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.
This fills all textboxes with random numbers. If you fill in random numbers yourself, it is very likely that those numbers do not have a solution. To avoid disappointment, use this button instead! It only uses random numbers that do have a solution.
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.
Do you want to use more equations? Go ahead and use this button. It adds another row that you can fill in. Not sure what numbers to put in this newly added row? Use the random numbers button again!
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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 |
|---|---|---|---|---|---|---|
| 125 | 503 | 0 | 125 | 0 | 1 | 0 |
| 503 | 125 | 4 | 3 | 1 | 0 | 1 |
| 125 | 3 | 41 | 2 | 0 | 1 | -41 |
| 3 | 2 | 1 | 1 | 1 | -41 | 42 |
| 2 | 1 | 2 | 0 | -41 | 42 | -125 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 533 | 752 | 0 | 533 | 0 | 1 | 0 |
| 752 | 533 | 1 | 219 | 1 | 0 | 1 |
| 533 | 219 | 2 | 95 | 0 | 1 | -2 |
| 219 | 95 | 2 | 29 | 1 | -2 | 5 |
| 95 | 29 | 3 | 8 | -2 | 5 | -17 |
| 29 | 8 | 3 | 5 | 5 | -17 | 56 |
| 8 | 5 | 1 | 3 | -17 | 56 | -73 |
| 5 | 3 | 1 | 2 | 56 | -73 | 129 |
| 3 | 2 | 1 | 1 | -73 | 129 | -202 |
| 2 | 1 | 2 | 0 | 129 | -202 | 533 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 109 | 73 | 1 | 36 | 0 | 1 | -1 |
| 73 | 36 | 2 | 1 | 1 | -1 | 3 |
| 36 | 1 | 36 | 0 | -1 | 3 | -109 |
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 ≡ 387 × 503-1 (mod 125) ≡ 387 × 42 (mod 125) ≡ 4 (mod 125)x ≡ 346 × 752-1 (mod 533) ≡ 346 × 331 (mod 533) ≡ 464 (mod 533)x ≡ 676 × 73-1 (mod 109) ≡ 676 × 3 (mod 109) ≡ 66 (mod 109)
So now we have the following equations of the form x ≡ a (mod m):
x ≡ 4 (mod 125)
x ≡ 464 (mod 533)
x ≡ 66 (mod 109)
Now the actual calculation
- Find the common modulus M
M = m1 × m2 × ... × mk = 125 × 533 × 109 = 7262125 - We calculate the numbers M1 to M3
M1=M/m1=7262125/125=58097,M2=M/m2=7262125/533=13625,M3=M/m3=7262125/109=66625 - 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 58 mod 125 ≡ 58n b q r t1 t2 t3 125 58097 0 125 0 1 0 58097 125 464 97 1 0 1 125 97 1 28 0 1 -1 97 28 3 13 1 -1 4 28 13 2 2 -1 4 -9 13 2 6 1 4 -9 58 2 1 2 0 -9 58 -125
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is 183 mod 533 ≡ 183n b q r t1 t2 t3 533 13625 0 533 0 1 0 13625 533 25 300 1 0 1 533 300 1 233 0 1 -1 300 233 1 67 1 -1 2 233 67 3 32 -1 2 -7 67 32 2 3 2 -7 16 32 3 10 2 -7 16 -167 3 2 1 1 16 -167 183 2 1 2 0 -167 183 -533
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is 21 mod 109 ≡ 21n b q r t1 t2 t3 109 66625 0 109 0 1 0 66625 109 611 26 1 0 1 109 26 4 5 0 1 -4 26 5 5 1 1 -4 21 5 1 5 0 -4 21 -109
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
= (4 × 58097 × 58 +
464 × 13625 × 183 +
66 × 66625 × 21) mod 7262125
= 6399129 (mod 7262125)
So our answer is 6399129 (mod 7262125).
Verification
So we found that x ≡ 6399129
If this is correct, then the following statements (i.e. the original equations) are true:
503x (mod 125) ≡ 387 (mod 125)
752x (mod 533) ≡ 346 (mod 533)
73x (mod 109) ≡ 676 (mod 109)
Let's see whether that's indeed the case if we use x ≡ 6399129.