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Chinese Remainder Theorem (31.27 cont.)

If \(a \;\leftrightarrow\; (a_1, a_2, \cdots, a_k)\) and \(b \;\leftrightarrow\; (b_1, b_2, \cdots, b_k)\)

Then
(a+b) mod n
$\leftrightarrow$ ((a1 + b1) mod n1, $\ldots$, (ak + bk) mod nk)
(a-b) mod n
$\leftrightarrow$ ((a1 - b1) mod n1, $\ldots$, (ak - bk) mod nk)
ab mod n
$\leftrightarrow$ (a1 b1 mod n1, $\ldots$, ak bk mod nk).



From \(a \;\longrightarrow\; (a_1, \cdots, a_k)\)
(a mod
ni, $\cdots$, a mod nk)



From \(a_1, \cdots, a_k \;\longrightarrow\; a\)
\(m_i \;=\; n/n_i\) for i = 1, $\ldots$, k
\(c_i \;=\; m_i (m_i^{-1} \;{\rm mod}\; n_i)\)
\(a \;\equiv\; (a_1 c_1 \;+\; \cdots \;+\; a_k c_k)\) (mod n)


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