Solve for $x$: $5x =1 \bmod12$
By hit and trial i see that $x=5$ satisfies the above. But what is the procedure for doind this
Thanks
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1Do you know the Euclidean algorithm? – user217285 Aug 01 '16 at 05:11
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@Nitin To find the G.C.D..yes – Gathdi Aug 01 '16 at 05:11
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See http://math.stackexchange.com/questions/407478/solving-a-linear-congruence – lab bhattacharjee Aug 01 '16 at 05:14
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So you know there exist integers $a,b$ such that $5a + 12b = 1$. What happens under mod 12? – user217285 Aug 01 '16 at 05:15
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In principle try to multiply mod $12$ by all numbers up to $11$. But anything divisible by $2$ or $3$ can't work, so don't bother with them. Then the first thing you really try works. – André Nicolas Aug 01 '16 at 05:47
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@Nitin $5a=1modn$ – Gathdi Aug 01 '16 at 08:05
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The solution is $x = 5^{-1} \pmod {12}$. Basically you want to calculate the modular inverse of $5 \pmod {12}$, which is represented as $5^{-1} \pmod{12}$. You can use the Extended Euclidean algorithm for that. But for small modulos, it's easier to just do trial and error. You want the unique number (modulo $12$) that when multiplied by $5$ will give you a result that is $1$ modulo $12$. You only have to test numbers in the range $1$ to $11$.
You can quite easily see that $(5)(5) = 25 \equiv 1 \pmod {12}$. So $5^{-1} \equiv 5 \pmod{12}$ and the solution is $x \equiv 5 \pmod{12}$.
Deepak
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1I agree that trial and error is the right way to do this, but it seems more efficient to test numbers of the form $12n+1$ until you find one divisible by $5$. (Efficient not only because in this example $n=2$ already works but mainly because I find it easier to mentally test divisibility by $5$ than by $12$.) – Andreas Blass Aug 01 '16 at 05:32
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@AndreasBlass Yes, this is true. In fact, I think I exploited this method somewhere on another question on this site, but I can't seem to find it. May have been a comment in response to someone else. – Deepak Aug 01 '16 at 05:50