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Example

Given \(a \;\equiv\; 2\) (mod 5)
\(a \;\equiv\; 3\) (mod 11)
Find
\(a \;\equiv\; x\) (mod 55)

a1 = 2, a2 = 3
m1 = 11, m2 = 5
n1 = 5, n2 = 11
m1-1 mod n1 = 11-1 mod 5 = 1
m2-1 mod n2 = 5-1 mod 11 = 9
\(c_1 \;=\; m_1(m_1^{-1} \;{\rm mod}\; n_1)\) = 11(1) = 11
\(c_2 \;=\; m_2(m_2^{-1} \;{\rm mod}\; n_2)\) = 5(9) = 45


a
$\equiv$ 2*11 + 3*45 (mod 55)
$\equiv$ 22 + 135 (mod 55)
$\equiv$ 157 (mod 55)
$\equiv$ 47 (mod 55)



Thus, we can work in modulo n or modulo ni.


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