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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
2383070238010
307238169101
2386933101-3
6931271-37
31743-37-31
73217-3169
3130-3169-238
So our multiplicative inverse is 69 mod 238 ≡ 69
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
719510120901-1
5102092921-13
20992225-13-7
92253173-724
251718-724-31
1782124-3186
8180-3186-719
So our multiplicative inverse is 86 mod 719 ≡ 86
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
7018020701010
8027011101101
70110169501-6
10195161-67
956155-67-111
65117-111118
5150-111118-701
So our multiplicative inverse is 118 mod 701 ≡ 118
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 ≡ 714 × 307-1 (mod 238) ≡ 714 × 69 (mod 238) ≡ 0 (mod 238)
x ≡ 387 × 510-1 (mod 719) ≡ 387 × 86 (mod 719) ≡ 208 (mod 719)
x ≡ 596 × 802-1 (mod 701) ≡ 596 × 118 (mod 701) ≡ 228 (mod 701)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 238 × 719 × 701 = 119956522
  2. We calculate the numbers M1 to M3
    M1=M/m1=119956522/238=504019,   M2=M/m2=119956522/719=166838,   M3=M/m3=119956522/701=171122
  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
    2385040190238010
    5040192382117173101
    23817316501-1
    173652431-13
    6543122-13-4
    43221213-47
    222111-47-11
    2112107-11238
    So our multiplicative inverse is -11 mod 238 ≡ 227
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    7191668380719010
    16683871923230101
    71930232901-23
    3029111-2324
    291290-2324-719
    So our multiplicative inverse is 24 mod 719 ≡ 24
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    7011711220701010
    17112270124478101
    7017887701-8
    7877111-89
    771770-89-701
    So our multiplicative inverse is 9 mod 701 ≡ 9
    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
    =  (0 × 504019 × 227 +
       208 × 166838 × 24 +
       228 × 171122 × 9)   mod 119956522
    = 104388942 (mod 119956522)


    So our answer is 104388942 (mod 119956522).


Verification

So we found that x ≡ 104388942
If this is correct, then the following statements (i.e. the original equations) are true:
307x (mod 238) ≡ 714 (mod 238)
510x (mod 719) ≡ 387 (mod 719)
802x (mod 701) ≡ 596 (mod 701)

Let's see whether that's indeed the case if we use x ≡ 104388942.