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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!

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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
37929418501-1
294853391-14
853927-14-9
397544-949
7413-949-58
431149-58107
3130-58107-379
So our multiplicative inverse is 107 mod 379 ≡ 107
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
99118258101-5
182812201-511
812041-511-49
20120011-49991
So our multiplicative inverse is -49 mod 991 ≡ 942
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1522390152010
239152187101
1528716501-1
87651221-12
6522221-12-5
2221112-57
211210-57-152
So our multiplicative inverse is 7 mod 152 ≡ 7
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 ≡ 446 × 294-1 (mod 379) ≡ 446 × 107 (mod 379) ≡ 347 (mod 379)
x ≡ 815 × 182-1 (mod 991) ≡ 815 × 942 (mod 991) ≡ 696 (mod 991)
x ≡ 450 × 239-1 (mod 152) ≡ 450 × 7 (mod 152) ≡ 110 (mod 152)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 379 × 991 × 152 = 57089528
  2. We calculate the numbers M1 to M3
    M1=M/m1=57089528/379=150632,   M2=M/m2=57089528/991=57608,   M3=M/m3=57089528/152=375589
  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
    3791506320379010
    150632379397169101
    37916924101-2
    16941451-29
    41581-29-74
    51509-74379
    So our multiplicative inverse is -74 mod 379 ≡ 305
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    991576080991010
    5760899158130101
    99113078101-7
    130811491-78
    8149132-78-15
    49321178-1523
    3217115-1523-38
    17151223-3861
    15271-3861-465
    212061-465991
    So our multiplicative inverse is -465 mod 991 ≡ 526
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1523755890152010
    3755891522470149101
    1521491301-1
    14934921-150
    3211-150-51
    212050-51152
    So our multiplicative inverse is -51 mod 152 ≡ 101
    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
    =  (347 × 150632 × 305 +
       696 × 57608 × 526 +
       110 × 375589 × 101)   mod 57089528
    = 43440190 (mod 57089528)


    So our answer is 43440190 (mod 57089528).


Verification

So we found that x ≡ 43440190
If this is correct, then the following statements (i.e. the original equations) are true:
294x (mod 379) ≡ 446 (mod 379)
182x (mod 991) ≡ 815 (mod 991)
239x (mod 152) ≡ 450 (mod 152)

Let's see whether that's indeed the case if we use x ≡ 43440190.