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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
3564090356010
409356153101
3565363801-6
53381151-67
381528-67-20
158177-2027
8711-2027-47
717027-47356
So our multiplicative inverse is -47 mod 356 ≡ 309
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3013310301010
331301130101
3013010101-10
3013001-10301
So our multiplicative inverse is -10 mod 301 ≡ 291
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
6436460643010
64664313101
6433214101-214
31301-214643
So our multiplicative inverse is -214 mod 643 ≡ 429
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 ≡ 120 × 409-1 (mod 356) ≡ 120 × 309 (mod 356) ≡ 56 (mod 356)
x ≡ 919 × 331-1 (mod 301) ≡ 919 × 291 (mod 301) ≡ 141 (mod 301)
x ≡ 8 × 646-1 (mod 643) ≡ 8 × 429 (mod 643) ≡ 217 (mod 643)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 356 × 301 × 643 = 68901308
  2. We calculate the numbers M1 to M3
    M1=M/m1=68901308/356=193543,   M2=M/m2=68901308/301=228908,   M3=M/m3=68901308/643=107156
  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
    3561935430356010
    193543356543235101
    356235112101-1
    23512111141-12
    12111417-12-3
    11471622-350
    7231-350-153
    212050-153356
    So our multiplicative inverse is -153 mod 356 ≡ 203
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    3012289080301010
    228908301760148101
    3011482501-2
    14852931-259
    5312-259-61
    321159-61120
    2120-61120-301
    So our multiplicative inverse is 120 mod 301 ≡ 120
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    6431071560643010
    107156643166418101
    643418122501-1
    41822511931-12
    225193132-12-3
    19332612-320
    321320-320-643
    So our multiplicative inverse is 20 mod 643 ≡ 20
    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
    =  (56 × 193543 × 203 +
       141 × 228908 × 120 +
       217 × 107156 × 20)   mod 68901308
    = 61654272 (mod 68901308)


    So our answer is 61654272 (mod 68901308).


Verification

So we found that x ≡ 61654272
If this is correct, then the following statements (i.e. the original equations) are true:
409x (mod 356) ≡ 120 (mod 356)
331x (mod 301) ≡ 919 (mod 301)
646x (mod 643) ≡ 8 (mod 643)

Let's see whether that's indeed the case if we use x ≡ 61654272.