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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!

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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
4017940401010
7944011393101
4013931801-1
39384911-150
8180-150-401
So our multiplicative inverse is 50 mod 401 ≡ 50
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
6139210613010
9216131308101
613308130501-1
308305131-12
30531012-12-203
32112-203205
2120-203205-613
So our multiplicative inverse is 205 mod 613 ≡ 205
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
54149414701-1
4944710241-111
4724123-111-12
24231111-1223
231230-1223-541
So our multiplicative inverse is 23 mod 541 ≡ 23
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 ≡ 41 × 794-1 (mod 401) ≡ 41 × 50 (mod 401) ≡ 45 (mod 401)
x ≡ 201 × 921-1 (mod 613) ≡ 201 × 205 (mod 613) ≡ 134 (mod 613)
x ≡ 680 × 494-1 (mod 541) ≡ 680 × 23 (mod 541) ≡ 492 (mod 541)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 401 × 613 × 541 = 132984833
  2. We calculate the numbers M1 to M3
    M1=M/m1=132984833/401=331633,   M2=M/m2=132984833/613=216941,   M3=M/m3=132984833/541=245813
  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
    4013316330401010
    3316334018276101
    401666501-66
    65111-6667
    5150-6667-401
    So our multiplicative inverse is 67 mod 401 ≡ 67
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    6132169410613010
    216941613353552101
    61355216101-1
    55261931-110
    613201-110-201
    313010-201613
    So our multiplicative inverse is -201 mod 613 ≡ 412
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5412458130541010
    245813541454199101
    541199214301-2
    1991431561-23
    14356231-23-8
    56311253-811
    312516-811-19
    2564111-1987
    6160-1987-541
    So our multiplicative inverse is 87 mod 541 ≡ 87
    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
    =  (45 × 331633 × 67 +
       134 × 216941 × 412 +
       492 × 245813 × 87)   mod 132984833
    = 93201267 (mod 132984833)


    So our answer is 93201267 (mod 132984833).


Verification

So we found that x ≡ 93201267
If this is correct, then the following statements (i.e. the original equations) are true:
794x (mod 401) ≡ 41 (mod 401)
921x (mod 613) ≡ 201 (mod 613)
494x (mod 541) ≡ 680 (mod 541)

Let's see whether that's indeed the case if we use x ≡ 93201267.