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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
53517053010
51753940101
534011301-1
4013311-14
131130-14-53
So our multiplicative inverse is 4 mod 53 ≡ 4
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3258870325010
8873252237101
32523718801-1
237882611-13
8861127-13-4
6127273-411
27736-411-37
761111-3748
6160-3748-325
So our multiplicative inverse is 48 mod 325 ≡ 48
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
40913231301-3
132131021-331
13261-331-189
212031-189409
So our multiplicative inverse is -189 mod 409 ≡ 220
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 ≡ 631 × 517-1 (mod 53) ≡ 631 × 4 (mod 53) ≡ 33 (mod 53)
x ≡ 60 × 887-1 (mod 325) ≡ 60 × 48 (mod 325) ≡ 280 (mod 325)
x ≡ 937 × 132-1 (mod 409) ≡ 937 × 220 (mod 409) ≡ 4 (mod 409)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 53 × 325 × 409 = 7045025
  2. We calculate the numbers M1 to M3
    M1=M/m1=7045025/53=132925,   M2=M/m2=7045025/325=21677,   M3=M/m3=7045025/409=17225
  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
    53132925053010
    1329255325081101
    53153001-53
    So our multiplicative inverse is 1 mod 53 ≡ 1
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    325216770325010
    2167732566227101
    32522719801-1
    227982311-13
    983135-13-10
    315613-1063
    5150-1063-325
    So our multiplicative inverse is 63 mod 325 ≡ 63
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    409172250409010
    172254094247101
    4094783301-8
    47331141-89
    331425-89-26
    145249-2661
    5411-2661-87
    414061-87409
    So our multiplicative inverse is -87 mod 409 ≡ 322
    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
    =  (33 × 132925 × 1 +
       280 × 21677 × 63 +
       4 × 17225 × 322)   mod 7045025
    = 343155 (mod 7045025)


    So our answer is 343155 (mod 7045025).


Verification

So we found that x ≡ 343155
If this is correct, then the following statements (i.e. the original equations) are true:
517x (mod 53) ≡ 631 (mod 53)
887x (mod 325) ≡ 60 (mod 325)
132x (mod 409) ≡ 937 (mod 409)

Let's see whether that's indeed the case if we use x ≡ 343155.