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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
7019360701010
9367011235101
701235223101-2
235231141-23
2314573-23-173
43113-173176
3130-173176-701
So our multiplicative inverse is 176 mod 701 ≡ 176
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
999700129901-1
70029921021-13
299102295-13-7
10295173-710
957134-710-137
741310-137147
4311-137147-284
3130147-284999
So our multiplicative inverse is -284 mod 999 ≡ 715
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1076514201-1
65421231-12
4223119-12-3
2319142-35
19443-35-23
43115-2328
3130-2328-107
So our multiplicative inverse is 28 mod 107 ≡ 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 ≡ 704 × 936-1 (mod 701) ≡ 704 × 176 (mod 701) ≡ 528 (mod 701)
x ≡ 36 × 700-1 (mod 999) ≡ 36 × 715 (mod 999) ≡ 765 (mod 999)
x ≡ 134 × 65-1 (mod 107) ≡ 134 × 28 (mod 107) ≡ 7 (mod 107)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 701 × 999 × 107 = 74931993
  2. We calculate the numbers M1 to M3
    M1=M/m1=74931993/701=106893,   M2=M/m2=74931993/999=75007,   M3=M/m3=74931993/107=700299
  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
    7011068930701010
    106893701152341101
    70134121901-2
    3411917181-235
    191811-235-37
    18118035-37701
    So our multiplicative inverse is -37 mod 701 ≡ 664
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    999750070999010
    750079997582101
    99982121501-12
    8215571-1261
    15721-1261-134
    717061-134999
    So our multiplicative inverse is -134 mod 999 ≡ 865
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1077002990107010
    700299107654491101
    1079111601-1
    91165111-16
    161115-16-7
    115216-720
    5150-720-107
    So our multiplicative inverse is 20 mod 107 ≡ 20
    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
    =  (528 × 106893 × 664 +
       765 × 75007 × 865 +
       7 × 700299 × 20)   mod 74931993
    = 61971732 (mod 74931993)


    So our answer is 61971732 (mod 74931993).


Verification

So we found that x ≡ 61971732
If this is correct, then the following statements (i.e. the original equations) are true:
936x (mod 701) ≡ 704 (mod 701)
700x (mod 999) ≡ 36 (mod 999)
65x (mod 107) ≡ 134 (mod 107)

Let's see whether that's indeed the case if we use x ≡ 61971732.