Using Chinese Remainder Theorem to Combine Modular equations

Given N modular equations: A ? x₁mod(m₁) . . A ? x_nmod(m_n) Find x in the equation A ? xmod(m₁*m₂*m₃..*m_n) where m_i is prime, or a power of a prime, and i takes values from 1 to n. The input is given as two arrays, the first being an array containing values of each x_i, and the second array containing the set of values of each prime. m_i Output an integer for the value of x in the final equation. 

Examples : 

Consider the two equations
A ? 2mod(3)
A ? 3mod(5)
Input : 
2 3
3 5
Output : 
8

Consider the four equations,
A ? 3mod(4)
A ? 4mod(7)
A ? 1mod(9) (32)
A ? 0mod(11)
Input :
3 4 1 0
4 7 9 11
Output :
1243

Explanation : We aim to solve these equations two at a time. We take the first two equations, combine it, and use that result to combine with the third equation, and so on. The process of combining two equations is explained as follows, by taking example 2 for reference:

1. A ? 3mod(4) and A ? 4mod(7) are the two equations that we are provided with at first. Let the resulting equation be some A₀ ? x₀ mod(m₁ * m₂).
   • A₀ is given by m₁' * m₁ * x₀ + m₀' * m₀ * x₁ where m₁' = modular inverse of m₁ modulo m₀ and m₀' = modular inverse of m₀ modulo m₁
   • We can calculate modular inverse using extended euclidean algorithm.
   • We find x₀ to be A₀ mod (m₁ * m₂)
   • We get our new equation to be A ? 11mod(28), where A is 95
2. We now try to combine this with equation 3, and by a similar method, we get A ? 235mod(252), where A = 2503
3. And finally, on combining this with equation 4, we get A ? 1243mod(2772) where A = 59455 and x = 1243

We observe that 2772 is rightly equal to 4 * 7 * 9 * 11. We have thus found the value of x for the final equation. You can refer to Extended Euclidean Algorithm and Modular multiplicative inverse for extra information on these topics. 

Output:

1243

Time Complexity : O(l) ,where l is the size of remainder list.

Space Complexity : O(1) ,as we are not using any extra space.

This theorem and algorithm has excellent applications. One very useful application is in calculating ⁿC_r % m where m is not a prime number, and Lucas Theorem cannot be directly applied. In such a case, we can calculate the prime factors of m, and use the prime factors one by one as a modulus in our ⁿC_r % m equation which we can calculate using Lucas Theorem, and then combine the resulting equations together using the above shown Chinese Remainder Theorem.