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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
404287111701-1
2871172531-13
11753211-13-7
5311493-731
11912-731-38
924131-38183
2120-38183-404
So our multiplicative inverse is 183 mod 404 ≡ 183
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
551416113501-1
4161353111-14
13511123-14-49
113324-49151
3211-49151-200
2120151-200551
So our multiplicative inverse is -200 mod 551 ≡ 351
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
7717790771010
77977118101
771896301-96
83221-96193
3211-96193-289
2120193-289771
So our multiplicative inverse is -289 mod 771 ≡ 482
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 ≡ 614 × 287-1 (mod 404) ≡ 614 × 183 (mod 404) ≡ 50 (mod 404)
x ≡ 524 × 416-1 (mod 551) ≡ 524 × 351 (mod 551) ≡ 441 (mod 551)
x ≡ 479 × 779-1 (mod 771) ≡ 479 × 482 (mod 771) ≡ 349 (mod 771)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 404 × 551 × 771 = 171627684
  2. We calculate the numbers M1 to M3
    M1=M/m1=171627684/404=424821,   M2=M/m2=171627684/551=311484,   M3=M/m3=171627684/771=222604
  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
    4044248210404010
    4248214041051217101
    404217118701-1
    2171871301-12
    1873067-12-13
    307422-1354
    7231-1354-175
    212054-175404
    So our multiplicative inverse is -175 mod 404 ≡ 229
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5513114840551010
    311484551565169101
    55116934401-3
    169443371-310
    443717-310-13
    3775210-1375
    7231-1375-238
    212075-238551
    So our multiplicative inverse is -238 mod 551 ≡ 313
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    7712226040771010
    222604771288556101
    771556121501-1
    55621521261-13
    215126189-13-4
    126891373-47
    8937215-47-18
    3715277-1843
    15721-1843-104
    717043-104771
    So our multiplicative inverse is -104 mod 771 ≡ 667
    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
    =  (50 × 424821 × 229 +
       441 × 311484 × 313 +
       349 × 222604 × 667)   mod 171627684
    = 133641634 (mod 171627684)


    So our answer is 133641634 (mod 171627684).


Verification

So we found that x ≡ 133641634
If this is correct, then the following statements (i.e. the original equations) are true:
287x (mod 404) ≡ 614 (mod 404)
416x (mod 551) ≡ 524 (mod 551)
779x (mod 771) ≡ 479 (mod 771)

Let's see whether that's indeed the case if we use x ≡ 133641634.