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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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 |
|---|---|---|---|---|---|---|
| 331 | 81 | 4 | 7 | 0 | 1 | -4 |
| 81 | 7 | 11 | 4 | 1 | -4 | 45 |
| 7 | 4 | 1 | 3 | -4 | 45 | -49 |
| 4 | 3 | 1 | 1 | 45 | -49 | 94 |
| 3 | 1 | 3 | 0 | -49 | 94 | -331 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 527 | 262 | 2 | 3 | 0 | 1 | -2 |
| 262 | 3 | 87 | 1 | 1 | -2 | 175 |
| 3 | 1 | 3 | 0 | -2 | 175 | -527 |
Source: ExtendedEuclideanAlgorithm.com
| n | b | q | r | t1 | t2 | t3 |
|---|---|---|---|---|---|---|
| 224 | 323 | 0 | 224 | 0 | 1 | 0 |
| 323 | 224 | 1 | 99 | 1 | 0 | 1 |
| 224 | 99 | 2 | 26 | 0 | 1 | -2 |
| 99 | 26 | 3 | 21 | 1 | -2 | 7 |
| 26 | 21 | 1 | 5 | -2 | 7 | -9 |
| 21 | 5 | 4 | 1 | 7 | -9 | 43 |
| 5 | 1 | 5 | 0 | -9 | 43 | -224 |
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 ≡ 89 × 81-1 (mod 331) ≡ 89 × 94 (mod 331) ≡ 91 (mod 331)x ≡ 205 × 262-1 (mod 527) ≡ 205 × 175 (mod 527) ≡ 39 (mod 527)x ≡ 774 × 323-1 (mod 224) ≡ 774 × 43 (mod 224) ≡ 130 (mod 224)
So now we have the following equations of the form x ≡ a (mod m):
x ≡ 91 (mod 331)
x ≡ 39 (mod 527)
x ≡ 130 (mod 224)
Now the actual calculation
- Find the common modulus M
M = m1 × m2 × ... × mk = 331 × 527 × 224 = 39073888 - We calculate the numbers M1 to M3
M1=M/m1=39073888/331=118048,M2=M/m2=39073888/527=74144,M3=M/m3=39073888/224=174437 - 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 89 mod 331 ≡ 89n b q r t1 t2 t3 331 118048 0 331 0 1 0 118048 331 356 212 1 0 1 331 212 1 119 0 1 -1 212 119 1 93 1 -1 2 119 93 1 26 -1 2 -3 93 26 3 15 2 -3 11 26 15 1 11 -3 11 -14 15 11 1 4 11 -14 25 11 4 2 3 -14 25 -64 4 3 1 1 25 -64 89 3 1 3 0 -64 89 -331
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is -97 mod 527 ≡ 430n b q r t1 t2 t3 527 74144 0 527 0 1 0 74144 527 140 364 1 0 1 527 364 1 163 0 1 -1 364 163 2 38 1 -1 3 163 38 4 11 -1 3 -13 38 11 3 5 3 -13 42 11 5 2 1 -13 42 -97 5 1 5 0 42 -97 527
Source: ExtendedEuclideanAlgorithm.com
So our multiplicative inverse is -19 mod 224 ≡ 205n b q r t1 t2 t3 224 174437 0 224 0 1 0 174437 224 778 165 1 0 1 224 165 1 59 0 1 -1 165 59 2 47 1 -1 3 59 47 1 12 -1 3 -4 47 12 3 11 3 -4 15 12 11 1 1 -4 15 -19 11 1 11 0 15 -19 224
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
= (91 × 118048 × 89 +
39 × 74144 × 430 +
130 × 174437 × 205) mod 39073888
= 10281282 (mod 39073888)
So our answer is 10281282 (mod 39073888).
Verification
So we found that x ≡ 10281282
If this is correct, then the following statements (i.e. the original equations) are true:
81x (mod 331) ≡ 89 (mod 331)
262x (mod 527) ≡ 205 (mod 527)
323x (mod 224) ≡ 774 (mod 224)
Let's see whether that's indeed the case if we use x ≡ 10281282.