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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
577402117501-1
4021752521-13
17552319-13-10
52192143-1023
191415-1023-33
1452423-3389
5411-3389-122
414089-122577
So our multiplicative inverse is -122 mod 577 ≡ 455
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
19642019010
642193315101
19151401-1
154331-14
4311-14-5
31304-519
So our multiplicative inverse is -5 mod 19 ≡ 14
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
59807059010
807591340101
594011901-1
4019221-13
19291-13-28
21203-2859
So our multiplicative inverse is -28 mod 59 ≡ 31
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 ≡ 137 × 402-1 (mod 577) ≡ 137 × 455 (mod 577) ≡ 19 (mod 577)
x ≡ 563 × 642-1 (mod 19) ≡ 563 × 14 (mod 19) ≡ 16 (mod 19)
x ≡ 250 × 807-1 (mod 59) ≡ 250 × 31 (mod 59) ≡ 21 (mod 59)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 577 × 19 × 59 = 646817
  2. We calculate the numbers M1 to M3
    M1=M/m1=646817/577=1121,   M2=M/m2=646817/19=34043,   M3=M/m3=646817/59=10963
  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
    57711210577010
    11215771544101
    57754413301-1
    5443316161-117
    331621-117-35
    16116017-35577
    So our multiplicative inverse is -35 mod 577 ≡ 542
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1934043019010
    3404319179114101
    19141501-1
    145241-13
    5411-13-4
    41403-419
    So our multiplicative inverse is -4 mod 19 ≡ 15
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5910963059010
    109635918548101
    594811101-1
    4811441-15
    11423-15-11
    43115-1116
    3130-1116-59
    So our multiplicative inverse is 16 mod 59 ≡ 16
    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
    =  (19 × 1121 × 542 +
       16 × 34043 × 15 +
       21 × 10963 × 16)   mod 646817
    = 112534 (mod 646817)


    So our answer is 112534 (mod 646817).


Verification

So we found that x ≡ 112534
If this is correct, then the following statements (i.e. the original equations) are true:
402x (mod 577) ≡ 137 (mod 577)
642x (mod 19) ≡ 563 (mod 19)
807x (mod 59) ≡ 250 (mod 59)

Let's see whether that's indeed the case if we use x ≡ 112534.