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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
659359130001-1
3593001591-12
3005955-12-11
5951142-11123
5411-11123-134
4140123-134659
So our multiplicative inverse is -134 mod 659 ≡ 525
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2177620217010
7622173111101
217111110601-1
111106151-12
1065211-12-43
51502-43217
So our multiplicative inverse is -43 mod 217 ≡ 174
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
551222210701-2
222107281-25
1078133-25-67
83225-67139
3211-67139-206
2120139-206551
So our multiplicative inverse is -206 mod 551 ≡ 345
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 ≡ 49 × 359-1 (mod 659) ≡ 49 × 525 (mod 659) ≡ 24 (mod 659)
x ≡ 616 × 762-1 (mod 217) ≡ 616 × 174 (mod 217) ≡ 203 (mod 217)
x ≡ 178 × 222-1 (mod 551) ≡ 178 × 345 (mod 551) ≡ 249 (mod 551)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 659 × 217 × 551 = 78794653
  2. We calculate the numbers M1 to M3
    M1=M/m1=78794653/659=119567,   M2=M/m2=78794653/217=363109,   M3=M/m3=78794653/551=143003
  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
    6591195670659010
    119567659181288101
    65928828301-2
    288833391-27
    833925-27-16
    395747-16119
    5411-16119-135
    4140119-135659
    So our multiplicative inverse is -135 mod 659 ≡ 524
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    2173631090217010
    363109217167368101
    2176831301-3
    6813531-316
    13341-316-67
    313016-67217
    So our multiplicative inverse is -67 mod 217 ≡ 150
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5511430030551010
    143003551259294101
    551294125701-1
    2942571371-12
    25737635-12-13
    3735122-1315
    352171-1315-268
    212015-268551
    So our multiplicative inverse is -268 mod 551 ≡ 283
    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
    =  (24 × 119567 × 524 +
       203 × 363109 × 150 +
       249 × 143003 × 283)   mod 78794653
    = 23270632 (mod 78794653)


    So our answer is 23270632 (mod 78794653).


Verification

So we found that x ≡ 23270632
If this is correct, then the following statements (i.e. the original equations) are true:
359x (mod 659) ≡ 49 (mod 659)
762x (mod 217) ≡ 616 (mod 217)
222x (mod 551) ≡ 178 (mod 551)

Let's see whether that's indeed the case if we use x ≡ 23270632.