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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
88184413701-1
8443722301-123
373017-123-24
3074223-24119
7231-24119-381
2120119-381881
So our multiplicative inverse is -381 mod 881 ≡ 500
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2657780265010
7782652248101
26524811701-1
2481714101-115
171017-115-16
1071315-1631
7321-1631-78
313031-78265
So our multiplicative inverse is -78 mod 265 ≡ 187
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
37436037010
436371129101
37291801-1
298351-14
8513-14-5
53124-59
3211-59-14
21209-1437
So our multiplicative inverse is -14 mod 37 ≡ 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 ≡ 428 × 844-1 (mod 881) ≡ 428 × 500 (mod 881) ≡ 798 (mod 881)
x ≡ 910 × 778-1 (mod 265) ≡ 910 × 187 (mod 265) ≡ 40 (mod 265)
x ≡ 808 × 436-1 (mod 37) ≡ 808 × 23 (mod 37) ≡ 10 (mod 37)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 881 × 265 × 37 = 8638205
  2. We calculate the numbers M1 to M3
    M1=M/m1=8638205/881=9805,   M2=M/m2=8638205/265=32597,   M3=M/m3=8638205/37=233465
  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
    88198050881010
    980588111114101
    88111478301-7
    114831311-78
    8331221-78-23
    31211108-2331
    211021-2331-85
    10110031-85881
    So our multiplicative inverse is -85 mod 881 ≡ 796
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    265325970265010
    325972651232101
    2652132101-132
    21201-132265
    So our multiplicative inverse is -132 mod 265 ≡ 133
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    37233465037010
    23346537630932101
    37321501-1
    325621-17
    5221-17-15
    21207-1537
    So our multiplicative inverse is -15 mod 37 ≡ 22
    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
    =  (798 × 9805 × 796 +
       40 × 32597 × 133 +
       10 × 233465 × 22)   mod 8638205
    = 253645 (mod 8638205)


    So our answer is 253645 (mod 8638205).


Verification

So we found that x ≡ 253645
If this is correct, then the following statements (i.e. the original equations) are true:
844x (mod 881) ≡ 428 (mod 881)
778x (mod 265) ≡ 910 (mod 265)
436x (mod 37) ≡ 808 (mod 37)

Let's see whether that's indeed the case if we use x ≡ 253645.