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Welcome to ChineseRemainderTheorem.com!

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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.

This fills all textboxes with random numbers. If you fill in random numbers yourself, it is very likely that those numbers do not have a solution. To avoid disappointment, use this button instead! It only uses random numbers that do have a solution.

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.

Do you want to use more equations? Go ahead and use this button. It adds another row that you can fill in. Not sure what numbers to put in this newly added row? Use the random numbers button again!

Do you have too many rows? Use one of these buttons to remove a row. You can always add a row again using the yellow "Add a row" button.

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?


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:

nbqr t1t2t3
69162716401-1
627649511-110
6451113-110-11
511331210-1143
131211-1143-54
12112043-54691
So our multiplicative inverse is -54 mod 691 ≡ 637
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1314690131010
469131376101
1317615501-1
76551211-12
5521213-12-5
2113182-57
13815-57-12
85137-1219
5312-1219-31
321119-3150
2120-3150-131
So our multiplicative inverse is 50 mod 131 ≡ 50
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
79254079010
25479317101
791741101-4
1711161-45
11615-45-9
65115-914
5150-914-79
So our multiplicative inverse is 14 mod 79 ≡ 14
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 ≡ 427 × 627-1 (mod 691) ≡ 427 × 637 (mod 691) ≡ 436 (mod 691)
x ≡ 558 × 469-1 (mod 131) ≡ 558 × 50 (mod 131) ≡ 128 (mod 131)
x ≡ 106 × 254-1 (mod 79) ≡ 106 × 14 (mod 79) ≡ 62 (mod 79)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 691 × 131 × 79 = 7151159
  2. We calculate the numbers M1 to M3
    M1=M/m1=7151159/691=10349,   M2=M/m2=7151159/131=54589,   M3=M/m3=7151159/79=90521
  3. 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:

    nbqr t1t2t3
    691103490691010
    1034969114675101
    69167511601-1
    675164231-143
    16351-143-216
    313043-216691
    So our multiplicative inverse is -216 mod 691 ≡ 475
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    131545890131010
    5458913141693101
    1319313801-1
    93382171-13
    381724-13-7
    174413-731
    4140-731-131
    So our multiplicative inverse is 31 mod 131 ≡ 31
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    7990521079010
    9052179114566101
    796611301-1
    6613511-16
    131130-16-79
    So our multiplicative inverse is 6 mod 79 ≡ 6
    Source: ExtendedEuclideanAlgorithm.com
  4. 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
    =  (436 × 10349 × 475 +
       128 × 54589 × 31 +
       62 × 90521 × 6)   mod 7151159
    = 5073758 (mod 7151159)


    So our answer is 5073758 (mod 7151159).


Verification

So we found that x ≡ 5073758
If this is correct, then the following statements (i.e. the original equations) are true:
627x (mod 691) ≡ 427 (mod 691)
469x (mod 131) ≡ 558 (mod 131)
254x (mod 79) ≡ 106 (mod 79)

Let's see whether that's indeed the case if we use x ≡ 5073758.