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" .
This removes all numbers from the textboxes, such that you can fill in your own.
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This button is similar to the "Clear everything" button, but only clears the left column.
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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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 |
|---|---|---|---|---|---|---|
| 619 | 2 | 309 | 1 | 0 | 1 | -309 |
| 2 | 1 | 2 | 0 | 1 | -309 | 619 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 571 | 144 | 3 | 139 | 0 | 1 | -3 |
| 144 | 139 | 1 | 5 | 1 | -3 | 4 |
| 139 | 5 | 27 | 4 | -3 | 4 | -111 |
| 5 | 4 | 1 | 1 | 4 | -111 | 115 |
| 4 | 1 | 4 | 0 | -111 | 115 | -571 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 191 | 33 | 5 | 26 | 0 | 1 | -5 |
| 33 | 26 | 1 | 7 | 1 | -5 | 6 |
| 26 | 7 | 3 | 5 | -5 | 6 | -23 |
| 7 | 5 | 1 | 2 | 6 | -23 | 29 |
| 5 | 2 | 2 | 1 | -23 | 29 | -81 |
| 2 | 1 | 2 | 0 | 29 | -81 | 191 |
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 ≡ 350 × 2-1 (mod 619) ≡ 350 × 310 (mod 619) ≡ 175 (mod 619)x ≡ 342 × 144-1 (mod 571) ≡ 342 × 115 (mod 571) ≡ 502 (mod 571)x ≡ 297 × 33-1 (mod 191) ≡ 297 × 110 (mod 191) ≡ 9 (mod 191)
So now we have the following equations of the form x ≡ a (mod m):
x ≡ 175 (mod 619)
x ≡ 502 (mod 571)
x ≡ 9 (mod 191)
Now the actual calculation
- Find the common modulus M
M = m1 × m2 × ... × mk = 619 × 571 × 191 = 67508759 - We calculate the numbers M1 to M3
M1=M/m1=67508759/619=109061,M2=M/m2=67508759/571=118229,M3=M/m3=67508759/191=353449 - 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 -164 mod 619 ≡ 455n b q r t1 t2 t3 619 109061 0 619 0 1 0 109061 619 176 117 1 0 1 619 117 5 34 0 1 -5 117 34 3 15 1 -5 16 34 15 2 4 -5 16 -37 15 4 3 3 16 -37 127 4 3 1 1 -37 127 -164 3 1 3 0 127 -164 619
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is 232 mod 571 ≡ 232n b q r t1 t2 t3 571 118229 0 571 0 1 0 118229 571 207 32 1 0 1 571 32 17 27 0 1 -17 32 27 1 5 1 -17 18 27 5 5 2 -17 18 -107 5 2 2 1 18 -107 232 2 1 2 0 -107 232 -571
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is -27 mod 191 ≡ 164n b q r t1 t2 t3 191 353449 0 191 0 1 0 353449 191 1850 99 1 0 1 191 99 1 92 0 1 -1 99 92 1 7 1 -1 2 92 7 13 1 -1 2 -27 7 1 7 0 2 -27 191
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
= (175 × 109061 × 455 +
502 × 118229 × 232 +
9 × 353449 × 164) mod 67508759
= 22117045 (mod 67508759)
So our answer is 22117045 (mod 67508759).
Verification
So we found that x ≡ 22117045
If this is correct, then the following statements (i.e. the original equations) are true:
2x (mod 619) ≡ 350 (mod 619)
144x (mod 571) ≡ 342 (mod 571)
33x (mod 191) ≡ 297 (mod 191)
Let's see whether that's indeed the case if we use x ≡ 22117045.