you know when is a number invertible $\pmod{n}$? That is, for which $a$ one has some $b$ such that $ab = 1 \pmod{n}$? – qualcuno Feb 08 '21 at 10:12
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1Observe that $11 \times 9 \equiv -1$ (mod 50). – João Víctor Melo Feb 08 '21 at 10:15
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$11x = 50k +4$
Multiplying the equation by 9,
$99x = 450k + 36$
$-x = 50(9k - 2x) + 36$
$x = 50(2x - 9k -1) + 14$
$x \equiv 14 \pmod {50}$
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Variant using the particularity of multiplication by $11$.
Generally $11\times \overline{ab}=\overline{asb}=100a+10s+b\, $ where $\, s=a+b$ (provided $s<10$).
Notice that when $s=5$ we have $\begin{cases}100a\equiv 0\pmod {50}\\10s\equiv 0\pmod{50}\\11\times \overline{ab}\equiv b\pmod{50}\end{cases}$
Since we want $b=4$ then $a=1$ is forced and $$11\times 14\equiv 4\pmod{50}$$
Note that it works too for finding the inverse of $11$ (this time $b=1$ and $a=4$), thus $11^{-1}\equiv 41\pmod{50}$.
zwim
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