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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
41243041010
24341538101
41381301-1
3831221-113
3211-113-14
212013-1441
So our multiplicative inverse is -14 mod 41 ≡ 27
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1892571401-7
25141111-78
141113-78-15
113328-1553
3211-1553-68
212053-68189
So our multiplicative inverse is -68 mod 189 ≡ 121
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
66130125901-2
30159561-211
59695-211-101
651111-101112
5150-101112-661
So our multiplicative inverse is 112 mod 661 ≡ 112
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 ≡ 139 × 243-1 (mod 41) ≡ 139 × 27 (mod 41) ≡ 22 (mod 41)
x ≡ 565 × 25-1 (mod 189) ≡ 565 × 121 (mod 189) ≡ 136 (mod 189)
x ≡ 381 × 301-1 (mod 661) ≡ 381 × 112 (mod 661) ≡ 368 (mod 661)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 41 × 189 × 661 = 5122089
  2. We calculate the numbers M1 to M3
    M1=M/m1=5122089/41=124929,   M2=M/m2=5122089/189=27101,   M3=M/m3=5122089/661=7749
  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
    41124929041010
    1249294130472101
    41220101-20
    21201-2041
    So our multiplicative inverse is -20 mod 41 ≡ 21
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    189271010189010
    2710118914374101
    1897424101-2
    74411331-23
    413318-23-5
    338413-523
    8180-523-189
    So our multiplicative inverse is 23 mod 189 ≡ 23
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    66177490661010
    774966111478101
    661478118301-1
    47818321121-13
    183112171-13-4
    112711413-47
    7141130-47-11
    41301117-1118
    301128-1118-47
    1181318-4765
    8322-4765-177
    321165-177242
    2120-177242-661
    So our multiplicative inverse is 242 mod 661 ≡ 242
    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
    =  (22 × 124929 × 21 +
       136 × 27101 × 23 +
       368 × 7749 × 242)   mod 5122089
    = 2805652 (mod 5122089)


    So our answer is 2805652 (mod 5122089).


Verification

So we found that x ≡ 2805652
If this is correct, then the following statements (i.e. the original equations) are true:
243x (mod 41) ≡ 139 (mod 41)
25x (mod 189) ≡ 565 (mod 189)
301x (mod 661) ≡ 381 (mod 661)

Let's see whether that's indeed the case if we use x ≡ 2805652.