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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
1352301-2
53121-23
3211-23-5
21203-513
So our multiplicative inverse is -5 mod 13 ≡ 8
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
5575501701-1
55077841-179
7413-179-80
431179-80159
3130-80159-557
So our multiplicative inverse is 159 mod 557 ≡ 159
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
701235223101-2
235231141-23
2314573-23-173
43113-173176
3130-173176-701
So our multiplicative inverse is 176 mod 701 ≡ 176
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 ≡ 368 × 5-1 (mod 13) ≡ 368 × 8 (mod 13) ≡ 6 (mod 13)
x ≡ 368 × 550-1 (mod 557) ≡ 368 × 159 (mod 557) ≡ 27 (mod 557)
x ≡ 38 × 235-1 (mod 701) ≡ 38 × 176 (mod 701) ≡ 379 (mod 701)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 13 × 557 × 701 = 5075941
  2. We calculate the numbers M1 to M3
    M1=M/m1=5075941/13=390457,   M2=M/m2=5075941/557=9113,   M3=M/m3=5075941/701=7241
  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
    13390457013010
    39045713300352101
    1326101-6
    21201-613
    So our multiplicative inverse is -6 mod 13 ≡ 7
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    55791130557010
    911355716201101
    557201215501-2
    2011551461-23
    15546317-23-11
    46172123-1125
    171215-1125-36
    1252225-3697
    5221-3697-230
    212097-230557
    So our multiplicative inverse is -230 mod 557 ≡ 327
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    70172410701010
    724170110231101
    7012313801-3
    23182871-385
    8711-385-88
    717085-88701
    So our multiplicative inverse is -88 mod 701 ≡ 613
    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
    =  (6 × 390457 × 7 +
       27 × 9113 × 327 +
       379 × 7241 × 613)   mod 5075941
    = 2558328 (mod 5075941)


    So our answer is 2558328 (mod 5075941).


Verification

So we found that x ≡ 2558328
If this is correct, then the following statements (i.e. the original equations) are true:
5x (mod 13) ≡ 368 (mod 13)
550x (mod 557) ≡ 368 (mod 557)
235x (mod 701) ≡ 38 (mod 701)

Let's see whether that's indeed the case if we use x ≡ 2558328.