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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!

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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
431323110801-1
32310821071-13
10810711-13-4
107110703-4431
So our multiplicative inverse is -4 mod 431 ≡ 427
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
41949041010
94941236101
4166501-6
65111-67
5150-67-41
So our multiplicative inverse is 7 mod 41 ≡ 7
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
887746114101-1
7461415411-16
14141318-16-19
4118256-1944
18533-1944-151
531244-151195
3211-151195-346
2120195-346887
So our multiplicative inverse is -346 mod 887 ≡ 541
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 ≡ 322 × 323-1 (mod 431) ≡ 322 × 427 (mod 431) ≡ 5 (mod 431)
x ≡ 771 × 949-1 (mod 41) ≡ 771 × 7 (mod 41) ≡ 26 (mod 41)
x ≡ 114 × 746-1 (mod 887) ≡ 114 × 541 (mod 887) ≡ 471 (mod 887)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 431 × 41 × 887 = 15674177
  2. We calculate the numbers M1 to M3
    M1=M/m1=15674177/431=36367,   M2=M/m2=15674177/41=382297,   M3=M/m3=15674177/887=17671
  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
    431363670431010
    3636743184163101
    431163210501-2
    1631051581-23
    10558147-23-5
    58471113-58
    471143-58-37
    113328-37119
    3211-37119-156
    2120119-156431
    So our multiplicative inverse is -156 mod 431 ≡ 275
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    41382297041010
    38229741932413101
    41133201-3
    132611-319
    2120-319-41
    So our multiplicative inverse is 19 mod 41 ≡ 19
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    887176710887010
    1767188719818101
    88781816901-1
    8186911591-112
    6959110-112-13
    59105912-1377
    10911-1377-90
    919077-90887
    So our multiplicative inverse is -90 mod 887 ≡ 797
    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
    =  (5 × 36367 × 275 +
       26 × 382297 × 19 +
       471 × 17671 × 797)   mod 15674177
    = 7033494 (mod 15674177)


    So our answer is 7033494 (mod 15674177).


Verification

So we found that x ≡ 7033494
If this is correct, then the following statements (i.e. the original equations) are true:
323x (mod 431) ≡ 322 (mod 431)
949x (mod 41) ≡ 771 (mod 41)
746x (mod 887) ≡ 114 (mod 887)

Let's see whether that's indeed the case if we use x ≡ 7033494.