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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
2718220271010
82227139101
271930101-30
91901-30271
So our multiplicative inverse is -30 mod 271 ≡ 241
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
743305213301-2
3051332391-25
13339316-25-17
3916275-1739
16722-1739-95
723139-95324
2120-95324-743
So our multiplicative inverse is 324 mod 743 ≡ 324
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2158790215010
879215419101
2151911601-11
196311-1134
6160-1134-215
So our multiplicative inverse is 34 mod 215 ≡ 34
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 ≡ 209 × 822-1 (mod 271) ≡ 209 × 241 (mod 271) ≡ 234 (mod 271)
x ≡ 606 × 305-1 (mod 743) ≡ 606 × 324 (mod 743) ≡ 192 (mod 743)
x ≡ 742 × 879-1 (mod 215) ≡ 742 × 34 (mod 215) ≡ 73 (mod 215)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 271 × 743 × 215 = 43290895
  2. We calculate the numbers M1 to M3
    M1=M/m1=43290895/271=159745,   M2=M/m2=43290895/743=58265,   M3=M/m3=43290895/215=201353
  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
    2711597450271010
    159745271589126101
    27112621901-2
    126196121-213
    191217-213-15
    1271513-1528
    7512-1528-43
    522128-43114
    2120-43114-271
    So our multiplicative inverse is 114 mod 271 ≡ 114
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    743582650743010
    5826574378311101
    743311212101-2
    3111212691-25
    12169152-25-7
    69521175-712
    521731-712-43
    17117012-43743
    So our multiplicative inverse is -43 mod 743 ≡ 700
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    2152013530215010
    201353215936113101
    215113110201-1
    1131021111-12
    1021193-12-19
    113322-1959
    3211-1959-78
    212059-78215
    So our multiplicative inverse is -78 mod 215 ≡ 137
    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
    =  (234 × 159745 × 114 +
       192 × 58265 × 700 +
       73 × 201353 × 137)   mod 43290895
    = 36364098 (mod 43290895)


    So our answer is 36364098 (mod 43290895).


Verification

So we found that x ≡ 36364098
If this is correct, then the following statements (i.e. the original equations) are true:
822x (mod 271) ≡ 209 (mod 271)
305x (mod 743) ≡ 606 (mod 743)
879x (mod 215) ≡ 742 (mod 215)

Let's see whether that's indeed the case if we use x ≡ 36364098.