Welcome to ChineseRemainderTheorem.com!
Use the calculator below to get a step-by-step calculation of the Chinese Remainder Theory. Just enter the numbers you would like and press "Calculate" .
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This is useful if you want your equations to be of the form x ≡ a (mod m) rather than bx ≡ a (mod m).
In that case, it can be especially useful after using the random numbers button.
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Are you ready to view a full step-by-step Chinese Remainder Theorem calculation for the numbers you have entered? Then use this button!
Want to know more?- Read about how to execute the Chinese Remainder Theorem
- Discover different ways to calculate multiplicative inverses
Transform the equations
You used one or more of the fields on the left, so your equations are of the form bx ≡ a mod m.
We want them to be of the form x ≡ a mod m, so we need to move the values on the left to the right side of the equation.
For a more detailed explanation about how this works, see this part of our page about how to execute the Chinese Remainder algorithm.
First, we calculate the inverses of the leftmost value on each row:
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 407 | 210 | 1 | 197 | 0 | 1 | -1 |
| 210 | 197 | 1 | 13 | 1 | -1 | 2 |
| 197 | 13 | 15 | 2 | -1 | 2 | -31 |
| 13 | 2 | 6 | 1 | 2 | -31 | 188 |
| 2 | 1 | 2 | 0 | -31 | 188 | -407 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 807 | 631 | 1 | 176 | 0 | 1 | -1 |
| 631 | 176 | 3 | 103 | 1 | -1 | 4 |
| 176 | 103 | 1 | 73 | -1 | 4 | -5 |
| 103 | 73 | 1 | 30 | 4 | -5 | 9 |
| 73 | 30 | 2 | 13 | -5 | 9 | -23 |
| 30 | 13 | 2 | 4 | 9 | -23 | 55 |
| 13 | 4 | 3 | 1 | -23 | 55 | -188 |
| 4 | 1 | 4 | 0 | 55 | -188 | 807 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 779 | 388 | 2 | 3 | 0 | 1 | -2 |
| 388 | 3 | 129 | 1 | 1 | -2 | 259 |
| 3 | 1 | 3 | 0 | -2 | 259 | -779 |
Source: ExtendedEuclideanAlgorithm.com
Click on any row to reveal a more detailed calculation of each multiplicative inverse.
Now that we now the inverses, let's move the leftmost value on each row to the right of the equation:
x ≡ 335 × 210-1 (mod 407) ≡ 335 × 188 (mod 407) ≡ 302 (mod 407)x ≡ 382 × 631-1 (mod 807) ≡ 382 × 619 (mod 807) ≡ 7 (mod 807)x ≡ 646 × 388-1 (mod 779) ≡ 646 × 259 (mod 779) ≡ 608 (mod 779)
So now we have the following equations of the form x ≡ a (mod m):
x ≡ 302 (mod 407)
x ≡ 7 (mod 807)
x ≡ 608 (mod 779)
Now the actual calculation
- Find the common modulus M
M = m1 × m2 × ... × mk = 407 × 807 × 779 = 255861771 - We calculate the numbers M1 to M3
M1=M/m1=255861771/407=628653,M2=M/m2=255861771/807=317053,M3=M/m3=255861771/779=328449 - We now calculate the modular multiplicative inverses M1-1 to M3-1
Have a look at the page that explains how to calculate modular multiplicative inverse.
Using, for example, the Extended Euclidean Algorithm, we will find that:
So our multiplicative inverse is 103 mod 407 ≡ 103n b q r t1 t2 t3 407 628653 0 407 0 1 0 628653 407 1544 245 1 0 1 407 245 1 162 0 1 -1 245 162 1 83 1 -1 2 162 83 1 79 -1 2 -3 83 79 1 4 2 -3 5 79 4 19 3 -3 5 -98 4 3 1 1 5 -98 103 3 1 3 0 -98 103 -407
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is -140 mod 807 ≡ 667n b q r t1 t2 t3 807 317053 0 807 0 1 0 317053 807 392 709 1 0 1 807 709 1 98 0 1 -1 709 98 7 23 1 -1 8 98 23 4 6 -1 8 -33 23 6 3 5 8 -33 107 6 5 1 1 -33 107 -140 5 1 5 0 107 -140 807
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is -62 mod 779 ≡ 717n b q r t1 t2 t3 779 328449 0 779 0 1 0 328449 779 421 490 1 0 1 779 490 1 289 0 1 -1 490 289 1 201 1 -1 2 289 201 1 88 -1 2 -3 201 88 2 25 2 -3 8 88 25 3 13 -3 8 -27 25 13 1 12 8 -27 35 13 12 1 1 -27 35 -62 12 1 12 0 35 -62 779
Source: ExtendedEuclideanAlgorithm.com
- Now we can calculate x with the equation we saw earlier
x = (a1 × M1 × M1-1 + a2 × M2 × M2-1 + ... + ak × Mk × Mk-1) mod M
= (302 × 628653 × 103 +
7 × 317053 × 667 +
608 × 328449 × 717) mod 255861771
= 210548728 (mod 255861771)
So our answer is 210548728 (mod 255861771).
Verification
So we found that x ≡ 210548728
If this is correct, then the following statements (i.e. the original equations) are true:
210x (mod 407) ≡ 335 (mod 407)
631x (mod 807) ≡ 382 (mod 807)
388x (mod 779) ≡ 646 (mod 779)
Let's see whether that's indeed the case if we use x ≡ 210548728.