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I have mutually coprime numbers $p_1,\ldots,p_n$ and three collections of numbers $i_1,\ldots,i_n$, $j_1,\ldots,j_n$, $r_1,\ldots,r_n$ such that $0 \leqslant i_k, r_k < p_k$ and $0 < j_k < p_k$ for any $k=1,\ldots,n$. Is it true that there exists a number $N \geqslant 1$ such that $$ i_1 + N j_1 = r_1 \mod p_1, \\ \cdots \\ i_n + N j_n = r_n \mod p_n ? $$ I tried to adapt the Chinese Remainder Theorem and to use the Little Fermat Theorem but without success. Please help me with it.

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  • $\begingroup$ The misleading part is that you use $p$ to denote an ordinary number. $\endgroup$ – J.H. Mar 18 '13 at 23:15
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$\rm\: j N \equiv r\!-\!i\ \ (mod\ p)\:$ has a solution $\rm\,N\,$ iff $\rm\:d = (j,p)\mid r\!-\!i.\:$ If so, then $\rm\:N\equiv (r-i)/j\ \ (mod\ p/d).\:$ Thus if every congruence is so solvable then the system will have a solution, computable by CRT

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