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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
2771027701-27
107131-2728
7321-2728-83
313028-83277
So our multiplicative inverse is -83 mod 277 ≡ 194
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1033400103010
340103331101
1033131001-3
3110311-310
101100-310-103
So our multiplicative inverse is 10 mod 103 ≡ 10
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1695760169010
576169369101
1696923101-2
6931271-25
31743-25-22
73215-2249
3130-2249-169
So our multiplicative inverse is 49 mod 169 ≡ 49
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 ≡ 603 × 10-1 (mod 277) ≡ 603 × 194 (mod 277) ≡ 88 (mod 277)
x ≡ 309 × 340-1 (mod 103) ≡ 309 × 10 (mod 103) ≡ 0 (mod 103)
x ≡ 458 × 576-1 (mod 169) ≡ 458 × 49 (mod 169) ≡ 134 (mod 169)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 277 × 103 × 169 = 4821739
  2. We calculate the numbers M1 to M3
    M1=M/m1=4821739/277=17407,   M2=M/m2=4821739/103=46813,   M3=M/m3=4821739/169=28531
  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
    277174070277010
    1740727762233101
    27723314401-1
    233445131-16
    441335-16-19
    135236-1944
    5312-1944-63
    321144-63107
    2120-63107-277
    So our multiplicative inverse is 107 mod 277 ≡ 107
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    103468130103010
    4681310345451101
    103512101-2
    5115101-2103
    So our multiplicative inverse is -2 mod 103 ≡ 101
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    169285310169010
    28531169168139101
    16913913001-1
    139304191-15
    3019111-15-6
    1911185-611
    11813-611-17
    832211-1745
    3211-1745-62
    212045-62169
    So our multiplicative inverse is -62 mod 169 ≡ 107
    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
    =  (88 × 17407 × 107 +
       0 × 46813 × 101 +
       134 × 28531 × 107)   mod 4821739
    = 4016588 (mod 4821739)


    So our answer is 4016588 (mod 4821739).


Verification

So we found that x ≡ 4016588
If this is correct, then the following statements (i.e. the original equations) are true:
10x (mod 277) ≡ 603 (mod 277)
340x (mod 103) ≡ 309 (mod 103)
576x (mod 169) ≡ 458 (mod 169)

Let's see whether that's indeed the case if we use x ≡ 4016588.