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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
417154210901-2
1541091451-23
10945219-23-8
4519273-819
19725-819-46
751219-4665
5221-4665-176
212065-176417
So our multiplicative inverse is -176 mod 417 ≡ 241
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
55727021701-2
2701715151-231
171512-231-33
1527131-33262
2120-33262-557
So our multiplicative inverse is 262 mod 557 ≡ 262
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
29026312701-1
263279201-110
272017-110-11
2072610-1132
7611-1132-43
616032-43290
So our multiplicative inverse is -43 mod 290 ≡ 247
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 ≡ 268 × 154-1 (mod 417) ≡ 268 × 241 (mod 417) ≡ 370 (mod 417)
x ≡ 741 × 270-1 (mod 557) ≡ 741 × 262 (mod 557) ≡ 306 (mod 557)
x ≡ 344 × 263-1 (mod 290) ≡ 344 × 247 (mod 290) ≡ 288 (mod 290)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 417 × 557 × 290 = 67358010
  2. We calculate the numbers M1 to M3
    M1=M/m1=67358010/417=161530,   M2=M/m2=67358010/557=120930,   M3=M/m3=67358010/290=232269
  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
    4171615300417010
    161530417387151101
    417151211501-2
    1511151361-23
    1153637-23-11
    367513-1158
    7170-1158-417
    So our multiplicative inverse is 58 mod 417 ≡ 58
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    5571209300557010
    12093055721761101
    557619801-9
    618751-964
    8513-964-73
    531264-73137
    3211-73137-210
    2120137-210557
    So our multiplicative inverse is -210 mod 557 ≡ 347
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    2902322690290010
    232269290800269101
    29026912101-1
    2692112171-113
    211714-113-14
    1744113-1469
    4140-1469-290
    So our multiplicative inverse is 69 mod 290 ≡ 69
    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
    =  (370 × 161530 × 58 +
       306 × 120930 × 347 +
       288 × 232269 × 69)   mod 67358010
    = 41689528 (mod 67358010)


    So our answer is 41689528 (mod 67358010).


Verification

So we found that x ≡ 41689528
If this is correct, then the following statements (i.e. the original equations) are true:
154x (mod 417) ≡ 268 (mod 417)
270x (mod 557) ≡ 741 (mod 557)
263x (mod 290) ≡ 344 (mod 290)

Let's see whether that's indeed the case if we use x ≡ 41689528.