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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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 |
|---|---|---|---|---|---|---|
| 830 | 823 | 1 | 7 | 0 | 1 | -1 |
| 823 | 7 | 117 | 4 | 1 | -1 | 118 |
| 7 | 4 | 1 | 3 | -1 | 118 | -119 |
| 4 | 3 | 1 | 1 | 118 | -119 | 237 |
| 3 | 1 | 3 | 0 | -119 | 237 | -830 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 559 | 502 | 1 | 57 | 0 | 1 | -1 |
| 502 | 57 | 8 | 46 | 1 | -1 | 9 |
| 57 | 46 | 1 | 11 | -1 | 9 | -10 |
| 46 | 11 | 4 | 2 | 9 | -10 | 49 |
| 11 | 2 | 5 | 1 | -10 | 49 | -255 |
| 2 | 1 | 2 | 0 | 49 | -255 | 559 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 593 | 440 | 1 | 153 | 0 | 1 | -1 |
| 440 | 153 | 2 | 134 | 1 | -1 | 3 |
| 153 | 134 | 1 | 19 | -1 | 3 | -4 |
| 134 | 19 | 7 | 1 | 3 | -4 | 31 |
| 19 | 1 | 19 | 0 | -4 | 31 | -593 |
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 ≡ 995 × 823-1 (mod 830) ≡ 995 × 237 (mod 830) ≡ 95 (mod 830)x ≡ 390 × 502-1 (mod 559) ≡ 390 × 304 (mod 559) ≡ 52 (mod 559)x ≡ 580 × 440-1 (mod 593) ≡ 580 × 31 (mod 593) ≡ 190 (mod 593)
So now we have the following equations of the form x ≡ a (mod m):
x ≡ 95 (mod 830)
x ≡ 52 (mod 559)
x ≡ 190 (mod 593)
Now the actual calculation
- Find the common modulus M
M = m1 × m2 × ... × mk = 830 × 559 × 593 = 275134210 - We calculate the numbers M1 to M3
M1=M/m1=275134210/830=331487,M2=M/m2=275134210/559=492190,M3=M/m3=275134210/593=463970 - 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 343 mod 830 ≡ 343n b q r t1 t2 t3 830 331487 0 830 0 1 0 331487 830 399 317 1 0 1 830 317 2 196 0 1 -2 317 196 1 121 1 -2 3 196 121 1 75 -2 3 -5 121 75 1 46 3 -5 8 75 46 1 29 -5 8 -13 46 29 1 17 8 -13 21 29 17 1 12 -13 21 -34 17 12 1 5 21 -34 55 12 5 2 2 -34 55 -144 5 2 2 1 55 -144 343 2 1 2 0 -144 343 -830
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is 147 mod 559 ≡ 147n b q r t1 t2 t3 559 492190 0 559 0 1 0 492190 559 880 270 1 0 1 559 270 2 19 0 1 -2 270 19 14 4 1 -2 29 19 4 4 3 -2 29 -118 4 3 1 1 29 -118 147 3 1 3 0 -118 147 -559
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is 192 mod 593 ≡ 192n b q r t1 t2 t3 593 463970 0 593 0 1 0 463970 593 782 244 1 0 1 593 244 2 105 0 1 -2 244 105 2 34 1 -2 5 105 34 3 3 -2 5 -17 34 3 11 1 5 -17 192 3 1 3 0 -17 192 -593
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
= (95 × 331487 × 343 +
52 × 492190 × 147 +
190 × 463970 × 192) mod 275134210
= 124129915 (mod 275134210)
So our answer is 124129915 (mod 275134210).
Verification
So we found that x ≡ 124129915
If this is correct, then the following statements (i.e. the original equations) are true:
823x (mod 830) ≡ 995 (mod 830)
502x (mod 559) ≡ 390 (mod 559)
440x (mod 593) ≡ 580 (mod 593)
Let's see whether that's indeed the case if we use x ≡ 124129915.