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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
163812101-2
8118101-2163
So our multiplicative inverse is -2 mod 163 ≡ 161
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2435620243010
562243276101
2437631501-3
7615511-316
151150-316-243
So our multiplicative inverse is 16 mod 243 ≡ 16
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4435580443010
5584431115101
44311539801-3
115981171-34
9817513-34-23
1713144-2327
13431-2327-104
414027-104443
So our multiplicative inverse is -104 mod 443 ≡ 339
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 ≡ 449 × 81-1 (mod 163) ≡ 449 × 161 (mod 163) ≡ 80 (mod 163)
x ≡ 955 × 562-1 (mod 243) ≡ 955 × 16 (mod 243) ≡ 214 (mod 243)
x ≡ 530 × 558-1 (mod 443) ≡ 530 × 339 (mod 443) ≡ 255 (mod 443)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 163 × 243 × 443 = 17546787
  2. We calculate the numbers M1 to M3
    M1=M/m1=17546787/163=107649,   M2=M/m2=17546787/243=72209,   M3=M/m3=17546787/443=39609
  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
    1631076490163010
    10764916366069101
    1636922501-2
    69252191-25
    251916-25-7
    196315-726
    6160-726-163
    So our multiplicative inverse is 26 mod 163 ≡ 26
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    243722090243010
    7220924329738101
    2433861501-6
    3815281-613
    15817-613-19
    871113-1932
    7170-1932-243
    So our multiplicative inverse is 32 mod 243 ≡ 32
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    443396090443010
    3960944389182101
    44318227901-2
    182792241-25
    792437-25-17
    247335-1756
    7321-1756-129
    313056-129443
    So our multiplicative inverse is -129 mod 443 ≡ 314
    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
    =  (80 × 107649 × 26 +
       214 × 72209 × 32 +
       255 × 39609 × 314)   mod 17546787
    = 12049855 (mod 17546787)


    So our answer is 12049855 (mod 17546787).


Verification

So we found that x ≡ 12049855
If this is correct, then the following statements (i.e. the original equations) are true:
81x (mod 163) ≡ 449 (mod 163)
562x (mod 243) ≡ 955 (mod 243)
558x (mod 443) ≡ 530 (mod 443)

Let's see whether that's indeed the case if we use x ≡ 12049855.