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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
4177330417010
7334171316101
417316110101-1
3161013131-14
10113710-14-29
1310134-2933
10331-2933-128
313033-128417
So our multiplicative inverse is -128 mod 417 ≡ 289
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4197370419010
7374191318101
419318110101-1
3181013151-14
10115611-14-25
1511144-2529
11423-2529-83
431129-83112
3130-83112-419
So our multiplicative inverse is 112 mod 419 ≡ 112
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2098180209010
8182093191101
20919111801-1
1911810111-111
181117-111-12
1171411-1223
7413-1223-35
431123-3558
3130-3558-209
So our multiplicative inverse is 58 mod 209 ≡ 58
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 ≡ 486 × 733-1 (mod 417) ≡ 486 × 289 (mod 417) ≡ 342 (mod 417)
x ≡ 966 × 737-1 (mod 419) ≡ 966 × 112 (mod 419) ≡ 90 (mod 419)
x ≡ 744 × 818-1 (mod 209) ≡ 744 × 58 (mod 209) ≡ 98 (mod 209)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 417 × 419 × 209 = 36517107
  2. We calculate the numbers M1 to M3
    M1=M/m1=36517107/417=87571,   M2=M/m2=36517107/419=87153,   M3=M/m3=36517107/209=174723
  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
    417875710417010
    875714172101101
    4171417001-417
    So our multiplicative inverse is 1 mod 417 ≡ 1
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    419871530419010
    871534192081101
    4191419001-419
    So our multiplicative inverse is 1 mod 419 ≡ 1
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    2091747230209010
    174723209835208101
    2092081101-1
    208120801-1209
    So our multiplicative inverse is -1 mod 209 ≡ 208
    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
    =  (342 × 87571 × 1 +
       90 × 87153 × 1 +
       98 × 174723 × 208)   mod 36517107
    = 20670198 (mod 36517107)


    So our answer is 20670198 (mod 36517107).


Verification

So we found that x ≡ 20670198
If this is correct, then the following statements (i.e. the original equations) are true:
733x (mod 417) ≡ 486 (mod 417)
737x (mod 419) ≡ 966 (mod 419)
818x (mod 209) ≡ 744 (mod 209)

Let's see whether that's indeed the case if we use x ≡ 20670198.