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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
3274130327010
413327186101
3278636901-3
86691171-34
691741-34-19
1711704-19327
So our multiplicative inverse is -19 mod 327 ≡ 308
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
7527530752010
75375211101
7521752001-752
So our multiplicative inverse is 1 mod 752 ≡ 1
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2158160215010
8162153171101
21517114401-1
171443391-14
443915-14-5
395744-539
5411-539-44
414039-44215
So our multiplicative inverse is -44 mod 215 ≡ 171
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 ≡ 217 × 413-1 (mod 327) ≡ 217 × 308 (mod 327) ≡ 128 (mod 327)
x ≡ 197 × 753-1 (mod 752) ≡ 197 × 1 (mod 752) ≡ 197 (mod 752)
x ≡ 505 × 816-1 (mod 215) ≡ 505 × 171 (mod 215) ≡ 140 (mod 215)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 327 × 752 × 215 = 52869360
  2. We calculate the numbers M1 to M3
    M1=M/m1=52869360/327=161680,   M2=M/m2=52869360/752=70305,   M3=M/m3=52869360/215=245904
  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
    3271616800327010
    161680327494142101
    32714224301-2
    142433131-27
    431334-27-23
    134317-2376
    4140-2376-327
    So our multiplicative inverse is 76 mod 327 ≡ 76
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    752703050752010
    7030575293369101
    75236921401-2
    369142651-253
    14524-253-108
    541153-108161
    4140-108161-752
    So our multiplicative inverse is 161 mod 752 ≡ 161
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    2152459040215010
    2459042151143159101
    21515915601-1
    159562471-13
    564719-13-4
    479523-423
    9241-423-96
    212023-96215
    So our multiplicative inverse is -96 mod 215 ≡ 119
    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
    =  (128 × 161680 × 76 +
       197 × 70305 × 161 +
       140 × 245904 × 119)   mod 52869360
    = 21912725 (mod 52869360)


    So our answer is 21912725 (mod 52869360).


Verification

So we found that x ≡ 21912725
If this is correct, then the following statements (i.e. the original equations) are true:
413x (mod 327) ≡ 217 (mod 327)
753x (mod 752) ≡ 197 (mod 752)
816x (mod 215) ≡ 505 (mod 215)

Let's see whether that's indeed the case if we use x ≡ 21912725.