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If the $m_1, ..., m_r$ are pairwise coprime, and if $a_1, \ldots, a_r$ are any integers, then system of $r$ conqruences

$$x \equiv a_ i \bmod m_ i \text{($ 1\le i \le r$)}$$

has a unique solution modulo $M= m_1 * m_2*....m_r$, which is given by

$$x=\Sigma_{i=1}^{r}\space a_iM_i y_i \bmod M,$$

where $M_i = \frac{M}{m_i}$ and $y_i =M_i^{-1} \bmod m_i$ for $ 1\le i \le r$.

Could you please explain CRT will give unique $x \bmod M$?

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    $\begingroup$ Because .... that's what the statement of CRT claims???? Go through the proof of CRT. That will tell you why the CRT is true. $\endgroup$ – fleablood Mar 28 at 19:55
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The existence of a unique solution follows directly by considering the kernel of the natural map $$ \mathbb Z \to \mathbb Z / m_1 \mathbb Z \times \cdots \times \mathbb Z / m_r \mathbb Z $$ given by $$ x \mapsto (x \bmod m_1, \dots, x \bmod m_r) $$

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Let x, y be integers both satisfying all those congruences.

Then $x\equiv a_i\equiv y\pmod {m_i}$ for every $i$.

Hence, $m_i$ divides $x-y$ for every $i$.

Since $m_1,\dots, m_r$ are pairwise coprime then...

Try to fill the gap.

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