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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
2507230250010
7232502223101
25022312701-1
22327871-19
27736-19-28
76119-2837
6160-2837-250
So our multiplicative inverse is 37 mod 250 ≡ 37
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
49110148701-4
101871141-45
871463-45-34
143425-34141
3211-34141-175
2120141-175491
So our multiplicative inverse is -175 mod 491 ≡ 316
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
941401213901-2
40113921231-25
139123116-25-7
123167115-754
161115-754-61
1152154-61176
5150-61176-941
So our multiplicative inverse is 176 mod 941 ≡ 176
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 ≡ 231 × 723-1 (mod 250) ≡ 231 × 37 (mod 250) ≡ 47 (mod 250)
x ≡ 292 × 101-1 (mod 491) ≡ 292 × 316 (mod 491) ≡ 455 (mod 491)
x ≡ 386 × 401-1 (mod 941) ≡ 386 × 176 (mod 941) ≡ 184 (mod 941)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 250 × 491 × 941 = 115507750
  2. We calculate the numbers M1 to M3
    M1=M/m1=115507750/250=462031,   M2=M/m2=115507750/491=235250,   M3=M/m3=115507750/941=122750
  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
    2504620310250010
    462031250184831101
    250318201-8
    3121511-8121
    2120-8121-250
    So our multiplicative inverse is 121 mod 250 ≡ 121
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    4912352500491010
    23525049147961101
    491618301-8
    6132011-8161
    3130-8161-491
    So our multiplicative inverse is 161 mod 491 ≡ 161
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    9411227500941010
    122750941130420101
    941420210101-2
    4201014161-29
    1011665-29-56
    165319-56177
    5150-56177-941
    So our multiplicative inverse is 177 mod 941 ≡ 177
    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
    =  (47 × 462031 × 121 +
       455 × 235250 × 161 +
       184 × 122750 × 177)   mod 115507750
    = 63934547 (mod 115507750)


    So our answer is 63934547 (mod 115507750).


Verification

So we found that x ≡ 63934547
If this is correct, then the following statements (i.e. the original equations) are true:
723x (mod 250) ≡ 231 (mod 250)
101x (mod 491) ≡ 292 (mod 491)
401x (mod 941) ≡ 386 (mod 941)

Let's see whether that's indeed the case if we use x ≡ 63934547.