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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
69712557201-5
125721531-56
7253119-56-11
53192156-1128
191514-1128-39
1543328-39145
4311-39145-184
3130145-184697
So our multiplicative inverse is -184 mod 697 ≡ 513
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2474250247010
4252471178101
24717816901-1
178692401-13
6940129-13-4
40291113-47
291127-47-18
117147-1825
7413-1825-43
431125-4368
3130-4368-247
So our multiplicative inverse is 68 mod 247 ≡ 68
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4074230407010
423407116101
4071625701-25
167221-2551
7231-2551-178
212051-178407
So our multiplicative inverse is -178 mod 407 ≡ 229
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 ≡ 259 × 125-1 (mod 697) ≡ 259 × 513 (mod 697) ≡ 437 (mod 697)
x ≡ 774 × 425-1 (mod 247) ≡ 774 × 68 (mod 247) ≡ 21 (mod 247)
x ≡ 787 × 423-1 (mod 407) ≡ 787 × 229 (mod 407) ≡ 329 (mod 407)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 697 × 247 × 407 = 70068713
  2. We calculate the numbers M1 to M3
    M1=M/m1=70068713/697=100529,   M2=M/m2=70068713/247=283679,   M3=M/m3=70068713/407=172159
  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
    6971005290697010
    100529697144161101
    69716145301-4
    16153321-413
    532261-413-342
    212013-342697
    So our multiplicative inverse is -342 mod 697 ≡ 355
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    2472836790247010
    2836792471148123101
    2471232101-2
    123112301-2247
    So our multiplicative inverse is -2 mod 247 ≡ 245
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4071721590407010
    172159407422405101
    4074051201-1
    405220211-1203
    2120-1203-407
    So our multiplicative inverse is 203 mod 407 ≡ 203
    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
    =  (437 × 100529 × 355 +
       21 × 283679 × 245 +
       329 × 172159 × 203)   mod 70068713
    = 35111812 (mod 70068713)


    So our answer is 35111812 (mod 70068713).


Verification

So we found that x ≡ 35111812
If this is correct, then the following statements (i.e. the original equations) are true:
125x (mod 697) ≡ 259 (mod 697)
425x (mod 247) ≡ 774 (mod 247)
423x (mod 407) ≡ 787 (mod 407)

Let's see whether that's indeed the case if we use x ≡ 35111812.