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I try to apply the chinese rest theorem. This is how I tried to do it. $$ \begin{cases} x \equiv 3(mod \quad 83)\\ x \equiv 3(mod \quad 47) \end{cases} $$ First I want to know if a solution exists, so I take $sgd(83,47)$. Applying Euklides algorithm.

$$ 83=47*1+36 \implies \\47 = 36*1 + 11 \implies \\36=11*3+3 \implies \\11=3*3 + 2 \implies \\3=2*1 + 1 \implies \\2 = 1*2 + 0 $$ So $sgd(83,47) = 1$, solution exist. Applying Euklides algorithm backwards to find $1=83u+47v$. $$ 1=3-2 \implies \\1=3-(11-3*3) \implies\\ 1=3*4-(47-36) \implies \\ ... \implies \\1=17(83)-30(47) $$ So $u=17$ and $v=-30$. We know that $x_{0} = a_{2}m_{1}u+a_{1}m_{2}v$, so $x_{0} = 3*83*17-3*47*30 \implies x_{0}=2823$.

Now $x$ should be $x=2823+(83*47)n \implies x=2823 + 3801n$.

So when $n=1$ the answer should be $x=6624$ but it is wrong, the answer is $x=3904$. Where have I done wrong?

Salviati
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2 Answers2

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Since $83 | x-3$ and $47|x-3$ we have $x-3=47l$ and $x-3=83k$ so $43l=83k$. Since $47$ and $83$ are prime the least candidate is $l=83$ and the result folows.

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Your answer: first, 83 and 47 are prime, so gcd(83,47) = 1.

Now, you get 1 = 83u + 47v and taking $x = 3*83u+3*47v$. It is OK, but why $x_0 = 3*83*17-3*47*30 = 3 \neq 2823.$

Short answer: $x-3 \equiv 0$ (mod 83) and $x-3\equiv 0$ (mod 47). Then $x-3 = 83*47 k = 3901k$. So, the lowest integer more than 100 satisfies equations is 3904.

GAVD
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  • Nice! Thank you. But can y maybe also point out where exactly it all went to hell with my first calculation? Because it looks like I did everything right, just took a long detour. – Salviati Sep 22 '17 at 10:25
  • Wow, that is what I get when using the calculator and not check if answer is feasible or not.. Thank you – Salviati Sep 22 '17 at 10:34