Existence of solution to linear congruence [closed]

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?

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

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• Something may be wrong. Is it $b$ that is coprime to all $p_i$? Is the modulus their product? – Bill Dubuque Dec 2 '18 at 17:38
• @BillDubuque Oh, you're right, messed up the LaTeX, the modulus it's their product, I'll fix it immediately! – Spasoje Durovic Dec 2 '18 at 20:46
• 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. – Carl Mummert Dec 6 '18 at 17:48

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}