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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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 |
|---|---|---|---|---|---|---|
| 757 | 899 | 0 | 757 | 0 | 1 | 0 |
| 899 | 757 | 1 | 142 | 1 | 0 | 1 |
| 757 | 142 | 5 | 47 | 0 | 1 | -5 |
| 142 | 47 | 3 | 1 | 1 | -5 | 16 |
| 47 | 1 | 47 | 0 | -5 | 16 | -757 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 113 | 158 | 0 | 113 | 0 | 1 | 0 |
| 158 | 113 | 1 | 45 | 1 | 0 | 1 |
| 113 | 45 | 2 | 23 | 0 | 1 | -2 |
| 45 | 23 | 1 | 22 | 1 | -2 | 3 |
| 23 | 22 | 1 | 1 | -2 | 3 | -5 |
| 22 | 1 | 22 | 0 | 3 | -5 | 113 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 691 | 78 | 8 | 67 | 0 | 1 | -8 |
| 78 | 67 | 1 | 11 | 1 | -8 | 9 |
| 67 | 11 | 6 | 1 | -8 | 9 | -62 |
| 11 | 1 | 11 | 0 | 9 | -62 | 691 |
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 ≡ 169 × 899-1 (mod 757) ≡ 169 × 16 (mod 757) ≡ 433 (mod 757)x ≡ 785 × 158-1 (mod 113) ≡ 785 × 108 (mod 113) ≡ 30 (mod 113)x ≡ 603 × 78-1 (mod 691) ≡ 603 × 629 (mod 691) ≡ 619 (mod 691)
So now we have the following equations of the form x ≡ a (mod m):
x ≡ 433 (mod 757)
x ≡ 30 (mod 113)
x ≡ 619 (mod 691)
Now the actual calculation
- Find the common modulus M
M = m1 × m2 × ... × mk = 757 × 113 × 691 = 59108831 - We calculate the numbers M1 to M3
M1=M/m1=59108831/757=78083,M2=M/m2=59108831/113=523087,M3=M/m3=59108831/691=85541 - 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 -196 mod 757 ≡ 561n b q r t1 t2 t3 757 78083 0 757 0 1 0 78083 757 103 112 1 0 1 757 112 6 85 0 1 -6 112 85 1 27 1 -6 7 85 27 3 4 -6 7 -27 27 4 6 3 7 -27 169 4 3 1 1 -27 169 -196 3 1 3 0 169 -196 757
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is 34 mod 113 ≡ 34n b q r t1 t2 t3 113 523087 0 113 0 1 0 523087 113 4629 10 1 0 1 113 10 11 3 0 1 -11 10 3 3 1 1 -11 34 3 1 3 0 -11 34 -113
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is -29 mod 691 ≡ 662n b q r t1 t2 t3 691 85541 0 691 0 1 0 85541 691 123 548 1 0 1 691 548 1 143 0 1 -1 548 143 3 119 1 -1 4 143 119 1 24 -1 4 -5 119 24 4 23 4 -5 24 24 23 1 1 -5 24 -29 23 1 23 0 24 -29 691
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
= (433 × 78083 × 561 +
30 × 523087 × 34 +
619 × 85541 × 662) mod 59108831
= 55402235 (mod 59108831)
So our answer is 55402235 (mod 59108831).
Verification
So we found that x ≡ 55402235
If this is correct, then the following statements (i.e. the original equations) are true:
899x (mod 757) ≡ 169 (mod 757)
158x (mod 113) ≡ 785 (mod 113)
78x (mod 691) ≡ 603 (mod 691)
Let's see whether that's indeed the case if we use x ≡ 55402235.