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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
2937835901-3
78591191-34
591932-34-15
192914-15139
2120-15139-293
So our multiplicative inverse is 139 mod 293 ≡ 139
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1743070174010
3071741133101
17413314101-1
133413101-14
411041-14-17
1011004-17174
So our multiplicative inverse is -17 mod 174 ≡ 157
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1035970103010
597103582101
1038212101-1
82213191-14
211912-14-5
192914-549
2120-549-103
So our multiplicative inverse is 49 mod 103 ≡ 49
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 ≡ 932 × 78-1 (mod 293) ≡ 932 × 139 (mod 293) ≡ 42 (mod 293)
x ≡ 330 × 307-1 (mod 174) ≡ 330 × 157 (mod 174) ≡ 132 (mod 174)
x ≡ 990 × 597-1 (mod 103) ≡ 990 × 49 (mod 103) ≡ 100 (mod 103)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 293 × 174 × 103 = 5251146
  2. We calculate the numbers M1 to M3
    M1=M/m1=5251146/293=17922,   M2=M/m2=5251146/174=30179,   M3=M/m3=5251146/103=50982
  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
    293179220293010
    179222936149101
    2934954801-5
    4948111-56
    481480-56-293
    So our multiplicative inverse is 6 mod 293 ≡ 6
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    174301790174010
    3017917417377101
    1747722001-2
    77203171-27
    201713-27-9
    173527-952
    3211-952-61
    212052-61174
    So our multiplicative inverse is -61 mod 174 ≡ 113
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    103509820103010
    50982103494100101
    1031001301-1
    10033311-134
    3130-134-103
    So our multiplicative inverse is 34 mod 103 ≡ 34
    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
    =  (42 × 17922 × 6 +
       132 × 30179 × 113 +
       100 × 50982 × 34)   mod 5251146
    = 3118734 (mod 5251146)


    So our answer is 3118734 (mod 5251146).


Verification

So we found that x ≡ 3118734
If this is correct, then the following statements (i.e. the original equations) are true:
78x (mod 293) ≡ 932 (mod 293)
307x (mod 174) ≡ 330 (mod 174)
597x (mod 103) ≡ 990 (mod 103)

Let's see whether that's indeed the case if we use x ≡ 3118734.