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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
5039060503010
9065031403101
503403110001-1
403100431-15
1003331-15-166
31305-166503
So our multiplicative inverse is -166 mod 503 ≡ 337
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
384103010
84132801101
313001-3
So our multiplicative inverse is 1 mod 3 ≡ 1
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
58726725301-2
26753521-211
532261-211-288
212011-288587
So our multiplicative inverse is -288 mod 587 ≡ 299
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 ≡ 544 × 906-1 (mod 503) ≡ 544 × 337 (mod 503) ≡ 236 (mod 503)
x ≡ 352 × 841-1 (mod 3) ≡ 352 × 1 (mod 3) ≡ 1 (mod 3)
x ≡ 367 × 267-1 (mod 587) ≡ 367 × 299 (mod 587) ≡ 551 (mod 587)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 503 × 3 × 587 = 885783
  2. We calculate the numbers M1 to M3
    M1=M/m1=885783/503=1761,   M2=M/m2=885783/3=295261,   M3=M/m3=885783/587=1509
  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
    50317610503010
    17615033252101
    503252125101-1
    252251111-12
    25112510-12-503
    So our multiplicative inverse is 2 mod 503 ≡ 2
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    329526103010
    2952613984201101
    313001-3
    So our multiplicative inverse is 1 mod 3 ≡ 1
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    58715090587010
    15095872335101
    587335125201-1
    3352521831-12
    2528333-12-7
    8332722-7191
    3211-7191-198
    2120191-198587
    So our multiplicative inverse is -198 mod 587 ≡ 389
    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
    =  (236 × 1761 × 2 +
       1 × 295261 × 1 +
       551 × 1509 × 389)   mod 885783
    = 367426 (mod 885783)


    So our answer is 367426 (mod 885783).


Verification

So we found that x ≡ 367426
If this is correct, then the following statements (i.e. the original equations) are true:
906x (mod 503) ≡ 544 (mod 503)
841x (mod 3) ≡ 352 (mod 3)
267x (mod 587) ≡ 367 (mod 587)

Let's see whether that's indeed the case if we use x ≡ 367426.