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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
2574460257010
4462571189101
25718916801-1
189682531-13
6853115-13-4
5315383-415
15817-415-19
871115-1934
7170-1934-257
So our multiplicative inverse is 34 mod 257 ≡ 34
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1734900173010
4901732144101
17314412901-1
144294281-15
292811-15-6
2812805-6173
So our multiplicative inverse is -6 mod 173 ≡ 167
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
6476520647010
65264715101
6475129201-129
52211-129259
2120-129259-647
So our multiplicative inverse is 259 mod 647 ≡ 259
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 ≡ 17 × 446-1 (mod 257) ≡ 17 × 34 (mod 257) ≡ 64 (mod 257)
x ≡ 889 × 490-1 (mod 173) ≡ 889 × 167 (mod 173) ≡ 29 (mod 173)
x ≡ 953 × 652-1 (mod 647) ≡ 953 × 259 (mod 647) ≡ 320 (mod 647)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 257 × 173 × 647 = 28766267
  2. We calculate the numbers M1 to M3
    M1=M/m1=28766267/257=111931,   M2=M/m2=28766267/173=166279,   M3=M/m3=28766267/647=44461
  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
    2571119310257010
    111931257435136101
    257136112101-1
    1361211151-12
    1211581-12-17
    1511502-17257
    So our multiplicative inverse is -17 mod 257 ≡ 240
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1731662790173010
    16627917396126101
    1732661701-6
    2617191-67
    17918-67-13
    98117-1320
    8180-1320-173
    So our multiplicative inverse is 20 mod 173 ≡ 20
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    647444610647010
    4446164768465101
    647465118201-1
    46518221011-13
    182101181-13-4
    101811203-47
    812041-47-32
    2012007-32647
    So our multiplicative inverse is -32 mod 647 ≡ 615
    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
    =  (64 × 111931 × 240 +
       29 × 166279 × 20 +
       320 × 44461 × 615)   mod 28766267
    = 8406791 (mod 28766267)


    So our answer is 8406791 (mod 28766267).


Verification

So we found that x ≡ 8406791
If this is correct, then the following statements (i.e. the original equations) are true:
446x (mod 257) ≡ 17 (mod 257)
490x (mod 173) ≡ 889 (mod 173)
652x (mod 647) ≡ 953 (mod 647)

Let's see whether that's indeed the case if we use x ≡ 8406791.