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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
53173053010
17353314101
531431101-3
1411131-34
11332-34-15
32114-1519
2120-1519-53
So our multiplicative inverse is 19 mod 53 ≡ 19
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
67510465101-6
10451221-613
512251-613-331
212013-331675
So our multiplicative inverse is -331 mod 675 ≡ 344
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4636270463010
6274631164101
463164213501-2
1641351291-23
13529419-23-14
29191103-1417
191019-1417-31
1091117-3148
9190-3148-463
So our multiplicative inverse is 48 mod 463 ≡ 48
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 ≡ 519 × 173-1 (mod 53) ≡ 519 × 19 (mod 53) ≡ 3 (mod 53)
x ≡ 731 × 104-1 (mod 675) ≡ 731 × 344 (mod 675) ≡ 364 (mod 675)
x ≡ 824 × 627-1 (mod 463) ≡ 824 × 48 (mod 463) ≡ 197 (mod 463)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 53 × 675 × 463 = 16563825
  2. We calculate the numbers M1 to M3
    M1=M/m1=16563825/53=312525,   M2=M/m2=16563825/675=24539,   M3=M/m3=16563825/463=35775
  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
    53312525053010
    31252553589637101
    533711601-1
    3716251-13
    16531-13-10
    51503-1053
    So our multiplicative inverse is -10 mod 53 ≡ 43
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    675245390675010
    2453967536239101
    675239219701-2
    2391971421-23
    19742429-23-14
    42291133-1417
    291323-1417-48
    1334117-48209
    3130-48209-675
    So our multiplicative inverse is 209 mod 675 ≡ 209
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    463357750463010
    3577546377124101
    46312439101-3
    124911331-34
    9133225-34-11
    3325184-1115
    25831-1115-56
    818015-56463
    So our multiplicative inverse is -56 mod 463 ≡ 407
    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
    =  (3 × 312525 × 43 +
       364 × 24539 × 209 +
       197 × 35775 × 407)   mod 16563825
    = 5166814 (mod 16563825)


    So our answer is 5166814 (mod 16563825).


Verification

So we found that x ≡ 5166814
If this is correct, then the following statements (i.e. the original equations) are true:
173x (mod 53) ≡ 519 (mod 53)
104x (mod 675) ≡ 731 (mod 675)
627x (mod 463) ≡ 824 (mod 463)

Let's see whether that's indeed the case if we use x ≡ 5166814.