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" .
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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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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 |
|---|---|---|---|---|---|---|
| 539 | 219 | 2 | 101 | 0 | 1 | -2 |
| 219 | 101 | 2 | 17 | 1 | -2 | 5 |
| 101 | 17 | 5 | 16 | -2 | 5 | -27 |
| 17 | 16 | 1 | 1 | 5 | -27 | 32 |
| 16 | 1 | 16 | 0 | -27 | 32 | -539 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 767 | 126 | 6 | 11 | 0 | 1 | -6 |
| 126 | 11 | 11 | 5 | 1 | -6 | 67 |
| 11 | 5 | 2 | 1 | -6 | 67 | -140 |
| 5 | 1 | 5 | 0 | 67 | -140 | 767 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 269 | 937 | 0 | 269 | 0 | 1 | 0 |
| 937 | 269 | 3 | 130 | 1 | 0 | 1 |
| 269 | 130 | 2 | 9 | 0 | 1 | -2 |
| 130 | 9 | 14 | 4 | 1 | -2 | 29 |
| 9 | 4 | 2 | 1 | -2 | 29 | -60 |
| 4 | 1 | 4 | 0 | 29 | -60 | 269 |
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 ≡ 261 × 219-1 (mod 539) ≡ 261 × 32 (mod 539) ≡ 267 (mod 539)x ≡ 477 × 126-1 (mod 767) ≡ 477 × 627 (mod 767) ≡ 716 (mod 767)x ≡ 14 × 937-1 (mod 269) ≡ 14 × 209 (mod 269) ≡ 236 (mod 269)
So now we have the following equations of the form x ≡ a (mod m):
x ≡ 267 (mod 539)
x ≡ 716 (mod 767)
x ≡ 236 (mod 269)
Now the actual calculation
- Find the common modulus M
M = m1 × m2 × ... × mk = 539 × 767 × 269 = 111208097 - We calculate the numbers M1 to M3
M1=M/m1=111208097/539=206323,M2=M/m2=111208097/767=144991,M3=M/m3=111208097/269=413413 - 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 52 mod 539 ≡ 52n b q r t1 t2 t3 539 206323 0 539 0 1 0 206323 539 382 425 1 0 1 539 425 1 114 0 1 -1 425 114 3 83 1 -1 4 114 83 1 31 -1 4 -5 83 31 2 21 4 -5 14 31 21 1 10 -5 14 -19 21 10 2 1 14 -19 52 10 1 10 0 -19 52 -539
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is 137 mod 767 ≡ 137n b q r t1 t2 t3 767 144991 0 767 0 1 0 144991 767 189 28 1 0 1 767 28 27 11 0 1 -27 28 11 2 6 1 -27 55 11 6 1 5 -27 55 -82 6 5 1 1 55 -82 137 5 1 5 0 -82 137 -767
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is -74 mod 269 ≡ 195n b q r t1 t2 t3 269 413413 0 269 0 1 0 413413 269 1536 229 1 0 1 269 229 1 40 0 1 -1 229 40 5 29 1 -1 6 40 29 1 11 -1 6 -7 29 11 2 7 6 -7 20 11 7 1 4 -7 20 -27 7 4 1 3 20 -27 47 4 3 1 1 -27 47 -74 3 1 3 0 47 -74 269
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
= (267 × 206323 × 52 +
716 × 144991 × 137 +
236 × 413413 × 195) mod 111208097
= 80888536 (mod 111208097)
So our answer is 80888536 (mod 111208097).
Verification
So we found that x ≡ 80888536
If this is correct, then the following statements (i.e. the original equations) are true:
219x (mod 539) ≡ 261 (mod 539)
126x (mod 767) ≡ 477 (mod 767)
937x (mod 269) ≡ 14 (mod 269)
Let's see whether that's indeed the case if we use x ≡ 80888536.