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Given that $a$ and $b$ are relatively prime positive integers, and that $a$ is relatively prime to all the following primes: $p_1,p_2,...,p_n$ then the following congruence:$${a+bx}\equiv 1\pmod{ p_1\cdot p_2\cdot p_3 \cdot...\cdot p_n}$$ Has solution in x, why can we claim this?

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closed as off-topic by Carl Mummert, Namaste, Gibbs, Dave, Shailesh Dec 7 '18 at 2:29

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  • $\begingroup$ Something may be wrong. Is it $b$ that is coprime to all $p_i$? Is the modulus their product? $\endgroup$ – Bill Dubuque Dec 2 '18 at 17:38
  • $\begingroup$ @BillDubuque Oh, you're right, messed up the LaTeX, the modulus it's their product, I'll fix it immediately! $\endgroup$ – Spasoje Durovic Dec 2 '18 at 20:46
  • $\begingroup$ This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc. $\endgroup$ – Carl Mummert Dec 6 '18 at 17:48
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It is solvable $\iff \gcd(n,b)\mid 1\!-\!a,\,$ where $n = \prod p_i.\,$ Indeed

$\qquad\qquad \begin{align} \!\bmod n\!:\ \exists x\!:\ bx\equiv&\, 1\!-\!a\\[.3em] \iff \exists x,y\!:\ ny+bx =&\, 1\!-\!a\\[.3em] \iff \gcd(n,\,b)\,\ \mid\ &\ 1\!-\!a\ \ {\rm by\ Bezout}\end{align}$

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