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 |
|---|---|---|---|---|---|---|
| 739 | 717 | 1 | 22 | 0 | 1 | -1 |
| 717 | 22 | 32 | 13 | 1 | -1 | 33 |
| 22 | 13 | 1 | 9 | -1 | 33 | -34 |
| 13 | 9 | 1 | 4 | 33 | -34 | 67 |
| 9 | 4 | 2 | 1 | -34 | 67 | -168 |
| 4 | 1 | 4 | 0 | 67 | -168 | 739 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 929 | 786 | 1 | 143 | 0 | 1 | -1 |
| 786 | 143 | 5 | 71 | 1 | -1 | 6 |
| 143 | 71 | 2 | 1 | -1 | 6 | -13 |
| 71 | 1 | 71 | 0 | 6 | -13 | 929 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 469 | 423 | 1 | 46 | 0 | 1 | -1 |
| 423 | 46 | 9 | 9 | 1 | -1 | 10 |
| 46 | 9 | 5 | 1 | -1 | 10 | -51 |
| 9 | 1 | 9 | 0 | 10 | -51 | 469 |
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 ≡ 572 × 717-1 (mod 739) ≡ 572 × 571 (mod 739) ≡ 713 (mod 739)x ≡ 949 × 786-1 (mod 929) ≡ 949 × 916 (mod 929) ≡ 669 (mod 929)x ≡ 570 × 423-1 (mod 469) ≡ 570 × 418 (mod 469) ≡ 8 (mod 469)
So now we have the following equations of the form x ≡ a (mod m):
x ≡ 713 (mod 739)
x ≡ 669 (mod 929)
x ≡ 8 (mod 469)
Now the actual calculation
- Find the common modulus M
M = m1 × m2 × ... × mk = 739 × 929 × 469 = 321983039 - We calculate the numbers M1 to M3
M1=M/m1=321983039/739=435701,M2=M/m2=321983039/929=346591,M3=M/m3=321983039/469=686531 - 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 -342 mod 739 ≡ 397n b q r t1 t2 t3 739 435701 0 739 0 1 0 435701 739 589 430 1 0 1 739 430 1 309 0 1 -1 430 309 1 121 1 -1 2 309 121 2 67 -1 2 -5 121 67 1 54 2 -5 7 67 54 1 13 -5 7 -12 54 13 4 2 7 -12 55 13 2 6 1 -12 55 -342 2 1 2 0 55 -342 739
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is 113 mod 929 ≡ 113n b q r t1 t2 t3 929 346591 0 929 0 1 0 346591 929 373 74 1 0 1 929 74 12 41 0 1 -12 74 41 1 33 1 -12 13 41 33 1 8 -12 13 -25 33 8 4 1 13 -25 113 8 1 8 0 -25 113 -929
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is 160 mod 469 ≡ 160n b q r t1 t2 t3 469 686531 0 469 0 1 0 686531 469 1463 384 1 0 1 469 384 1 85 0 1 -1 384 85 4 44 1 -1 5 85 44 1 41 -1 5 -6 44 41 1 3 5 -6 11 41 3 13 2 -6 11 -149 3 2 1 1 11 -149 160 2 1 2 0 -149 160 -469
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
= (713 × 435701 × 397 +
669 × 346591 × 113 +
8 × 686531 × 160) mod 321983039
= 43881055 (mod 321983039)
So our answer is 43881055 (mod 321983039).
Verification
So we found that x ≡ 43881055
If this is correct, then the following statements (i.e. the original equations) are true:
717x (mod 739) ≡ 572 (mod 739)
786x (mod 929) ≡ 949 (mod 929)
423x (mod 469) ≡ 570 (mod 469)
Let's see whether that's indeed the case if we use x ≡ 43881055.