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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
8879410887010
941887154101
88754162301-16
5423281-1633
23827-1633-82
871133-82115
7170-82115-887
So our multiplicative inverse is 115 mod 887 ≡ 115
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1097550109010
7551096101101
1091011801-1
10181251-113
8513-113-14
531213-1427
3211-1427-41
212027-41109
So our multiplicative inverse is -41 mod 109 ≡ 68
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2871287001-287
So our multiplicative inverse is 1 mod 287 ≡ 1
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 ≡ 305 × 941-1 (mod 887) ≡ 305 × 115 (mod 887) ≡ 482 (mod 887)
x ≡ 249 × 755-1 (mod 109) ≡ 249 × 68 (mod 109) ≡ 37 (mod 109)
x ≡ 665 × 1-1 (mod 287) ≡ 665 × 1 (mod 287) ≡ 91 (mod 287)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 887 × 109 × 287 = 27748021
  2. We calculate the numbers M1 to M3
    M1=M/m1=27748021/887=31283,   M2=M/m2=27748021/109=254569,   M3=M/m3=27748021/287=96683
  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
    887312830887010
    3128388735238101
    887238317301-3
    2381731651-34
    17365243-34-11
    65431224-1115
    4322121-1115-26
    22211115-2641
    211210-2641-887
    So our multiplicative inverse is 41 mod 887 ≡ 41
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1092545690109010
    254569109233554101
    109542101-2
    5415401-2109
    So our multiplicative inverse is -2 mod 109 ≡ 107
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    287966830287010
    96683287336251101
    28725113601-1
    251366351-17
    363511-17-8
    3513507-8287
    So our multiplicative inverse is -8 mod 287 ≡ 279
    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
    =  (482 × 31283 × 41 +
       37 × 254569 × 107 +
       91 × 96683 × 279)   mod 27748021
    = 1778917 (mod 27748021)


    So our answer is 1778917 (mod 27748021).


Verification

So we found that x ≡ 1778917
If this is correct, then the following statements (i.e. the original equations) are true:
941x (mod 887) ≡ 305 (mod 887)
755x (mod 109) ≡ 249 (mod 109)
1x (mod 287) ≡ 665 (mod 287)

Let's see whether that's indeed the case if we use x ≡ 1778917.