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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
11596011010
59611542101
1125101-5
21201-511
So our multiplicative inverse is -5 mod 11 ≡ 6
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1365110136010
5111363103101
13610313301-1
10333341-14
33481-14-33
41404-33136
So our multiplicative inverse is -33 mod 136 ≡ 103
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
4535150453010
515453162101
4536271901-7
6219351-722
19534-722-73
541122-7395
4140-7395-453
So our multiplicative inverse is 95 mod 453 ≡ 95
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 ≡ 902 × 596-1 (mod 11) ≡ 902 × 6 (mod 11) ≡ 0 (mod 11)
x ≡ 485 × 511-1 (mod 136) ≡ 485 × 103 (mod 136) ≡ 43 (mod 136)
x ≡ 2 × 515-1 (mod 453) ≡ 2 × 95 (mod 453) ≡ 190 (mod 453)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 11 × 136 × 453 = 677688
  2. We calculate the numbers M1 to M3
    M1=M/m1=677688/11=61608,   M2=M/m2=677688/136=4983,   M3=M/m3=677688/453=1496
  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
    1161608011010
    616081156008101
    1181301-1
    83221-13
    3211-13-4
    21203-411
    So our multiplicative inverse is -4 mod 11 ≡ 7
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    13649830136010
    49831363687101
    1368714901-1
    87491381-12
    4938111-12-3
    3811352-311
    11521-311-25
    515011-25136
    So our multiplicative inverse is -25 mod 136 ≡ 111
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    45314960453010
    14964533137101
    45313734201-3
    137423111-310
    421139-310-33
    1191210-3343
    9241-3343-205
    212043-205453
    So our multiplicative inverse is -205 mod 453 ≡ 248
    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
    =  (0 × 61608 × 7 +
       43 × 4983 × 111 +
       190 × 1496 × 248)   mod 677688
    = 76747 (mod 677688)


    So our answer is 76747 (mod 677688).


Verification

So we found that x ≡ 76747
If this is correct, then the following statements (i.e. the original equations) are true:
596x (mod 11) ≡ 902 (mod 11)
511x (mod 136) ≡ 485 (mod 136)
515x (mod 453) ≡ 2 (mod 453)

Let's see whether that's indeed the case if we use x ≡ 76747.