$6247x \equiv 1139 \pmod{9461}$, where $9461$ is a prime number
I just started learning modular equations and I have no idea how to solve this advanced questions.
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Martin Sleziak
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Matthew
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Do you know Euclid's division algorithm to find the gcd of two numbers ? – Peter Oct 16 '16 at 09:26
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yes. Use it to find the gcd of 6247 and 1139? – Matthew Oct 16 '16 at 09:37
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No, the gcd of $6247$ and $9461$, which is $1$ in this case because $9461$ is a prime. The extended algorithm (working the equations backwards) finds integers $a,b$ with $6247a+9461b=1$. – Peter Oct 16 '16 at 09:38
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Then, you know that $a$ is a multiplicative inverse of $6247$ in the ring $\mathbb Z_{9461}$ (which is even a field because $9461$ is prime). – Peter Oct 16 '16 at 09:41
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$62471934 + 9461 -1277 = 1$ So, $a$ is 1934 which is the multiplicative inverse of 6247. – Matthew Oct 16 '16 at 09:52
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2http://www.cse.cuhk.edu.hk/~andrejb/engg2440/hw/16H03.pdf <--- Remember to give credit to this question. – Asaf Karagila Oct 16 '16 at 09:53
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So, $1934$ is the multiplicative inverse. Multiply it with $1139$ and reduce modulo $9461$. The result is $7874$. – Peter Oct 16 '16 at 09:54
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1@Matthew Why the deletion? – Did Oct 16 '16 at 15:13
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A convenient method to compute modular inverses is this version of the extended Euclidean algorithm. – Bill Dubuque Oct 21 '16 at 02:24
1 Answers
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I will show the method step by step for a much easier example :
$$33x\equiv 41\mod 101$$
$$101=3\cdot 33+2$$
$$33=16\cdot 2+1$$
$$2=2\cdot 1+0$$
Therefore, $gcd(33,101)=1$ and the extended algorithm gives
$$1=33-16\cdot 2=33-16\cdot (101-3\cdot 33)=49\cdot 33-16\cdot 101$$
Therefore $33^{-1}=49$ in $\mathbb Z_{101}$
So, the solution is $$x=41\cdot 49=2009\equiv 90\mod 101$$
If you get a negative number $a$ , you can choose $p+a$ instead to get a positive solution. (Here $p=101$ , so if the result would have been $-40$, we could have used $-40+101=61$ instead).
The modulus $p$ need not be a prime number. The method works always , if the $gcd$ is $1$.
Peter
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