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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
847479136801-1
47936811111-12
368111335-12-7
11135362-723
35655-723-122
651123-122145
5150-122145-847
So our multiplicative inverse is 145 mod 847 ≡ 145
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
74691501-1
6951341-114
5411-114-15
414014-1574
So our multiplicative inverse is -15 mod 74 ≡ 59
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
73317742501-4
17725721-429
252121-429-352
212029-352733
So our multiplicative inverse is -352 mod 733 ≡ 381
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 ≡ 856 × 479-1 (mod 847) ≡ 856 × 145 (mod 847) ≡ 458 (mod 847)
x ≡ 447 × 69-1 (mod 74) ≡ 447 × 59 (mod 74) ≡ 29 (mod 74)
x ≡ 416 × 177-1 (mod 733) ≡ 416 × 381 (mod 733) ≡ 168 (mod 733)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 847 × 74 × 733 = 45942974
  2. We calculate the numbers M1 to M3
    M1=M/m1=45942974/847=54242,   M2=M/m2=45942974/74=620851,   M3=M/m3=45942974/733=62678
  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
    847542420847010
    542428476434101
    84734243101-24
    3431131-2425
    313101-2425-274
    313025-274847
    So our multiplicative inverse is -274 mod 847 ≡ 573
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    74620851074010
    62085174838965101
    74651901-1
    659721-18
    9241-18-33
    21208-3374
    So our multiplicative inverse is -33 mod 74 ≡ 41
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    733626780733010
    6267873385373101
    733373136001-1
    3733601131-12
    36013279-12-55
    139142-5557
    9421-5557-169
    414057-169733
    So our multiplicative inverse is -169 mod 733 ≡ 564
    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
    =  (458 × 54242 × 573 +
       29 × 620851 × 41 +
       168 × 62678 × 564)   mod 45942974
    = 7949553 (mod 45942974)


    So our answer is 7949553 (mod 45942974).


Verification

So we found that x ≡ 7949553
If this is correct, then the following statements (i.e. the original equations) are true:
479x (mod 847) ≡ 856 (mod 847)
69x (mod 74) ≡ 447 (mod 74)
177x (mod 733) ≡ 416 (mod 733)

Let's see whether that's indeed the case if we use x ≡ 7949553.