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Chinese Remainder Theorem (33.27 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, $\ldots$ , (a_k + b_k) mod n_k)
(a-b) mod n $\leftrightarrow$ ((a₁ - b₁) mod n₁, $\ldots$ , (a_k - b_k) mod n_k)
ab mod n $\leftrightarrow$ (a₁ b₁ mod n₁, $\ldots$ , a_k b_k mod n_k).

From $a \;\longrightarrow\; (a_1, \cdots, a_k)$
(a mod n_i, $\cdots$ , a mod n_k)

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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