Why do we only need to check at most m values when searching for a solution to a congruence modulo m in the iterative CRT method?
Conditions
- The step size of the current arithmetic progression is coprime to the target modulus m.
- The search is bounded by the modulus m.
Reasoning, step by step
- Observe that adding the step size repeatedly modulo m generates a sequence of remainders.
- Recognize that because the step size and m are coprime, this sequence cycles through all m possible residues exactly once.
- Conclude that if the target remainder is not found within m steps, the cycle has completed without hitting the target, so no solution exists.
Example
The video states, 'Because the step size (60) and the modulus (7) are coprime, these remainders form a complete cycle covering all residues from 0 to 6... you only need to check at most m values. If the target remainder hasn't appeared by then, no solution exists for that congruence.'
Common misconceptions
- Thinking the search could go on indefinitely.
- Assuming the cycle length depends on the starting value rather than the modulus and step size.
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The step size updates to the least common multiple (LCM) of the moduli already satisfied because adding the LCM to a current solution preserves the remainders for all those moduli. Any increment divisible by all the current moduli will not change the remainder modulo any of them, and the LCM is the smallest positive such increment.
Conditions: A number currently satisfies a set of congruences.; The new step size must preserve the remainders for all those congruences.
If two congruences x ≡ r (mod m) and x ≡ r (mod n) share the same remainder r, then is an immediate particular solution. The full combined solution set is then r plus multiples of the LCM of the two moduli: * LCM(m, n).
Conditions: The two congruences have the identical remainder r.; The moduli m and n are positive integers.
The method solves the congruences one by one by maintaining an arithmetic progression of candidate solutions. It starts with the sequence of numbers satisfying the first congruence, tests them against the second to find a common solution, and then updates the step size to the least common multiple (LCM) of the first two moduli.
Conditions: The unknown x is an integer.; The system consists of the congruences x ≡ 3 (mod 4), x ≡ 5 (mod 6), and x ≡ 2 (mod 5).
Instead of computing large terms in the arithmetic progression, the trick computes the sequence of remainders modulo the new target modulus. It starts with the current solution's remainder and repeatedly adds the step size's remainder modulo the target, stopping when the target remainder is reached.
Conditions: You have an arithmetic progression of candidates satisfying previous congruences.; You need to find the first term satisfying a new congruence x ≡ r (mod m).
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