How this calculation works
Start with every number satisfying the first condition. Choose its step multiplier so it satisfies the second. Coprime moduli guarantee one answer modulo their product.
x = r₁ + m₁k; solve m₁k ≡ r₂−r₁ (mod m₂)A worked example
x ≡ 2 (mod 3) and x ≡ 3 (mod 5) combine to x ≡ 8 (mod 15).
Good to know
This version handles exactly two coprime moduli. It rejects non-coprime inputs rather than claiming a generalized result. Each modulus is at most one billion; the exact combined answer may exceed JavaScript safe integers and is shown as text.
Reference: MIT OpenCourseWare: Theory of Numbers lecture notes (opens in a new tab) ↗
Combine two clocks into one
Suppose a number leaves remainder 2 after division by 3 and remainder 3 after division by 5. The first condition lets us write x = 2 + 3k. Substitute into the second: 3k ≡ 1 (mod 5), so k ≡ 2 and x ≡ 8 (mod 15).
Because 3 and 5 are coprime, every pair of remainders matches exactly one residue class modulo their product 15. Adding 15 to 8 preserves both conditions. The two verification rows make that claim inspectable.
The linear congruence calculator solves the coefficient step that arises while combining the two conditions.
Use a solver for the right shape
The linear congruence calculator solves a coefficient times one unknown modulo one modulus. This page instead combines two already stated residue conditions. Both use modular inverses internally, but their inputs describe different tasks.
You can apply this two-modulus result again with a third modulus if that third modulus is coprime to the combined modulus. The page itself deliberately stops at two conditions so every step stays visible.
Coprime is a real condition
Non-coprime moduli can sometimes have a compatible combined answer and sometimes conflict. That is a generalized CRT problem, not the claim implemented here. The input is rejected instead of giving a result under a theorem whose prerequisite failed.
Remainders are entered in their standard range from zero through modulus minus one. This makes the displayed smallest non-negative answer unambiguous. Large products are retained as exact whole-number text.
A few questions, answered
Must the moduli be prime?
No. They only need gcd(m₁,m₂)=1.
Is 8 the only integer answer in the example?
No. Every integer congruent to 8 modulo 15 also works.
Can I supply three remainders at once?
This page combines exactly two; a third coprime condition can be combined with the result in a second step.