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 |
|---|---|---|---|---|---|---|
| 991 | 42 | 23 | 25 | 0 | 1 | -23 |
| 42 | 25 | 1 | 17 | 1 | -23 | 24 |
| 25 | 17 | 1 | 8 | -23 | 24 | -47 |
| 17 | 8 | 2 | 1 | 24 | -47 | 118 |
| 8 | 1 | 8 | 0 | -47 | 118 | -991 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 701 | 551 | 1 | 150 | 0 | 1 | -1 |
| 551 | 150 | 3 | 101 | 1 | -1 | 4 |
| 150 | 101 | 1 | 49 | -1 | 4 | -5 |
| 101 | 49 | 2 | 3 | 4 | -5 | 14 |
| 49 | 3 | 16 | 1 | -5 | 14 | -229 |
| 3 | 1 | 3 | 0 | 14 | -229 | 701 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 928 | 705 | 1 | 223 | 0 | 1 | -1 |
| 705 | 223 | 3 | 36 | 1 | -1 | 4 |
| 223 | 36 | 6 | 7 | -1 | 4 | -25 |
| 36 | 7 | 5 | 1 | 4 | -25 | 129 |
| 7 | 1 | 7 | 0 | -25 | 129 | -928 |
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 ≡ 778 × 42-1 (mod 991) ≡ 778 × 118 (mod 991) ≡ 632 (mod 991)x ≡ 104 × 551-1 (mod 701) ≡ 104 × 472 (mod 701) ≡ 18 (mod 701)x ≡ 74 × 705-1 (mod 928) ≡ 74 × 129 (mod 928) ≡ 266 (mod 928)
So now we have the following equations of the form x ≡ a (mod m):
x ≡ 632 (mod 991)
x ≡ 18 (mod 701)
x ≡ 266 (mod 928)
Now the actual calculation
- Find the common modulus M
M = m1 × m2 × ... × mk = 991 × 701 × 928 = 644673248 - We calculate the numbers M1 to M3
M1=M/m1=644673248/991=650528,M2=M/m2=644673248/701=919648,M3=M/m3=644673248/928=694691 - 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 39 mod 991 ≡ 39n b q r t1 t2 t3 991 650528 0 991 0 1 0 650528 991 656 432 1 0 1 991 432 2 127 0 1 -2 432 127 3 51 1 -2 7 127 51 2 25 -2 7 -16 51 25 2 1 7 -16 39 25 1 25 0 -16 39 -991
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is 230 mod 701 ≡ 230n b q r t1 t2 t3 701 919648 0 701 0 1 0 919648 701 1311 637 1 0 1 701 637 1 64 0 1 -1 637 64 9 61 1 -1 10 64 61 1 3 -1 10 -11 61 3 20 1 10 -11 230 3 1 3 0 -11 230 -701
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is -341 mod 928 ≡ 587n b q r t1 t2 t3 928 694691 0 928 0 1 0 694691 928 748 547 1 0 1 928 547 1 381 0 1 -1 547 381 1 166 1 -1 2 381 166 2 49 -1 2 -5 166 49 3 19 2 -5 17 49 19 2 11 -5 17 -39 19 11 1 8 17 -39 56 11 8 1 3 -39 56 -95 8 3 2 2 56 -95 246 3 2 1 1 -95 246 -341 2 1 2 0 246 -341 928
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
= (632 × 650528 × 39 +
18 × 919648 × 230 +
266 × 694691 × 587) mod 644673248
= 22022634 (mod 644673248)
So our answer is 22022634 (mod 644673248).
Verification
So we found that x ≡ 22022634
If this is correct, then the following statements (i.e. the original equations) are true:
42x (mod 991) ≡ 778 (mod 991)
551x (mod 701) ≡ 104 (mod 701)
705x (mod 928) ≡ 74 (mod 928)
Let's see whether that's indeed the case if we use x ≡ 22022634.