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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
5239300523010
9305231407101
523407111601-1
4071163591-14
11659157-14-5
5957124-59
572281-59-257
21209-257523
So our multiplicative inverse is -257 mod 523 ≡ 266
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
56954412501-1
5442521191-122
251916-122-23
1963122-2391
6160-2391-569
So our multiplicative inverse is 91 mod 569 ≡ 91
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
25933025010
93325378101
2583101-3
81801-325
So our multiplicative inverse is -3 mod 25 ≡ 22
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 ≡ 63 × 930-1 (mod 523) ≡ 63 × 266 (mod 523) ≡ 22 (mod 523)
x ≡ 55 × 544-1 (mod 569) ≡ 55 × 91 (mod 569) ≡ 453 (mod 569)
x ≡ 847 × 933-1 (mod 25) ≡ 847 × 22 (mod 25) ≡ 9 (mod 25)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 523 × 569 × 25 = 7439675
  2. We calculate the numbers M1 to M3
    M1=M/m1=7439675/523=14225,   M2=M/m2=7439675/569=13075,   M3=M/m3=7439675/25=297587
  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
    523142250523010
    1422552327104101
    5231045301-5
    10433421-5171
    3211-5171-176
    2120171-176523
    So our multiplicative inverse is -176 mod 523 ≡ 347
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    569130750569010
    1307556922557101
    56955711201-1
    557124651-147
    12522-147-95
    522147-95237
    2120-95237-569
    So our multiplicative inverse is 237 mod 569 ≡ 237
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    25297587025010
    297587251190312101
    25122101-2
    1211201-225
    So our multiplicative inverse is -2 mod 25 ≡ 23
    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
    =  (22 × 14225 × 347 +
       453 × 13075 × 237 +
       9 × 297587 × 23)   mod 7439675
    = 4167809 (mod 7439675)


    So our answer is 4167809 (mod 7439675).


Verification

So we found that x ≡ 4167809
If this is correct, then the following statements (i.e. the original equations) are true:
930x (mod 523) ≡ 63 (mod 523)
544x (mod 569) ≡ 55 (mod 569)
933x (mod 25) ≡ 847 (mod 25)

Let's see whether that's indeed the case if we use x ≡ 4167809.