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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
1739810173010
9811735116101
17311615701-1
11657221-13
572281-13-85
21203-85173
So our multiplicative inverse is -85 mod 173 ≡ 88
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2978150297010
8152972221101
29722117601-1
221762691-13
766917-13-4
697963-439
7611-439-43
616039-43297
So our multiplicative inverse is -43 mod 297 ≡ 254
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
53192053010
19253333101
533312001-1
33201131-12
201317-12-3
137162-35
7611-35-8
61605-853
So our multiplicative inverse is -8 mod 53 ≡ 45
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 ≡ 910 × 981-1 (mod 173) ≡ 910 × 88 (mod 173) ≡ 154 (mod 173)
x ≡ 405 × 815-1 (mod 297) ≡ 405 × 254 (mod 297) ≡ 108 (mod 297)
x ≡ 983 × 192-1 (mod 53) ≡ 983 × 45 (mod 53) ≡ 33 (mod 53)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 173 × 297 × 53 = 2723193
  2. We calculate the numbers M1 to M3
    M1=M/m1=2723193/173=15741,   M2=M/m2=2723193/297=9169,   M3=M/m3=2723193/53=51381
  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
    173157410173010
    1574117390171101
    1731711201-1
    17128511-186
    2120-186-173
    So our multiplicative inverse is 86 mod 173 ≡ 86
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    29791690297010
    916929730259101
    29725913801-1
    259386311-17
    383117-17-8
    317437-839
    7321-839-86
    313039-86297
    So our multiplicative inverse is -86 mod 297 ≡ 211
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5351381053010
    513815396924101
    53242501-2
    245441-29
    5411-29-11
    41409-1153
    So our multiplicative inverse is -11 mod 53 ≡ 42
    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
    =  (154 × 15741 × 86 +
       108 × 9169 × 211 +
       33 × 51381 × 42)   mod 2723193
    = 1179495 (mod 2723193)


    So our answer is 1179495 (mod 2723193).


Verification

So we found that x ≡ 1179495
If this is correct, then the following statements (i.e. the original equations) are true:
981x (mod 173) ≡ 910 (mod 173)
815x (mod 297) ≡ 405 (mod 297)
192x (mod 53) ≡ 983 (mod 53)

Let's see whether that's indeed the case if we use x ≡ 1179495.