Chinese Remainder Theorem Calculator

When working with systems of congruences, the Chinese Remainder Theorem provides a direct way to find solutions. This theorem applies to problems where you need to solve for a number that leaves specific remainders when divided by different moduli, as long as those moduli are pairwise coprime. Use this Chinese Remainder Theorem Calculator to input your equations and get the unique solution modulo the product of the moduli.

Start by entering the number of equations, then provide the remainders (a values) and moduli (n values). The tool checks if the moduli are coprime and handles the calculations step by step, showing the result like x ≡ some value (mod N). If inputs are invalid, it displays an error message to guide corrections.

Understanding Modulo Operations and Congruences

Modulo operations form the foundation for solving congruence systems. The modulo operation finds the remainder after division. For any integers a and n (where n > 0), a mod n is the remainder r such that 0 ≤ r < n and a = q * n + r for some integer q.

For example, 17 mod 5 = 2 because 17 = 3 * 5 + 2. This remainder tells you how much is left over.

Congruences express this idea symbolically: a ≡ b (mod n) means a and b have the same remainder when divided by n, or equivalently, n divides (a – b). You can add, subtract, or multiply both sides of a congruence by the same integer without changing its validity, but division requires care—only if the divisor is coprime with the modulus.

In practice, congruences appear in scheduling, cryptography, and puzzles. Suppose you need a number x that is 2 more than a multiple of 3 and 1 more than a multiple of 4. That’s x ≡ 2 (mod 3) and x ≡ 1 (mod 4). The Chinese Remainder Theorem helps combine these into one solution.

To solve such problems manually or verify calculator results, first ensure the moduli are pairwise coprime (their greatest common divisor, or gcd, is 1 for every pair). If not, no unique solution exists under the theorem.

Applying the Euclidean Algorithm for GCD

The Euclidean algorithm computes the gcd of two numbers efficiently, which is key for checking coprimality in the Chinese Remainder Theorem.

To find gcd(x, y) where x > y:

1. Divide x by y to get quotient k and remainder r: x = k * y + r.
2. Replace x with y and y with r.
3. Repeat until r = 0. The last non-zero remainder is the gcd.

Example: gcd(1785, 546).

• 1785 = 3 * 546 + 147
• 546 = 3 * 147 + 105
• 147 = 1 * 105 + 42
• 105 = 2 * 42 + 21
• 42 = 2 * 21 + 0

So gcd = 21.

This process not only finds the gcd but also generates equations used in the next step. If gcd is not 1 for any pair of moduli in your system, adjust the problem or note that the theorem doesn’t apply directly.

In the Chinese Remainder Theorem Calculator, it automatically verifies pairwise coprimality. If moduli like 4 and 6 (gcd=2) are entered, it shows an error: “Moduli must be pairwise coprime.” This prevents incorrect solutions.

Using Bézout’s Identity for Modular Inverses

Bézout’s identity states that for any integers a and b with gcd(a, b) = d, there exist integers k and l such that k * a + l * b = d.

When d = 1 (a and b coprime), this gives the modular inverse: l is the inverse of b modulo a, since l * b ≡ 1 (mod a).

To find k and l, use the extended Euclidean algorithm, which back-substitutes the remainders from the gcd steps.

From the earlier example with 1785 and 546 (gcd=21):

Start from the penultimate equation and substitute backward:

21 = 105 – 2 * 42

42 = 147 – 1 * 105 → 21 = 105 – 2*(147 – 1105) = 3105 – 2*147

105 = 546 – 3147 → 21 = 3(546 – 3147) – 2147 = 3546 – 11147

147 = 1785 – 3546 → 21 = 3546 – 11*(1785 – 3546) = 36546 – 11*1785

So k = -11, l = 36 (signs can be negative; adjust by adding multiples if needed).

In Chinese Remainder Theorem applications, you need inverses for each partial modulus. For a system with moduli n1, n2, …, nk, compute N = product of all ni, then mi = N / ni, and find the inverse yi of mi modulo ni (yi * mi ≡ 1 mod ni) using Bézout’s method.

The calculator handles these inverses internally, saving time on manual back-substitution, especially for larger numbers.

The Chinese Remainder Theorem: Step-by-Step Solution

The Chinese Remainder Theorem guarantees a unique solution modulo N (product of pairwise coprime moduli) for a system:

x ≡ a1 (mod n1)

x ≡ a2 (mod n2)

…

x ≡ ak (mod nk)

Steps to solve:

1. Compute N = n1 * n2 * … * nk.
2. For each i, mi = N / ni.
3. Find yi such that yi * mi ≡ 1 (mod ni) using Bézout’s identity.
4. Compute x = sum over i (ai * mi * yi) mod N.

This x satisfies all congruences.

For negative intermediate results, add multiples of N to get a positive equivalent.

The theorem’s power lies in combining separate congruences into one. In computing, it’s used in RSA cryptography for efficient modular arithmetic with large primes.

If moduli aren’t coprime, solutions may exist but aren’t unique modulo N. For example, x ≡ 1 (mod 2), x ≡ 1 (mod 4) works (x ≡ 1 mod 4), but x ≡ 1 (mod 2), x ≡ 0 (mod 4) has no solution since 1 ≠ 0 mod 2.

The Chinese Remainder Theorem Calculator implements these steps precisely. Select up to 6 equations from the dropdown, input ai and ni, and click calculate. It outputs the solution or errors like “No modular inverse exists” if coprimality fails.

Example: Solving a Real-World Congruence System

Consider a problem: Find x such that:

x ≡ 1 (mod 3)

x ≡ 2 (mod 4)

x ≡ 3 (mod 5)

This models scenarios like cyclic scheduling where remainders represent offsets.

First, check coprimality: gcd(3,4)=1, gcd(3,5)=1, gcd(4,5)=1. Good.

N = 345 = 60

m1 = 60/3 = 20

m2 = 60/4 = 15

m3 = 60/5 = 12

Now inverses:

For y1: y1 * 20 ≡ 1 (mod 3). 20 mod 3 = 2, so y1 * 2 ≡ 1 mod 3. Try y1=2: 4 ≡ 1 mod 3. Yes.

For y2: y2 * 15 ≡ 1 (mod 4). 15 mod 4 = 3, y2 * 3 ≡ 1 mod 4. y2=3: 9 ≡ 1 mod 4. Yes.

For y3: y3 * 12 ≡ 1 (mod 5). 12 mod 5 = 2, y3 * 2 ≡ 1 mod 5. y3=3: 6 ≡ 1 mod 5. Yes.

x = 1202 + 2153 + 3123 = 40 + 90 + 108 = 238

238 mod 60 = 238 – 3*60 = 238 – 180 = 58

Verify: 58 / 3 = 19*3 + 1 → remainder 1

58 / 4 = 14*4 + 2 → 2

58 / 5 = 11*5 + 3 → 3

Perfect.

In the calculator, input 3 equations, a1=1 n1=3, a2=2 n2=4, a3=3 n3=5. Result: x ≡ 58 (mod 60).

For more equations, say add x ≡ 4 (mod 7). New N=60*7=420. Recalculate mis and yis. The tool scales up easily.

Handling Edge Cases and Errors

Common issues:

• Non-coprime moduli: Solution may not exist or be unique. Example: x ≡ 0 (mod 2), x ≡ 1 (mod 4). 4 divides x-1, but x even → x-1 odd, contradiction. Calculator flags this.
• Negative remainders: Adjust by adding ni. -1 mod 5 = 4, since -1 + 5=4.
• Zero modulus: Invalid; moduli must be positive integers >1 typically.
• Large numbers: Calculator handles big integers, but manual computation needs care with overflows.

If ai >= ni, reduce ai mod ni first.

For non-unique cases (non-coprime but consistent), use general methods like successive substitution.

Advanced Applications in Number Theory

Beyond basics, the Chinese Remainder Theorem enables ring isomorphisms in abstract algebra, mapping Z/NZ to product of Z/niZ when ni coprime.

In cryptography, it speeds up computations by breaking large moduli into smaller coprime factors.

For programming, implement CRT to solve Diophantine systems or hash functions.

Example with four equations:

x ≡ 0 (mod 2)

x ≡ 1 (mod 3)

x ≡ 5 (mod 7)

x ≡ 3 (mod 11)

Coprime check: All pairs gcd=1.

N=237*11=462

m1=231, y1: 231 mod 2=1, inverse 1.

m2=154, y2: 154 mod 3=1, inverse 1.

m3=66, y3: 66 mod 7=3, 3*y3≡1 mod7 → y3=5 (15≡1 mod7).

m4=42, y4: 42 mod11=9, 9*y4≡1 mod11 → y4=5 (45≡1 mod11).

x=02311 +11541 +5665 +3425 =0+154+1650+630=2434

2434 mod462: 2434-5*462=2434-2310=124

Verify each congruence.

The calculator processes this quickly—select 4 equations and input.

Tips for Using the Calculator Effectively

• Start with small systems to test.
• Ensure ai < ni by reducing if needed.
• For learning, compute manually first, then verify.
• If error, check coprimality with gcd tool mentally.
• Extend to non-coprime by solving pairwise if consistent.

This Chinese Remainder Theorem Calculator simplifies solving congruence systems, from basic puzzles to advanced math. Input your data, get results, and understand the process through these explanations.