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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
3958380395010
838395248101
3954881101-8
4811441-833
11423-833-74
431133-74107
3130-74107-395
So our multiplicative inverse is 107 mod 395 ≡ 107
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2638260263010
826263337101
263377401-7
374911-764
4140-764-263
So our multiplicative inverse is 64 mod 263 ≡ 64
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
163812101-2
8118101-2163
So our multiplicative inverse is -2 mod 163 ≡ 161
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 ≡ 521 × 838-1 (mod 395) ≡ 521 × 107 (mod 395) ≡ 52 (mod 395)
x ≡ 719 × 826-1 (mod 263) ≡ 719 × 64 (mod 263) ≡ 254 (mod 263)
x ≡ 519 × 81-1 (mod 163) ≡ 519 × 161 (mod 163) ≡ 103 (mod 163)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 395 × 263 × 163 = 16933255
  2. We calculate the numbers M1 to M3
    M1=M/m1=16933255/395=42869,   M2=M/m2=16933255/263=64385,   M3=M/m3=16933255/163=103885
  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
    395428690395010
    42869395108209101
    395209118601-1
    2091861231-12
    1862382-12-17
    2321112-17189
    2120-17189-395
    So our multiplicative inverse is 189 mod 395 ≡ 189
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    263643850263010
    64385263244213101
    26321315001-1
    213504131-15
    5013311-15-16
    1311125-1621
    11251-1621-121
    212021-121263
    So our multiplicative inverse is -121 mod 263 ≡ 142
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1631038850163010
    10388516363754101
    163543101-3
    5415401-3163
    So our multiplicative inverse is -3 mod 163 ≡ 160
    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
    =  (52 × 42869 × 189 +
       254 × 64385 × 142 +
       103 × 103885 × 160)   mod 16933255
    = 2133447 (mod 16933255)


    So our answer is 2133447 (mod 16933255).


Verification

So we found that x ≡ 2133447
If this is correct, then the following statements (i.e. the original equations) are true:
838x (mod 395) ≡ 521 (mod 395)
826x (mod 263) ≡ 719 (mod 263)
81x (mod 163) ≡ 519 (mod 163)

Let's see whether that's indeed the case if we use x ≡ 2133447.