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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
63111555601-5
11556231-511
563182-511-203
321111-203214
2120-203214-631
So our multiplicative inverse is 214 mod 631 ≡ 214
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
47723201701-20
2317161-2021
17625-2021-62
651121-6283
5150-6283-477
So our multiplicative inverse is 83 mod 477 ≡ 83
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3116270311010
62731125101
311562101-62
51501-62311
So our multiplicative inverse is -62 mod 311 ≡ 249
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 ≡ 399 × 115-1 (mod 631) ≡ 399 × 214 (mod 631) ≡ 201 (mod 631)
x ≡ 818 × 23-1 (mod 477) ≡ 818 × 83 (mod 477) ≡ 160 (mod 477)
x ≡ 967 × 627-1 (mod 311) ≡ 967 × 249 (mod 311) ≡ 69 (mod 311)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 631 × 477 × 311 = 93606957
  2. We calculate the numbers M1 to M3
    M1=M/m1=93606957/631=148347,   M2=M/m2=93606957/477=196241,   M3=M/m3=93606957/311=300987
  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
    6311483470631010
    14834763123562101
    63162101101-10
    6211571-1051
    11714-1051-61
    741351-61112
    4311-61112-173
    3130112-173631
    So our multiplicative inverse is -173 mod 631 ≡ 458
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4771962410477010
    196241477411194101
    47719428901-2
    194892161-25
    891659-25-27
    169175-2732
    9712-2732-59
    723132-59209
    2120-59209-477
    So our multiplicative inverse is 209 mod 477 ≡ 209
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    3113009870311010
    300987311967250101
    31125016101-1
    25061461-15
    616101-15-51
    61605-51311
    So our multiplicative inverse is -51 mod 311 ≡ 260
    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
    =  (201 × 148347 × 458 +
       160 × 196241 × 209 +
       69 × 300987 × 260)   mod 93606957
    = 63834685 (mod 93606957)


    So our answer is 63834685 (mod 93606957).


Verification

So we found that x ≡ 63834685
If this is correct, then the following statements (i.e. the original equations) are true:
115x (mod 631) ≡ 399 (mod 631)
23x (mod 477) ≡ 818 (mod 477)
627x (mod 311) ≡ 967 (mod 311)

Let's see whether that's indeed the case if we use x ≡ 63834685.