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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
58312747501-4
127751521-45
7552123-45-9
5223265-923
23635-923-78
651123-78101
5150-78101-583
So our multiplicative inverse is 101 mod 583 ≡ 101
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
120501010
20512050101
So our multiplicative inverse is 0 mod 1 ≡ 0
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
943516142701-1
5164271891-12
42789471-12-9
89711182-911
7118317-911-42
18171111-4253
171170-4253-943
So our multiplicative inverse is 53 mod 943 ≡ 53
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 ≡ 546 × 127-1 (mod 583) ≡ 546 × 101 (mod 583) ≡ 344 (mod 583)
x ≡ 367 × 205-1 (mod 1) ≡ 367 × 0 (mod 1) ≡ 0 (mod 1)
x ≡ 457 × 516-1 (mod 943) ≡ 457 × 53 (mod 943) ≡ 646 (mod 943)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 583 × 1 × 943 = 549769
  2. We calculate the numbers M1 to M3
    M1=M/m1=549769/583=943,   M2=M/m2=549769/1=549769,   M3=M/m3=549769/943=583
  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
    5839430583010
    9435831360101
    583360122301-1
    36022311371-12
    223137186-12-3
    137861512-35
    8651135-35-8
    51351165-813
    351623-813-34
    1635113-34183
    3130-34183-583
    So our multiplicative inverse is 183 mod 583 ≡ 183
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    154976901010
    54976915497690101
    So our multiplicative inverse is 0 mod 1 ≡ 0
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    943583136001-1
    58336012231-12
    3602231137-12-3
    2231371862-35
    13786151-35-8
    86511355-813
    5135116-813-21
    35162313-2155
    16351-2155-296
    313055-296943
    So our multiplicative inverse is -296 mod 943 ≡ 647
    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
    =  (344 × 943 × 183 +
       0 × 549769 × 0 +
       646 × 583 × 647)   mod 549769
    = 112863 (mod 549769)


    So our answer is 112863 (mod 549769).


Verification

So we found that x ≡ 112863
If this is correct, then the following statements (i.e. the original equations) are true:
127x (mod 583) ≡ 546 (mod 583)
205x (mod 1) ≡ 367 (mod 1)
516x (mod 943) ≡ 457 (mod 943)

Let's see whether that's indeed the case if we use x ≡ 112863.