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 |
|---|---|---|---|---|---|---|
| 319 | 849 | 0 | 319 | 0 | 1 | 0 |
| 849 | 319 | 2 | 211 | 1 | 0 | 1 |
| 319 | 211 | 1 | 108 | 0 | 1 | -1 |
| 211 | 108 | 1 | 103 | 1 | -1 | 2 |
| 108 | 103 | 1 | 5 | -1 | 2 | -3 |
| 103 | 5 | 20 | 3 | 2 | -3 | 62 |
| 5 | 3 | 1 | 2 | -3 | 62 | -65 |
| 3 | 2 | 1 | 1 | 62 | -65 | 127 |
| 2 | 1 | 2 | 0 | -65 | 127 | -319 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 211 | 529 | 0 | 211 | 0 | 1 | 0 |
| 529 | 211 | 2 | 107 | 1 | 0 | 1 |
| 211 | 107 | 1 | 104 | 0 | 1 | -1 |
| 107 | 104 | 1 | 3 | 1 | -1 | 2 |
| 104 | 3 | 34 | 2 | -1 | 2 | -69 |
| 3 | 2 | 1 | 1 | 2 | -69 | 71 |
| 2 | 1 | 2 | 0 | -69 | 71 | -211 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 225 | 403 | 0 | 225 | 0 | 1 | 0 |
| 403 | 225 | 1 | 178 | 1 | 0 | 1 |
| 225 | 178 | 1 | 47 | 0 | 1 | -1 |
| 178 | 47 | 3 | 37 | 1 | -1 | 4 |
| 47 | 37 | 1 | 10 | -1 | 4 | -5 |
| 37 | 10 | 3 | 7 | 4 | -5 | 19 |
| 10 | 7 | 1 | 3 | -5 | 19 | -24 |
| 7 | 3 | 2 | 1 | 19 | -24 | 67 |
| 3 | 1 | 3 | 0 | -24 | 67 | -225 |
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 ≡ 21 × 849-1 (mod 319) ≡ 21 × 127 (mod 319) ≡ 115 (mod 319)x ≡ 472 × 529-1 (mod 211) ≡ 472 × 71 (mod 211) ≡ 174 (mod 211)x ≡ 257 × 403-1 (mod 225) ≡ 257 × 67 (mod 225) ≡ 119 (mod 225)
So now we have the following equations of the form x ≡ a (mod m):
x ≡ 115 (mod 319)
x ≡ 174 (mod 211)
x ≡ 119 (mod 225)
Now the actual calculation
- Find the common modulus M
M = m1 × m2 × ... × mk = 319 × 211 × 225 = 15144525 - We calculate the numbers M1 to M3
M1=M/m1=15144525/319=47475,M2=M/m2=15144525/211=71775,M3=M/m3=15144525/225=67309 - 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 131 mod 319 ≡ 131n b q r t1 t2 t3 319 47475 0 319 0 1 0 47475 319 148 263 1 0 1 319 263 1 56 0 1 -1 263 56 4 39 1 -1 5 56 39 1 17 -1 5 -6 39 17 2 5 5 -6 17 17 5 3 2 -6 17 -57 5 2 2 1 17 -57 131 2 1 2 0 -57 131 -319
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is -6 mod 211 ≡ 205n b q r t1 t2 t3 211 71775 0 211 0 1 0 71775 211 340 35 1 0 1 211 35 6 1 0 1 -6 35 1 35 0 1 -6 211
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is -86 mod 225 ≡ 139n b q r t1 t2 t3 225 67309 0 225 0 1 0 67309 225 299 34 1 0 1 225 34 6 21 0 1 -6 34 21 1 13 1 -6 7 21 13 1 8 -6 7 -13 13 8 1 5 7 -13 20 8 5 1 3 -13 20 -33 5 3 1 2 20 -33 53 3 2 1 1 -33 53 -86 2 1 2 0 53 -86 225
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
= (115 × 47475 × 131 +
174 × 71775 × 205 +
119 × 67309 × 139) mod 15144525
= 12015569 (mod 15144525)
So our answer is 12015569 (mod 15144525).
Verification
So we found that x ≡ 12015569
If this is correct, then the following statements (i.e. the original equations) are true:
849x (mod 319) ≡ 21 (mod 319)
529x (mod 211) ≡ 472 (mod 211)
403x (mod 225) ≡ 257 (mod 225)
Let's see whether that's indeed the case if we use x ≡ 12015569.