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" .
This removes all numbers from the textboxes, such that you can fill in your own.
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This button is similar to the "Clear everything" button, but only clears the left column.
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 |
|---|---|---|---|---|---|---|
| 163 | 591 | 0 | 163 | 0 | 1 | 0 |
| 591 | 163 | 3 | 102 | 1 | 0 | 1 |
| 163 | 102 | 1 | 61 | 0 | 1 | -1 |
| 102 | 61 | 1 | 41 | 1 | -1 | 2 |
| 61 | 41 | 1 | 20 | -1 | 2 | -3 |
| 41 | 20 | 2 | 1 | 2 | -3 | 8 |
| 20 | 1 | 20 | 0 | -3 | 8 | -163 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 949 | 22 | 43 | 3 | 0 | 1 | -43 |
| 22 | 3 | 7 | 1 | 1 | -43 | 302 |
| 3 | 1 | 3 | 0 | -43 | 302 | -949 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 449 | 221 | 2 | 7 | 0 | 1 | -2 |
| 221 | 7 | 31 | 4 | 1 | -2 | 63 |
| 7 | 4 | 1 | 3 | -2 | 63 | -65 |
| 4 | 3 | 1 | 1 | 63 | -65 | 128 |
| 3 | 1 | 3 | 0 | -65 | 128 | -449 |
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 ≡ 298 × 591-1 (mod 163) ≡ 298 × 8 (mod 163) ≡ 102 (mod 163)x ≡ 896 × 22-1 (mod 949) ≡ 896 × 302 (mod 949) ≡ 127 (mod 949)x ≡ 593 × 221-1 (mod 449) ≡ 593 × 128 (mod 449) ≡ 23 (mod 449)
So now we have the following equations of the form x ≡ a (mod m):
x ≡ 102 (mod 163)
x ≡ 127 (mod 949)
x ≡ 23 (mod 449)
Now the actual calculation
- Find the common modulus M
M = m1 × m2 × ... × mk = 163 × 949 × 449 = 69454463 - We calculate the numbers M1 to M3
M1=M/m1=69454463/163=426101,M2=M/m2=69454463/949=73187,M3=M/m3=69454463/449=154687 - 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 -60 mod 163 ≡ 103n b q r t1 t2 t3 163 426101 0 163 0 1 0 426101 163 2614 19 1 0 1 163 19 8 11 0 1 -8 19 11 1 8 1 -8 9 11 8 1 3 -8 9 -17 8 3 2 2 9 -17 43 3 2 1 1 -17 43 -60 2 1 2 0 43 -60 163
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is -308 mod 949 ≡ 641n b q r t1 t2 t3 949 73187 0 949 0 1 0 73187 949 77 114 1 0 1 949 114 8 37 0 1 -8 114 37 3 3 1 -8 25 37 3 12 1 -8 25 -308 3 1 3 0 25 -308 949
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is -138 mod 449 ≡ 311n b q r t1 t2 t3 449 154687 0 449 0 1 0 154687 449 344 231 1 0 1 449 231 1 218 0 1 -1 231 218 1 13 1 -1 2 218 13 16 10 -1 2 -33 13 10 1 3 2 -33 35 10 3 3 1 -33 35 -138 3 1 3 0 35 -138 449
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
= (102 × 426101 × 103 +
127 × 73187 × 641 +
23 × 154687 × 311) mod 69454463
= 11586468 (mod 69454463)
So our answer is 11586468 (mod 69454463).
Verification
So we found that x ≡ 11586468
If this is correct, then the following statements (i.e. the original equations) are true:
591x (mod 163) ≡ 298 (mod 163)
22x (mod 949) ≡ 896 (mod 949)
221x (mod 449) ≡ 593 (mod 449)
Let's see whether that's indeed the case if we use x ≡ 11586468.