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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
49821049010
821491637101
493711201-1
3712311-14
121120-14-49
So our multiplicative inverse is 4 mod 49 ≡ 4
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
439112310301-3
112103191-34
1039114-34-47
94214-4798
4140-4798-439
So our multiplicative inverse is 98 mod 439 ≡ 98
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1933240193010
3241931131101
19313116201-1
13162271-13
62786-13-25
76113-2528
6160-2528-193
So our multiplicative inverse is 28 mod 193 ≡ 28
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 ≡ 362 × 821-1 (mod 49) ≡ 362 × 4 (mod 49) ≡ 27 (mod 49)
x ≡ 202 × 112-1 (mod 439) ≡ 202 × 98 (mod 439) ≡ 41 (mod 439)
x ≡ 258 × 324-1 (mod 193) ≡ 258 × 28 (mod 193) ≡ 83 (mod 193)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 49 × 439 × 193 = 4151623
  2. We calculate the numbers M1 to M3
    M1=M/m1=4151623/49=84727,   M2=M/m2=4151623/439=9457,   M3=M/m3=4151623/193=21511
  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
    4984727049010
    847274917296101
    4968101-8
    61601-849
    So our multiplicative inverse is -8 mod 49 ≡ 41
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    43994570439010
    945743921238101
    439238120101-1
    2382011371-12
    20137516-12-11
    3716252-1124
    16531-1124-83
    515024-83439
    So our multiplicative inverse is -83 mod 439 ≡ 356
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    193215110193010
    2151119311188101
    1938821701-2
    8817531-211
    17352-211-57
    321111-5768
    2120-5768-193
    So our multiplicative inverse is 68 mod 193 ≡ 68
    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
    =  (27 × 84727 × 41 +
       41 × 9457 × 356 +
       83 × 21511 × 68)   mod 4151623
    = 347290 (mod 4151623)


    So our answer is 347290 (mod 4151623).


Verification

So we found that x ≡ 347290
If this is correct, then the following statements (i.e. the original equations) are true:
821x (mod 49) ≡ 362 (mod 49)
112x (mod 439) ≡ 202 (mod 439)
324x (mod 193) ≡ 258 (mod 193)

Let's see whether that's indeed the case if we use x ≡ 347290.