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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
1426670142010
667142499101
1429914301-1
99432131-13
431334-13-10
134313-1033
4140-1033-142
So our multiplicative inverse is 33 mod 142 ≡ 33
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
5358860535010
8865351351101
535351118401-1
35118411671-12
184167117-12-3
167179142-329
171413-329-32
1434229-32157
3211-32157-189
2120157-189535
So our multiplicative inverse is -189 mod 535 ≡ 346
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
25910325301-2
103531501-23
535013-23-5
5031623-583
3211-583-88
212083-88259
So our multiplicative inverse is -88 mod 259 ≡ 171
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 ≡ 311 × 667-1 (mod 142) ≡ 311 × 33 (mod 142) ≡ 39 (mod 142)
x ≡ 341 × 886-1 (mod 535) ≡ 341 × 346 (mod 535) ≡ 286 (mod 535)
x ≡ 971 × 103-1 (mod 259) ≡ 971 × 171 (mod 259) ≡ 22 (mod 259)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 142 × 535 × 259 = 19676230
  2. We calculate the numbers M1 to M3
    M1=M/m1=19676230/142=138565,   M2=M/m2=19676230/535=36778,   M3=M/m3=19676230/259=75970
  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
    1421385650142010
    138565142975115101
    14211512701-1
    11527471-15
    27736-15-16
    76115-1621
    6160-1621-142
    So our multiplicative inverse is 21 mod 142 ≡ 21
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    535367780535010
    3677853568398101
    535398113701-1
    39813721241-13
    137124113-13-4
    12413973-439
    13716-439-43
    761139-4382
    6160-4382-535
    So our multiplicative inverse is 82 mod 535 ≡ 82
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    259759700259010
    7597025929383101
    2598331001-3
    8310831-325
    10331-325-78
    313025-78259
    So our multiplicative inverse is -78 mod 259 ≡ 181
    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
    =  (39 × 138565 × 21 +
       286 × 36778 × 82 +
       22 × 75970 × 181)   mod 19676230
    = 19236211 (mod 19676230)


    So our answer is 19236211 (mod 19676230).


Verification

So we found that x ≡ 19236211
If this is correct, then the following statements (i.e. the original equations) are true:
667x (mod 142) ≡ 311 (mod 142)
886x (mod 535) ≡ 341 (mod 535)
103x (mod 259) ≡ 971 (mod 259)

Let's see whether that's indeed the case if we use x ≡ 19236211.