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 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 |
|---|---|---|---|---|---|---|
| 856 | 517 | 1 | 339 | 0 | 1 | -1 |
| 517 | 339 | 1 | 178 | 1 | -1 | 2 |
| 339 | 178 | 1 | 161 | -1 | 2 | -3 |
| 178 | 161 | 1 | 17 | 2 | -3 | 5 |
| 161 | 17 | 9 | 8 | -3 | 5 | -48 |
| 17 | 8 | 2 | 1 | 5 | -48 | 101 |
| 8 | 1 | 8 | 0 | -48 | 101 | -856 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 937 | 886 | 1 | 51 | 0 | 1 | -1 |
| 886 | 51 | 17 | 19 | 1 | -1 | 18 |
| 51 | 19 | 2 | 13 | -1 | 18 | -37 |
| 19 | 13 | 1 | 6 | 18 | -37 | 55 |
| 13 | 6 | 2 | 1 | -37 | 55 | -147 |
| 6 | 1 | 6 | 0 | 55 | -147 | 937 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 293 | 4 | 73 | 1 | 0 | 1 | -73 |
| 4 | 1 | 4 | 0 | 1 | -73 | 293 |
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 ≡ 998 × 517-1 (mod 856) ≡ 998 × 101 (mod 856) ≡ 646 (mod 856)x ≡ 191 × 886-1 (mod 937) ≡ 191 × 790 (mod 937) ≡ 33 (mod 937)x ≡ 890 × 4-1 (mod 293) ≡ 890 × 220 (mod 293) ≡ 76 (mod 293)
So now we have the following equations of the form x ≡ a (mod m):
x ≡ 646 (mod 856)
x ≡ 33 (mod 937)
x ≡ 76 (mod 293)
Now the actual calculation
- Find the common modulus M
M = m1 × m2 × ... × mk = 856 × 937 × 293 = 235007096 - We calculate the numbers M1 to M3
M1=M/m1=235007096/856=274541,M2=M/m2=235007096/937=250808,M3=M/m3=235007096/293=802072 - 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 -51 mod 856 ≡ 805n b q r t1 t2 t3 856 274541 0 856 0 1 0 274541 856 320 621 1 0 1 856 621 1 235 0 1 -1 621 235 2 151 1 -1 3 235 151 1 84 -1 3 -4 151 84 1 67 3 -4 7 84 67 1 17 -4 7 -11 67 17 3 16 7 -11 40 17 16 1 1 -11 40 -51 16 1 16 0 40 -51 856
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is -216 mod 937 ≡ 721n b q r t1 t2 t3 937 250808 0 937 0 1 0 250808 937 267 629 1 0 1 937 629 1 308 0 1 -1 629 308 2 13 1 -1 3 308 13 23 9 -1 3 -70 13 9 1 4 3 -70 73 9 4 2 1 -70 73 -216 4 1 4 0 73 -216 937
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is 85 mod 293 ≡ 85n b q r t1 t2 t3 293 802072 0 293 0 1 0 802072 293 2737 131 1 0 1 293 131 2 31 0 1 -2 131 31 4 7 1 -2 9 31 7 4 3 -2 9 -38 7 3 2 1 9 -38 85 3 1 3 0 -38 85 -293
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
= (646 × 274541 × 805 +
33 × 250808 × 721 +
76 × 802072 × 85) mod 235007096
= 223775310 (mod 235007096)
So our answer is 223775310 (mod 235007096).
Verification
So we found that x ≡ 223775310
If this is correct, then the following statements (i.e. the original equations) are true:
517x (mod 856) ≡ 998 (mod 856)
886x (mod 937) ≡ 191 (mod 937)
4x (mod 293) ≡ 890 (mod 293)
Let's see whether that's indeed the case if we use x ≡ 223775310.