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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!

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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
4571523101-3
152115201-3457
So our multiplicative inverse is -3 mod 457 ≡ 454
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
601317128401-1
3172841331-12
28433820-12-17
33201132-1719
201317-1719-36
1371619-3655
7611-3655-91
616055-91601
So our multiplicative inverse is -91 mod 601 ≡ 510
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
73711167101-6
111711401-67
7140131-67-13
4031197-1320
31934-1320-73
942120-73166
4140-73166-737
So our multiplicative inverse is 166 mod 737 ≡ 166
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 ≡ 119 × 152-1 (mod 457) ≡ 119 × 454 (mod 457) ≡ 100 (mod 457)
x ≡ 843 × 317-1 (mod 601) ≡ 843 × 510 (mod 601) ≡ 215 (mod 601)
x ≡ 25 × 111-1 (mod 737) ≡ 25 × 166 (mod 737) ≡ 465 (mod 737)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 457 × 601 × 737 = 202422209
  2. We calculate the numbers M1 to M3
    M1=M/m1=202422209/457=442937,   M2=M/m2=202422209/601=336809,   M3=M/m3=202422209/737=274657
  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
    4574429370457010
    442937457969104101
    45710444101-4
    104412221-49
    4122119-49-13
    2219139-1322
    19361-1322-145
    313022-145457
    So our multiplicative inverse is -145 mod 457 ≡ 312
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    6013368090601010
    336809601560249101
    601249210301-2
    2491032431-25
    10343217-25-12
    4317295-1229
    17918-1229-41
    981129-4170
    8180-4170-601
    So our multiplicative inverse is 70 mod 601 ≡ 70
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    7372746570737010
    274657737372493101
    737493124401-1
    493244251-13
    2445484-13-145
    54113-145148
    4140-145148-737
    So our multiplicative inverse is 148 mod 737 ≡ 148
    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
    =  (100 × 442937 × 312 +
       215 × 336809 × 70 +
       465 × 274657 × 148)   mod 202422209
    = 139973716 (mod 202422209)


    So our answer is 139973716 (mod 202422209).


Verification

So we found that x ≡ 139973716
If this is correct, then the following statements (i.e. the original equations) are true:
152x (mod 457) ≡ 119 (mod 457)
317x (mod 601) ≡ 843 (mod 601)
111x (mod 737) ≡ 25 (mod 737)

Let's see whether that's indeed the case if we use x ≡ 139973716.