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.
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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 |
|---|---|---|---|---|---|---|
| 31 | 253 | 0 | 31 | 0 | 1 | 0 |
| 253 | 31 | 8 | 5 | 1 | 0 | 1 |
| 31 | 5 | 6 | 1 | 0 | 1 | -6 |
| 5 | 1 | 5 | 0 | 1 | -6 | 31 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 169 | 773 | 0 | 169 | 0 | 1 | 0 |
| 773 | 169 | 4 | 97 | 1 | 0 | 1 |
| 169 | 97 | 1 | 72 | 0 | 1 | -1 |
| 97 | 72 | 1 | 25 | 1 | -1 | 2 |
| 72 | 25 | 2 | 22 | -1 | 2 | -5 |
| 25 | 22 | 1 | 3 | 2 | -5 | 7 |
| 22 | 3 | 7 | 1 | -5 | 7 | -54 |
| 3 | 1 | 3 | 0 | 7 | -54 | 169 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 536 | 587 | 0 | 536 | 0 | 1 | 0 |
| 587 | 536 | 1 | 51 | 1 | 0 | 1 |
| 536 | 51 | 10 | 26 | 0 | 1 | -10 |
| 51 | 26 | 1 | 25 | 1 | -10 | 11 |
| 26 | 25 | 1 | 1 | -10 | 11 | -21 |
| 25 | 1 | 25 | 0 | 11 | -21 | 536 |
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 ≡ 411 × 253-1 (mod 31) ≡ 411 × 25 (mod 31) ≡ 14 (mod 31)x ≡ 192 × 773-1 (mod 169) ≡ 192 × 115 (mod 169) ≡ 110 (mod 169)x ≡ 591 × 587-1 (mod 536) ≡ 591 × 515 (mod 536) ≡ 453 (mod 536)
So now we have the following equations of the form x ≡ a (mod m):
x ≡ 14 (mod 31)
x ≡ 110 (mod 169)
x ≡ 453 (mod 536)
Now the actual calculation
- Find the common modulus M
M = m1 × m2 × ... × mk = 31 × 169 × 536 = 2808104 - We calculate the numbers M1 to M3
M1=M/m1=2808104/31=90584,M2=M/m2=2808104/169=16616,M3=M/m3=2808104/536=5239 - 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 -15 mod 31 ≡ 16n b q r t1 t2 t3 31 90584 0 31 0 1 0 90584 31 2922 2 1 0 1 31 2 15 1 0 1 -15 2 1 2 0 1 -15 31
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is 72 mod 169 ≡ 72n b q r t1 t2 t3 169 16616 0 169 0 1 0 16616 169 98 54 1 0 1 169 54 3 7 0 1 -3 54 7 7 5 1 -3 22 7 5 1 2 -3 22 -25 5 2 2 1 22 -25 72 2 1 2 0 -25 72 -169
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is 31 mod 536 ≡ 31n b q r t1 t2 t3 536 5239 0 536 0 1 0 5239 536 9 415 1 0 1 536 415 1 121 0 1 -1 415 121 3 52 1 -1 4 121 52 2 17 -1 4 -9 52 17 3 1 4 -9 31 17 1 17 0 -9 31 -536
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
= (14 × 90584 × 16 +
110 × 16616 × 72 +
453 × 5239 × 31) mod 2808104
= 812493 (mod 2808104)
So our answer is 812493 (mod 2808104).
Verification
So we found that x ≡ 812493
If this is correct, then the following statements (i.e. the original equations) are true:
253x (mod 31) ≡ 411 (mod 31)
773x (mod 169) ≡ 192 (mod 169)
587x (mod 536) ≡ 591 (mod 536)
Let's see whether that's indeed the case if we use x ≡ 812493.