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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
1014240101010
424101420101
101205101-5
2012001-5101
So our multiplicative inverse is -5 mod 101 ≡ 96
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
3777000377010
7003771323101
37732315401-1
323545531-16
545311-16-7
5315306-7377
So our multiplicative inverse is -7 mod 377 ≡ 370
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
2396360239010
6362392158101
23915818101-1
158811771-12
817714-12-3
7741912-359
4140-359-239
So our multiplicative inverse is 59 mod 239 ≡ 59
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 ≡ 497 × 424-1 (mod 101) ≡ 497 × 96 (mod 101) ≡ 40 (mod 101)
x ≡ 672 × 700-1 (mod 377) ≡ 672 × 370 (mod 377) ≡ 197 (mod 377)
x ≡ 671 × 636-1 (mod 239) ≡ 671 × 59 (mod 239) ≡ 154 (mod 239)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 101 × 377 × 239 = 9100403
  2. We calculate the numbers M1 to M3
    M1=M/m1=9100403/101=90103,   M2=M/m2=9100403/377=24139,   M3=M/m3=9100403/239=38077
  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
    101901030101010
    9010310189211101
    101119201-9
    112511-946
    2120-946-101
    So our multiplicative inverse is 46 mod 101 ≡ 46
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    377241390377010
    241393776411101
    3771134301-34
    113321-34103
    3211-34103-137
    2120103-137377
    So our multiplicative inverse is -137 mod 377 ≡ 240
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    239380770239010
    3807723915976101
    2397631101-3
    76116101-319
    111011-319-22
    10110019-22239
    So our multiplicative inverse is -22 mod 239 ≡ 217
    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
    =  (40 × 90103 × 46 +
       197 × 24139 × 240 +
       154 × 38077 × 217)   mod 9100403
    = 4124577 (mod 9100403)


    So our answer is 4124577 (mod 9100403).


Verification

So we found that x ≡ 4124577
If this is correct, then the following statements (i.e. the original equations) are true:
424x (mod 101) ≡ 497 (mod 101)
700x (mod 377) ≡ 672 (mod 377)
636x (mod 239) ≡ 671 (mod 239)

Let's see whether that's indeed the case if we use x ≡ 4124577.