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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
35566035010
56635166101
3565501-5
65111-56
5150-56-35
So our multiplicative inverse is 6 mod 35 ≡ 6
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
70412756901-5
127691581-56
6958111-56-11
5811536-1161
11332-1161-194
321161-194255
2120-194255-704
So our multiplicative inverse is 255 mod 704 ≡ 255
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2639800263010
9802633191101
26319117201-1
191722471-13
7247125-13-4
47251223-47
252213-47-11
223717-1184
3130-1184-263
So our multiplicative inverse is 84 mod 263 ≡ 84
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 ≡ 797 × 566-1 (mod 35) ≡ 797 × 6 (mod 35) ≡ 22 (mod 35)
x ≡ 746 × 127-1 (mod 704) ≡ 746 × 255 (mod 704) ≡ 150 (mod 704)
x ≡ 1 × 980-1 (mod 263) ≡ 1 × 84 (mod 263) ≡ 84 (mod 263)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 35 × 704 × 263 = 6480320
  2. We calculate the numbers M1 to M3
    M1=M/m1=6480320/35=185152,   M2=M/m2=6480320/704=9205,   M3=M/m3=6480320/263=24640
  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
    35185152035010
    1851523552902101
    35217101-17
    21201-1735
    So our multiplicative inverse is -17 mod 35 ≡ 18
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    70492050704010
    92057041353101
    70453131501-13
    5315381-1340
    15817-1340-53
    871140-5393
    7170-5393-704
    So our multiplicative inverse is 93 mod 704 ≡ 93
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    263246400263010
    2464026393181101
    26318118201-1
    181822171-13
    8217414-13-13
    1714133-1316
    14342-1316-77
    321116-7793
    2120-7793-263
    So our multiplicative inverse is 93 mod 263 ≡ 93
    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
    =  (22 × 185152 × 18 +
       150 × 9205 × 93 +
       84 × 24640 × 93)   mod 6480320
    = 5398422 (mod 6480320)


    So our answer is 5398422 (mod 6480320).


Verification

So we found that x ≡ 5398422
If this is correct, then the following statements (i.e. the original equations) are true:
566x (mod 35) ≡ 797 (mod 35)
127x (mod 704) ≡ 746 (mod 704)
980x (mod 263) ≡ 1 (mod 263)

Let's see whether that's indeed the case if we use x ≡ 5398422.