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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
83734242101-24
34211131-2425
211318-2425-49
1381525-4974
8513-4974-123
531274-123197
3211-123197-320
2120197-320837
So our multiplicative inverse is -320 mod 837 ≡ 517
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
45741414301-1
414439271-110
4327116-110-11
271611110-1121
161115-1121-32
1152121-3285
5150-3285-457
So our multiplicative inverse is 85 mod 457 ≡ 85
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2954280295010
4282951133101
29513322901-2
133294171-29
2917112-29-11
1712159-1120
12522-1120-51
522120-51122
2120-51122-295
So our multiplicative inverse is 122 mod 295 ≡ 122
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 ≡ 286 × 34-1 (mod 837) ≡ 286 × 517 (mod 837) ≡ 550 (mod 837)
x ≡ 528 × 414-1 (mod 457) ≡ 528 × 85 (mod 457) ≡ 94 (mod 457)
x ≡ 86 × 428-1 (mod 295) ≡ 86 × 122 (mod 295) ≡ 167 (mod 295)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 837 × 457 × 295 = 112840155
  2. We calculate the numbers M1 to M3
    M1=M/m1=112840155/837=134815,   M2=M/m2=112840155/457=246915,   M3=M/m3=112840155/295=382509
  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
    8371348150837010
    13481583716158101
    83758142501-14
    5825281-1429
    25831-1429-101
    818029-101837
    So our multiplicative inverse is -101 mod 837 ≡ 736
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4572469150457010
    246915457540135101
    45713535201-3
    135522311-37
    5231121-37-10
    31211107-1017
    211021-1017-44
    10110017-44457
    So our multiplicative inverse is -44 mod 457 ≡ 413
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    2953825090295010
    3825092951296189101
    295189110601-1
    1891061831-12
    10683123-12-3
    83233142-311
    231419-311-14
    1491511-1425
    9514-1425-39
    541125-3964
    4140-3964-295
    So our multiplicative inverse is 64 mod 295 ≡ 64
    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
    =  (550 × 134815 × 736 +
       94 × 246915 × 413 +
       167 × 382509 × 64)   mod 112840155
    = 91648702 (mod 112840155)


    So our answer is 91648702 (mod 112840155).


Verification

So we found that x ≡ 91648702
If this is correct, then the following statements (i.e. the original equations) are true:
34x (mod 837) ≡ 286 (mod 837)
414x (mod 457) ≡ 528 (mod 457)
428x (mod 295) ≡ 86 (mod 295)

Let's see whether that's indeed the case if we use x ≡ 91648702.