I have this equation $x \equiv (2003^{-1}) \mod 1511$ and am confused with how to try solving it. Since $2003^{-1} < 1511$, you can't use the extended Euclidean algorithm with regards to the inverse. So I'm looking for some tips on where to start with solving this equation. Thanks.
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solve $1511 x + 2003 y = 1.$ Then $2003y \equiv 1 \pmod{1511}.$ If you find $y> 1511,$ or $y < 0$ use $2003 (y \pm 1511) \equiv 1 \pmod {1511}$ – Will Jagy Nov 03 '20 at 23:04
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$2003\equiv492\bmod1511$ – J. W. Tanner Nov 03 '20 at 23:05
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The extended Euclidean algorithm works fine - see the links for that and many other ways to compute modular inverses – Bill Dubuque Nov 03 '20 at 23:55
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since $gcd(2003,1511)= 1$ it has one and only solution
$\Rightarrow $ $2003x\equiv 1mod1511$
($2003mod1511=492)$$\Rightarrow $
$492x\equiv1mod1511$$\Rightarrow $
find the inverse of $492$ in $mod1511$ and you will see its $777$ so
$777\cdot492x\equiv777mod1511 $ $\Rightarrow $
$x\equiv777mod1511$
領域展開
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I am maybe wrong here cause you propanly mean the inverse of 2003 , i will fix this later. – 領域展開 Nov 03 '20 at 23:54