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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
3134980313010
4983131185101
313185112801-1
1851281571-12
12857214-12-5
5714412-522
141140-522-313
So our multiplicative inverse is 22 mod 313 ≡ 22
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
26922414501-1
224454441-15
454411-15-6
4414405-6269
So our multiplicative inverse is -6 mod 269 ≡ 263
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3064030306010
403306197101
3069731501-3
9715671-319
15721-319-41
717019-41306
So our multiplicative inverse is -41 mod 306 ≡ 265
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 ≡ 336 × 498-1 (mod 313) ≡ 336 × 22 (mod 313) ≡ 193 (mod 313)
x ≡ 478 × 224-1 (mod 269) ≡ 478 × 263 (mod 269) ≡ 91 (mod 269)
x ≡ 351 × 403-1 (mod 306) ≡ 351 × 265 (mod 306) ≡ 297 (mod 306)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 313 × 269 × 306 = 25764282
  2. We calculate the numbers M1 to M3
    M1=M/m1=25764282/313=82314,   M2=M/m2=25764282/269=95778,   M3=M/m3=25764282/306=84197
  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
    313823140313010
    82314313262308101
    3133081501-1
    30856131-162
    5312-162-63
    321162-63125
    2120-63125-313
    So our multiplicative inverse is 125 mod 313 ≡ 125
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    269957780269010
    9577826935614101
    2691419301-19
    143421-1977
    3211-1977-96
    212077-96269
    So our multiplicative inverse is -96 mod 269 ≡ 173
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    306841970306010
    8419730627547101
    3064762401-6
    47241231-67
    242311-67-13
    2312307-13306
    So our multiplicative inverse is -13 mod 306 ≡ 293
    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
    =  (193 × 82314 × 125 +
       91 × 95778 × 173 +
       297 × 84197 × 293)   mod 25764282
    = 25331283 (mod 25764282)


    So our answer is 25331283 (mod 25764282).


Verification

So we found that x ≡ 25331283
If this is correct, then the following statements (i.e. the original equations) are true:
498x (mod 313) ≡ 336 (mod 313)
224x (mod 269) ≡ 478 (mod 269)
403x (mod 306) ≡ 351 (mod 306)

Let's see whether that's indeed the case if we use x ≡ 25331283.