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 |
|---|---|---|---|---|---|---|
| 293 | 298 | 0 | 293 | 0 | 1 | 0 |
| 298 | 293 | 1 | 5 | 1 | 0 | 1 |
| 293 | 5 | 58 | 3 | 0 | 1 | -58 |
| 5 | 3 | 1 | 2 | 1 | -58 | 59 |
| 3 | 2 | 1 | 1 | -58 | 59 | -117 |
| 2 | 1 | 2 | 0 | 59 | -117 | 293 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 263 | 134 | 1 | 129 | 0 | 1 | -1 |
| 134 | 129 | 1 | 5 | 1 | -1 | 2 |
| 129 | 5 | 25 | 4 | -1 | 2 | -51 |
| 5 | 4 | 1 | 1 | 2 | -51 | 53 |
| 4 | 1 | 4 | 0 | -51 | 53 | -263 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 355 | 487 | 0 | 355 | 0 | 1 | 0 |
| 487 | 355 | 1 | 132 | 1 | 0 | 1 |
| 355 | 132 | 2 | 91 | 0 | 1 | -2 |
| 132 | 91 | 1 | 41 | 1 | -2 | 3 |
| 91 | 41 | 2 | 9 | -2 | 3 | -8 |
| 41 | 9 | 4 | 5 | 3 | -8 | 35 |
| 9 | 5 | 1 | 4 | -8 | 35 | -43 |
| 5 | 4 | 1 | 1 | 35 | -43 | 78 |
| 4 | 1 | 4 | 0 | -43 | 78 | -355 |
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 ≡ 772 × 298-1 (mod 293) ≡ 772 × 176 (mod 293) ≡ 213 (mod 293)x ≡ 708 × 134-1 (mod 263) ≡ 708 × 53 (mod 263) ≡ 178 (mod 263)x ≡ 122 × 487-1 (mod 355) ≡ 122 × 78 (mod 355) ≡ 286 (mod 355)
So now we have the following equations of the form x ≡ a (mod m):
x ≡ 213 (mod 293)
x ≡ 178 (mod 263)
x ≡ 286 (mod 355)
Now the actual calculation
- Find the common modulus M
M = m1 × m2 × ... × mk = 293 × 263 × 355 = 27355945 - We calculate the numbers M1 to M3
M1=M/m1=27355945/293=93365,M2=M/m2=27355945/263=104015,M3=M/m3=27355945/355=77059 - 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 135 mod 293 ≡ 135n b q r t1 t2 t3 293 93365 0 293 0 1 0 93365 293 318 191 1 0 1 293 191 1 102 0 1 -1 191 102 1 89 1 -1 2 102 89 1 13 -1 2 -3 89 13 6 11 2 -3 20 13 11 1 2 -3 20 -23 11 2 5 1 20 -23 135 2 1 2 0 -23 135 -293
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is 87 mod 263 ≡ 87n b q r t1 t2 t3 263 104015 0 263 0 1 0 104015 263 395 130 1 0 1 263 130 2 3 0 1 -2 130 3 43 1 1 -2 87 3 1 3 0 -2 87 -263
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is 74 mod 355 ≡ 74n b q r t1 t2 t3 355 77059 0 355 0 1 0 77059 355 217 24 1 0 1 355 24 14 19 0 1 -14 24 19 1 5 1 -14 15 19 5 3 4 -14 15 -59 5 4 1 1 15 -59 74 4 1 4 0 -59 74 -355
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
= (213 × 93365 × 135 +
178 × 104015 × 87 +
286 × 77059 × 74) mod 27355945
= 17479421 (mod 27355945)
So our answer is 17479421 (mod 27355945).
Verification
So we found that x ≡ 17479421
If this is correct, then the following statements (i.e. the original equations) are true:
298x (mod 293) ≡ 772 (mod 293)
134x (mod 263) ≡ 708 (mod 263)
487x (mod 355) ≡ 122 (mod 355)
Let's see whether that's indeed the case if we use x ≡ 17479421.