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 |
|---|---|---|---|---|---|---|
| 701 | 734 | 0 | 701 | 0 | 1 | 0 |
| 734 | 701 | 1 | 33 | 1 | 0 | 1 |
| 701 | 33 | 21 | 8 | 0 | 1 | -21 |
| 33 | 8 | 4 | 1 | 1 | -21 | 85 |
| 8 | 1 | 8 | 0 | -21 | 85 | -701 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 997 | 426 | 2 | 145 | 0 | 1 | -2 |
| 426 | 145 | 2 | 136 | 1 | -2 | 5 |
| 145 | 136 | 1 | 9 | -2 | 5 | -7 |
| 136 | 9 | 15 | 1 | 5 | -7 | 110 |
| 9 | 1 | 9 | 0 | -7 | 110 | -997 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 969 | 733 | 1 | 236 | 0 | 1 | -1 |
| 733 | 236 | 3 | 25 | 1 | -1 | 4 |
| 236 | 25 | 9 | 11 | -1 | 4 | -37 |
| 25 | 11 | 2 | 3 | 4 | -37 | 78 |
| 11 | 3 | 3 | 2 | -37 | 78 | -271 |
| 3 | 2 | 1 | 1 | 78 | -271 | 349 |
| 2 | 1 | 2 | 0 | -271 | 349 | -969 |
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 ≡ 489 × 734-1 (mod 701) ≡ 489 × 85 (mod 701) ≡ 206 (mod 701)x ≡ 26 × 426-1 (mod 997) ≡ 26 × 110 (mod 997) ≡ 866 (mod 997)x ≡ 24 × 733-1 (mod 969) ≡ 24 × 349 (mod 969) ≡ 624 (mod 969)
So now we have the following equations of the form x ≡ a (mod m):
x ≡ 206 (mod 701)
x ≡ 866 (mod 997)
x ≡ 624 (mod 969)
Now the actual calculation
- Find the common modulus M
M = m1 × m2 × ... × mk = 701 × 997 × 969 = 677231193 - We calculate the numbers M1 to M3
M1=M/m1=677231193/701=966093,M2=M/m2=677231193/997=679269,M3=M/m3=677231193/969=698897 - 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 -128 mod 701 ≡ 573n b q r t1 t2 t3 701 966093 0 701 0 1 0 966093 701 1378 115 1 0 1 701 115 6 11 0 1 -6 115 11 10 5 1 -6 61 11 5 2 1 -6 61 -128 5 1 5 0 61 -128 701
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is -425 mod 997 ≡ 572n b q r t1 t2 t3 997 679269 0 997 0 1 0 679269 997 681 312 1 0 1 997 312 3 61 0 1 -3 312 61 5 7 1 -3 16 61 7 8 5 -3 16 -131 7 5 1 2 16 -131 147 5 2 2 1 -131 147 -425 2 1 2 0 147 -425 997
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is -379 mod 969 ≡ 590n b q r t1 t2 t3 969 698897 0 969 0 1 0 698897 969 721 248 1 0 1 969 248 3 225 0 1 -3 248 225 1 23 1 -3 4 225 23 9 18 -3 4 -39 23 18 1 5 4 -39 43 18 5 3 3 -39 43 -168 5 3 1 2 43 -168 211 3 2 1 1 -168 211 -379 2 1 2 0 211 -379 969
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
= (206 × 966093 × 573 +
866 × 679269 × 572 +
624 × 698897 × 590) mod 677231193
= 112266057 (mod 677231193)
So our answer is 112266057 (mod 677231193).
Verification
So we found that x ≡ 112266057
If this is correct, then the following statements (i.e. the original equations) are true:
734x (mod 701) ≡ 489 (mod 701)
426x (mod 997) ≡ 26 (mod 997)
733x (mod 969) ≡ 24 (mod 969)
Let's see whether that's indeed the case if we use x ≡ 112266057.