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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
8038280803010
828803125101
8032532301-32
253811-32257
3130-32257-803
So our multiplicative inverse is 257 mod 803 ≡ 257
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1316490131010
6491314125101
1311251601-1
12562051-121
6511-121-22
515021-22131
So our multiplicative inverse is -22 mod 131 ≡ 109
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1571720157010
172157115101
1571510701-10
157211-1021
7170-1021-157
So our multiplicative inverse is 21 mod 157 ≡ 21
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 ≡ 871 × 828-1 (mod 803) ≡ 871 × 257 (mod 803) ≡ 613 (mod 803)
x ≡ 245 × 649-1 (mod 131) ≡ 245 × 109 (mod 131) ≡ 112 (mod 131)
x ≡ 321 × 172-1 (mod 157) ≡ 321 × 21 (mod 157) ≡ 147 (mod 157)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 803 × 131 × 157 = 16515301
  2. We calculate the numbers M1 to M3
    M1=M/m1=16515301/803=20567,   M2=M/m2=16515301/131=126071,   M3=M/m3=16515301/157=105193
  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
    803205670803010
    2056780325492101
    803492131101-1
    49231111811-12
    3111811130-12-3
    1811301512-35
    13051228-35-13
    51281235-1318
    282315-1318-31
    2354318-31142
    5312-31142-173
    3211142-173315
    2120-173315-803
    So our multiplicative inverse is 315 mod 803 ≡ 315
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1311260710131010
    12607113196249101
    1314923301-2
    49331161-23
    331621-23-8
    1611603-8131
    So our multiplicative inverse is -8 mod 131 ≡ 123
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1571051930157010
    1051931576703101
    157352101-52
    31301-52157
    So our multiplicative inverse is -52 mod 157 ≡ 105
    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
    =  (613 × 20567 × 315 +
       112 × 126071 × 123 +
       147 × 105193 × 105)   mod 16515301
    = 15514573 (mod 16515301)


    So our answer is 15514573 (mod 16515301).


Verification

So we found that x ≡ 15514573
If this is correct, then the following statements (i.e. the original equations) are true:
828x (mod 803) ≡ 871 (mod 803)
649x (mod 131) ≡ 245 (mod 131)
172x (mod 157) ≡ 321 (mod 157)

Let's see whether that's indeed the case if we use x ≡ 15514573.