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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
699257218501-2
2571851721-23
18572241-23-8
72411313-811
4131110-811-19
31103111-1968
101100-1968-699
So our multiplicative inverse is 68 mod 699 ≡ 68
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
469337113201-1
3371322731-13
13273159-13-4
73591143-47
591443-47-32
143427-32135
3211-32135-167
2120135-167469
So our multiplicative inverse is -167 mod 469 ≡ 302
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4879030487010
9034871416101
48741617101-1
416715611-16
7161110-16-7
6110616-748
101100-748-487
So our multiplicative inverse is 48 mod 487 ≡ 48
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 ≡ 361 × 257-1 (mod 699) ≡ 361 × 68 (mod 699) ≡ 83 (mod 699)
x ≡ 903 × 337-1 (mod 469) ≡ 903 × 302 (mod 469) ≡ 217 (mod 469)
x ≡ 739 × 903-1 (mod 487) ≡ 739 × 48 (mod 487) ≡ 408 (mod 487)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 699 × 469 × 487 = 159653697
  2. We calculate the numbers M1 to M3
    M1=M/m1=159653697/699=228403,   M2=M/m2=159653697/469=340413,   M3=M/m3=159653697/487=327831
  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
    6992284030699010
    228403699326529101
    699529117001-1
    5291703191-14
    17019818-14-33
    1918114-3337
    181180-3337-699
    So our multiplicative inverse is 37 mod 699 ≡ 37
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4693404130469010
    340413469725388101
    46938818101-1
    388814641-15
    8164117-15-6
    64173135-623
    171314-623-29
    1343123-29110
    4140-29110-469
    So our multiplicative inverse is 110 mod 469 ≡ 110
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4873278310487010
    32783148767380101
    487806701-6
    8071131-667
    7321-667-140
    313067-140487
    So our multiplicative inverse is -140 mod 487 ≡ 347
    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
    =  (83 × 228403 × 37 +
       217 × 340413 × 110 +
       408 × 327831 × 347)   mod 159653697
    = 159560114 (mod 159653697)


    So our answer is 159560114 (mod 159653697).


Verification

So we found that x ≡ 159560114
If this is correct, then the following statements (i.e. the original equations) are true:
257x (mod 699) ≡ 361 (mod 699)
337x (mod 469) ≡ 903 (mod 469)
903x (mod 487) ≡ 739 (mod 487)

Let's see whether that's indeed the case if we use x ≡ 159560114.