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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
759413134601-1
4133461671-12
34667511-12-11
6711612-1168
111110-1168-759
So our multiplicative inverse is 68 mod 759 ≡ 68
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1999430199010
9431994147101
19914715201-1
147522431-13
524319-13-4
439473-419
9712-419-23
723119-2388
2120-2388-199
So our multiplicative inverse is 88 mod 199 ≡ 88
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
59206059010
20659329101
59292101-2
2912901-259
So our multiplicative inverse is -2 mod 59 ≡ 57
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 ≡ 412 × 413-1 (mod 759) ≡ 412 × 68 (mod 759) ≡ 692 (mod 759)
x ≡ 841 × 943-1 (mod 199) ≡ 841 × 88 (mod 199) ≡ 179 (mod 199)
x ≡ 919 × 206-1 (mod 59) ≡ 919 × 57 (mod 59) ≡ 50 (mod 59)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 759 × 199 × 59 = 8911419
  2. We calculate the numbers M1 to M3
    M1=M/m1=8911419/759=11741,   M2=M/m2=8911419/199=44781,   M3=M/m3=8911419/59=151041
  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
    759117410759010
    1174175915356101
    75935624701-2
    356477271-215
    4727120-215-17
    27201715-1732
    20726-1732-81
    761132-81113
    6160-81113-759
    So our multiplicative inverse is 113 mod 759 ≡ 113
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    199447810199010
    447811992256101
    199633101-33
    61601-33199
    So our multiplicative inverse is -33 mod 199 ≡ 166
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    59151041059010
    1510415925601101
    59159001-59
    So our multiplicative inverse is 1 mod 59 ≡ 1
    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
    =  (692 × 11741 × 113 +
       179 × 44781 × 166 +
       50 × 151041 × 1)   mod 8911419
    = 1684913 (mod 8911419)


    So our answer is 1684913 (mod 8911419).


Verification

So we found that x ≡ 1684913
If this is correct, then the following statements (i.e. the original equations) are true:
413x (mod 759) ≡ 412 (mod 759)
943x (mod 199) ≡ 841 (mod 199)
206x (mod 59) ≡ 919 (mod 59)

Let's see whether that's indeed the case if we use x ≡ 1684913.