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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!

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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
1474150147010
4151472121101
14712112601-1
121264171-15
261719-15-6
179185-611
9811-611-17
818011-17147
So our multiplicative inverse is -17 mod 147 ≡ 130
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
971732123901-1
7322393151-14
239151514-14-61
1514114-6165
141140-6165-971
So our multiplicative inverse is 65 mod 971 ≡ 65
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
17801017010
80117472101
1728101-8
21201-817
So our multiplicative inverse is -8 mod 17 ≡ 9
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 ≡ 430 × 415-1 (mod 147) ≡ 430 × 130 (mod 147) ≡ 40 (mod 147)
x ≡ 769 × 732-1 (mod 971) ≡ 769 × 65 (mod 971) ≡ 464 (mod 971)
x ≡ 902 × 801-1 (mod 17) ≡ 902 × 9 (mod 17) ≡ 9 (mod 17)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 147 × 971 × 17 = 2426529
  2. We calculate the numbers M1 to M3
    M1=M/m1=2426529/147=16507,   M2=M/m2=2426529/971=2499,   M3=M/m3=2426529/17=142737
  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
    147165070147010
    1650714711243101
    1474331801-3
    4318271-37
    18724-37-17
    74137-1724
    4311-1724-41
    313024-41147
    So our multiplicative inverse is -41 mod 147 ≡ 106
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    97124990971010
    24999712557101
    971557141401-1
    55741411431-12
    4141432128-12-5
    1431281152-57
    1281588-57-61
    158177-6168
    8711-6168-129
    717068-129971
    So our multiplicative inverse is -129 mod 971 ≡ 842
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    17142737017010
    1427371783965101
    1753201-3
    52211-37
    2120-37-17
    So our multiplicative inverse is 7 mod 17 ≡ 7
    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
    =  (40 × 16507 × 106 +
       464 × 2499 × 842 +
       9 × 142737 × 7)   mod 2426529
    = 2197837 (mod 2426529)


    So our answer is 2197837 (mod 2426529).


Verification

So we found that x ≡ 2197837
If this is correct, then the following statements (i.e. the original equations) are true:
415x (mod 147) ≡ 430 (mod 147)
732x (mod 971) ≡ 769 (mod 971)
801x (mod 17) ≡ 902 (mod 17)

Let's see whether that's indeed the case if we use x ≡ 2197837.