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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
93925371401-37
25141111-3738
141113-3738-75
1133238-75263
3211-75263-338
2120263-338939
So our multiplicative inverse is -338 mod 939 ≡ 601
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
46439516901-1
395695501-16
6950119-16-7
50192126-720
191217-720-27
1271520-2747
7512-2747-74
522147-74195
2120-74195-464
So our multiplicative inverse is 195 mod 464 ≡ 195
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
5475680547010
568547121101
5472126101-26
2112101-26547
So our multiplicative inverse is -26 mod 547 ≡ 521
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 ≡ 4 × 25-1 (mod 939) ≡ 4 × 601 (mod 939) ≡ 526 (mod 939)
x ≡ 399 × 395-1 (mod 464) ≡ 399 × 195 (mod 464) ≡ 317 (mod 464)
x ≡ 253 × 568-1 (mod 547) ≡ 253 × 521 (mod 547) ≡ 533 (mod 547)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 939 × 464 × 547 = 238325712
  2. We calculate the numbers M1 to M3
    M1=M/m1=238325712/939=253808,   M2=M/m2=238325712/464=513633,   M3=M/m3=238325712/547=435696
  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
    9392538080939010
    253808939270278101
    939278310501-3
    2781052681-37
    10568137-37-10
    68371317-1017
    373116-1017-27
    3165117-27152
    6160-27152-939
    So our multiplicative inverse is 152 mod 939 ≡ 152
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4645136330464010
    5136334641106449101
    46444911501-1
    4491529141-130
    151411-130-31
    14114030-31464
    So our multiplicative inverse is -31 mod 464 ≡ 433
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5474356960547010
    435696547796284101
    547284126301-1
    2842631211-12
    263211211-12-25
    21111102-2527
    111011-2527-52
    10110027-52547
    So our multiplicative inverse is -52 mod 547 ≡ 495
    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
    =  (526 × 253808 × 152 +
       317 × 513633 × 433 +
       533 × 435696 × 495)   mod 238325712
    = 71001133 (mod 238325712)


    So our answer is 71001133 (mod 238325712).


Verification

So we found that x ≡ 71001133
If this is correct, then the following statements (i.e. the original equations) are true:
25x (mod 939) ≡ 4 (mod 939)
395x (mod 464) ≡ 399 (mod 464)
568x (mod 547) ≡ 253 (mod 547)

Let's see whether that's indeed the case if we use x ≡ 71001133.