Finding The Solution With Basic Algebra and Modular Arithmetic
For example, consider the problem of finding an integer x such that
A brute-force approach converts these congruences into sets and writes the elements out to the product of 3×4×5 = 60 (the solutions modulo 60 for each congruence):
- x ∈ {2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, …}
- x ∈ {3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, …}
- x ∈ {1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, …}.
To find an x that satisfies all three congruences, intersect the three sets to get:
- x ∈ {11, …}.
Which can be expressed as
Another way to find a solution is with basic algebra, modular arithmetic, and stepwise substitution.
We start by translating these equivalences into equations for some t, s, and u:
- Equation 1: x = 2 + 3 × t (mod 3)
- Equation 2: x = 3 + 4 × s (mod 4)
- Equation 3: x = 1 + 5 × u (mod 5).
Start by substituting the x from equation 1 into equivalence 2: 2 + 3 × t = 3 (mod 4), hence 3 × t = 1 (mod 4), or t = (1/3) (mod 4) = 3 (mod 4), meaning that t = 3 + 4 × s for integer s.
Plug t into equation 1: x = 2 + 3 × t (mod 3) = 2 + 3 × (3 + 4 × s) (mod 3) = 11 + 12 × s (mod 3).
Plug this x into equivalence 3: 11 + 12 × s = 1 (mod 5). Casting out 5s, we get 1 + 2 × s = 1 (mod 5), or 2 × s = 0 (mod 5), meaning that s = 0 + 5 × u for integer u.
Finally, x = 11 + 12 × s = 11 + 12 × (5 × u) = 11 + (60 × u). Since 60 = lcm(3, 4, 5), we have solutions 11, 71, 131, 191, …
Read more about this topic: Chinese Remainder Theorem
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