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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
439762501-62
75121-6263
5221-6263-188
212063-188439
So our multiplicative inverse is -188 mod 439 ≡ 251
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3497300349010
730349232101
34932102901-10
3229131-1011
29392-1011-109
321111-109120
2120-109120-349
So our multiplicative inverse is 120 mod 349 ≡ 120
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
8639540863010
954863191101
8639194401-9
9144231-919
443142-919-275
321119-275294
2120-275294-863
So our multiplicative inverse is 294 mod 863 ≡ 294
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 ≡ 880 × 7-1 (mod 439) ≡ 880 × 251 (mod 439) ≡ 63 (mod 439)
x ≡ 516 × 730-1 (mod 349) ≡ 516 × 120 (mod 349) ≡ 147 (mod 349)
x ≡ 302 × 954-1 (mod 863) ≡ 302 × 294 (mod 863) ≡ 762 (mod 863)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 439 × 349 × 863 = 132221093
  2. We calculate the numbers M1 to M3
    M1=M/m1=132221093/439=301187,   M2=M/m2=132221093/349=378857,   M3=M/m3=132221093/863=153211
  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
    4393011870439010
    30118743968633101
    43933131001-13
    3310331-1340
    10331-1340-133
    313040-133439
    So our multiplicative inverse is -133 mod 439 ≡ 306
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    3493788570349010
    3788573491085192101
    349192115701-1
    1921571351-12
    15735417-12-9
    3517212-920
    171170-920-349
    So our multiplicative inverse is 20 mod 349 ≡ 20
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    8631532110863010
    153211863177460101
    863460140301-1
    4604031571-12
    4035774-12-15
    5741412-15212
    4140-15212-863
    So our multiplicative inverse is 212 mod 863 ≡ 212
    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
    =  (63 × 301187 × 306 +
       147 × 378857 × 20 +
       762 × 153211 × 212)   mod 132221093
    = 69599123 (mod 132221093)


    So our answer is 69599123 (mod 132221093).


Verification

So we found that x ≡ 69599123
If this is correct, then the following statements (i.e. the original equations) are true:
7x (mod 439) ≡ 880 (mod 439)
730x (mod 349) ≡ 516 (mod 349)
954x (mod 863) ≡ 302 (mod 863)

Let's see whether that's indeed the case if we use x ≡ 69599123.