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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
1512730151010
2731511122101
15112212901-1
12229461-15
29645-15-21
65115-2126
5150-2126-151
So our multiplicative inverse is 26 mod 151 ≡ 26
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
685277213101-2
2771312151-25
13115811-25-42
1511145-4247
11423-4247-136
431147-136183
3130-136183-685
So our multiplicative inverse is 183 mod 685 ≡ 183
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
484697101-7
6916901-7484
So our multiplicative inverse is -7 mod 484 ≡ 477
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 ≡ 166 × 273-1 (mod 151) ≡ 166 × 26 (mod 151) ≡ 88 (mod 151)
x ≡ 881 × 277-1 (mod 685) ≡ 881 × 183 (mod 685) ≡ 248 (mod 685)
x ≡ 112 × 69-1 (mod 484) ≡ 112 × 477 (mod 484) ≡ 184 (mod 484)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 151 × 685 × 484 = 50062540
  2. We calculate the numbers M1 to M3
    M1=M/m1=50062540/151=331540,   M2=M/m2=50062540/685=73084,   M3=M/m3=50062540/484=103435
  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
    1513315400151010
    331540151219595101
    1519515601-1
    95561391-12
    5639117-12-3
    3917252-38
    17532-38-27
    52218-2762
    2120-2762-151
    So our multiplicative inverse is 62 mod 151 ≡ 62
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    685730840685010
    73084685106474101
    685474121101-1
    4742112521-13
    2115243-13-13
    5231713-13224
    3130-13224-685
    So our multiplicative inverse is 224 mod 685 ≡ 224
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4841034350484010
    103435484213343101
    484343114101-1
    3431412611-13
    14161219-13-7
    6119343-724
    19443-724-103
    431124-103127
    3130-103127-484
    So our multiplicative inverse is 127 mod 484 ≡ 127
    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 × 331540 × 62 +
       248 × 73084 × 224 +
       184 × 103435 × 127)   mod 50062540
    = 25594588 (mod 50062540)


    So our answer is 25594588 (mod 50062540).


Verification

So we found that x ≡ 25594588
If this is correct, then the following statements (i.e. the original equations) are true:
273x (mod 151) ≡ 166 (mod 151)
277x (mod 685) ≡ 881 (mod 685)
69x (mod 484) ≡ 112 (mod 484)

Let's see whether that's indeed the case if we use x ≡ 25594588.