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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
8032653801-3
26583311-3100
8180-3100-803
So our multiplicative inverse is 100 mod 803 ≡ 100
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
829729110001-1
7291007291-18
10029313-18-25
2913238-2558
13341-2558-257
313058-257829
So our multiplicative inverse is -257 mod 829 ≡ 572
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3373820337010
382337145101
3374572201-7
4522211-715
221220-715-337
So our multiplicative inverse is 15 mod 337 ≡ 15
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 ≡ 586 × 265-1 (mod 803) ≡ 586 × 100 (mod 803) ≡ 784 (mod 803)
x ≡ 929 × 729-1 (mod 829) ≡ 929 × 572 (mod 829) ≡ 828 (mod 829)
x ≡ 297 × 382-1 (mod 337) ≡ 297 × 15 (mod 337) ≡ 74 (mod 337)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 803 × 829 × 337 = 224336519
  2. We calculate the numbers M1 to M3
    M1=M/m1=224336519/803=279373,   M2=M/m2=224336519/829=270611,   M3=M/m3=224336519/337=665687
  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
    8032793730803010
    279373803347732101
    80373217101-1
    7327110221-111
    712235-111-34
    2254211-34147
    5221-34147-328
    2120147-328803
    So our multiplicative inverse is -328 mod 803 ≡ 475
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    8292706110829010
    270611829326357101
    829357211501-2
    3571153121-27
    1151297-27-65
    127157-6572
    7512-6572-137
    522172-137346
    2120-137346-829
    So our multiplicative inverse is 346 mod 829 ≡ 346
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    3376656870337010
    6656873371975112101
    3371123101-3
    112111201-3337
    So our multiplicative inverse is -3 mod 337 ≡ 334
    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
    =  (784 × 279373 × 475 +
       828 × 270611 × 346 +
       74 × 665687 × 334)   mod 224336519
    = 153619502 (mod 224336519)


    So our answer is 153619502 (mod 224336519).


Verification

So we found that x ≡ 153619502
If this is correct, then the following statements (i.e. the original equations) are true:
265x (mod 803) ≡ 586 (mod 803)
729x (mod 829) ≡ 929 (mod 829)
382x (mod 337) ≡ 297 (mod 337)

Let's see whether that's indeed the case if we use x ≡ 153619502.