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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
1583530158010
353158237101
1583741001-4
3710371-413
10713-413-17
732113-1747
3130-1747-158
So our multiplicative inverse is 47 mod 158 ≡ 47
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
5113415101-15
3413401-15511
So our multiplicative inverse is -15 mod 511 ≡ 496
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1912770191010
277191186101
1918621901-2
86194101-29
191019-29-11
109119-1120
9190-1120-191
So our multiplicative inverse is 20 mod 191 ≡ 20
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 ≡ 524 × 353-1 (mod 158) ≡ 524 × 47 (mod 158) ≡ 138 (mod 158)
x ≡ 352 × 34-1 (mod 511) ≡ 352 × 496 (mod 511) ≡ 341 (mod 511)
x ≡ 739 × 277-1 (mod 191) ≡ 739 × 20 (mod 191) ≡ 73 (mod 191)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 158 × 511 × 191 = 15420958
  2. We calculate the numbers M1 to M3
    M1=M/m1=15420958/158=97601,   M2=M/m2=15420958/511=30178,   M3=M/m3=15420958/191=80738
  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
    158976010158010
    97601158617115101
    15811514301-1
    115432291-13
    4329114-13-4
    2914213-411
    141140-411-158
    So our multiplicative inverse is 11 mod 158 ≡ 11
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    511301780511010
    301785115929101
    51129171801-17
    29181111-1718
    181117-1718-35
    1171418-3553
    7413-3553-88
    431153-88141
    3130-88141-511
    So our multiplicative inverse is 141 mod 511 ≡ 141
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    191807380191010
    80738191422136101
    19113615501-1
    136552261-13
    552623-13-7
    263823-759
    3211-759-66
    212059-66191
    So our multiplicative inverse is -66 mod 191 ≡ 125
    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
    =  (138 × 97601 × 11 +
       341 × 30178 × 141 +
       73 × 80738 × 125)   mod 15420958
    = 7316328 (mod 15420958)


    So our answer is 7316328 (mod 15420958).


Verification

So we found that x ≡ 7316328
If this is correct, then the following statements (i.e. the original equations) are true:
353x (mod 158) ≡ 524 (mod 158)
34x (mod 511) ≡ 352 (mod 511)
277x (mod 191) ≡ 739 (mod 191)

Let's see whether that's indeed the case if we use x ≡ 7316328.