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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
6176131401-1
613415311-1154
4140-1154-617
So our multiplicative inverse is 154 mod 617 ≡ 154
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3134980313010
4983131185101
313185112801-1
1851281571-12
12857214-12-5
5714412-522
141140-522-313
So our multiplicative inverse is 22 mod 313 ≡ 22
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
5877058010
7758119101
58193101-3
1911901-358
So our multiplicative inverse is -3 mod 58 ≡ 55
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 ≡ 870 × 613-1 (mod 617) ≡ 870 × 154 (mod 617) ≡ 91 (mod 617)
x ≡ 596 × 498-1 (mod 313) ≡ 596 × 22 (mod 313) ≡ 279 (mod 313)
x ≡ 575 × 77-1 (mod 58) ≡ 575 × 55 (mod 58) ≡ 15 (mod 58)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 617 × 313 × 58 = 11201018
  2. We calculate the numbers M1 to M3
    M1=M/m1=11201018/617=18154,   M2=M/m2=11201018/313=35786,   M3=M/m3=11201018/58=193121
  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
    617181540617010
    1815461729261101
    61726129501-2
    261952711-25
    9571124-25-7
    71242235-719
    242311-719-26
    23123019-26617
    So our multiplicative inverse is -26 mod 617 ≡ 591
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    313357860313010
    35786313114104101
    3131043101-3
    104110401-3313
    So our multiplicative inverse is -3 mod 313 ≡ 310
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    58193121058010
    19312158332939101
    583911901-1
    3919211-13
    191190-13-58
    So our multiplicative inverse is 3 mod 58 ≡ 3
    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
    =  (91 × 18154 × 591 +
       279 × 35786 × 310 +
       15 × 193121 × 3)   mod 11201018
    = 2991307 (mod 11201018)


    So our answer is 2991307 (mod 11201018).


Verification

So we found that x ≡ 2991307
If this is correct, then the following statements (i.e. the original equations) are true:
613x (mod 617) ≡ 870 (mod 617)
498x (mod 313) ≡ 596 (mod 313)
77x (mod 58) ≡ 575 (mod 58)

Let's see whether that's indeed the case if we use x ≡ 2991307.