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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
4017552601-5
75262231-511
262313-511-16
2337211-16123
3211-16123-139
2120123-139401
So our multiplicative inverse is -139 mod 401 ≡ 262
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1134890113010
489113437101
113373201-3
3721811-355
2120-355-113
So our multiplicative inverse is 55 mod 113 ≡ 55
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
887343220101-2
34320111421-23
201142159-23-5
142592243-513
5924211-513-31
24112213-3175
11251-3175-406
212075-406887
So our multiplicative inverse is -406 mod 887 ≡ 481
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 ≡ 305 × 75-1 (mod 401) ≡ 305 × 262 (mod 401) ≡ 111 (mod 401)
x ≡ 190 × 489-1 (mod 113) ≡ 190 × 55 (mod 113) ≡ 54 (mod 113)
x ≡ 401 × 343-1 (mod 887) ≡ 401 × 481 (mod 887) ≡ 402 (mod 887)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 401 × 113 × 887 = 40192631
  2. We calculate the numbers M1 to M3
    M1=M/m1=40192631/401=100231,   M2=M/m2=40192631/113=355687,   M3=M/m3=40192631/887=45313
  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
    4011002310401010
    100231401249382101
    40138211901-1
    382192021-121
    19291-121-190
    212021-190401
    So our multiplicative inverse is -190 mod 401 ≡ 211
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1133556870113010
    355687113314776101
    1137613701-1
    7637221-13
    372181-13-55
    21203-55113
    So our multiplicative inverse is -55 mod 113 ≡ 58
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    887453130887010
    453138875176101
    88776115101-11
    76511251-1112
    512521-1112-35
    25125012-35887
    So our multiplicative inverse is -35 mod 887 ≡ 852
    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
    =  (111 × 100231 × 211 +
       54 × 355687 × 58 +
       402 × 45313 × 852)   mod 40192631
    = 10483855 (mod 40192631)


    So our answer is 10483855 (mod 40192631).


Verification

So we found that x ≡ 10483855
If this is correct, then the following statements (i.e. the original equations) are true:
75x (mod 401) ≡ 305 (mod 401)
489x (mod 113) ≡ 190 (mod 113)
343x (mod 887) ≡ 401 (mod 887)

Let's see whether that's indeed the case if we use x ≡ 10483855.