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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
671421125001-1
42125011711-12
250171179-12-3
171792132-38
791361-38-51
1311308-51671
So our multiplicative inverse is -51 mod 671 ≡ 620
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
52950212701-1
5022718161-119
2716111-119-20
16111519-2039
11521-2039-98
515039-98529
So our multiplicative inverse is -98 mod 529 ≡ 431
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1937650193010
7651933186101
1931861701-1
18672641-127
7413-127-28
431127-2855
3130-2855-193
So our multiplicative inverse is 55 mod 193 ≡ 55
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 ≡ 677 × 421-1 (mod 671) ≡ 677 × 620 (mod 671) ≡ 365 (mod 671)
x ≡ 765 × 502-1 (mod 529) ≡ 765 × 431 (mod 529) ≡ 148 (mod 529)
x ≡ 497 × 765-1 (mod 193) ≡ 497 × 55 (mod 193) ≡ 122 (mod 193)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 671 × 529 × 193 = 68507087
  2. We calculate the numbers M1 to M3
    M1=M/m1=68507087/671=102097,   M2=M/m2=68507087/529=129503,   M3=M/m3=68507087/193=354959
  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
    6711020970671010
    102097671152105101
    67110564101-6
    105412231-613
    4123118-613-19
    23181513-1932
    18533-1932-115
    531232-115147
    3211-115147-262
    2120147-262671
    So our multiplicative inverse is -262 mod 671 ≡ 409
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5291295030529010
    129503529244427101
    529427110201-1
    4271024191-15
    1021957-15-26
    197255-2657
    7512-2657-83
    522157-83223
    2120-83223-529
    So our multiplicative inverse is 223 mod 529 ≡ 223
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1933549590193010
    354959193183932101
    193326101-6
    3213201-6193
    So our multiplicative inverse is -6 mod 193 ≡ 187
    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
    =  (365 × 102097 × 409 +
       148 × 129503 × 223 +
       122 × 354959 × 187)   mod 68507087
    = 5346222 (mod 68507087)


    So our answer is 5346222 (mod 68507087).


Verification

So we found that x ≡ 5346222
If this is correct, then the following statements (i.e. the original equations) are true:
421x (mod 671) ≡ 677 (mod 671)
502x (mod 529) ≡ 765 (mod 529)
765x (mod 193) ≡ 497 (mod 193)

Let's see whether that's indeed the case if we use x ≡ 5346222.