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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
647273210101-2
2731012711-25
10171130-25-7
71302115-719
301128-719-45
1181319-4564
8322-4564-173
321164-173237
2120-173237-647
So our multiplicative inverse is 237 mod 647 ≡ 237
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1679650167010
9651675130101
16713013701-1
130373191-14
3719118-14-5
1918114-59
181180-59-167
So our multiplicative inverse is 9 mod 167 ≡ 9
Source: ExtendedEuclideanAlgorithm.com

nbqr t1t2t3
1739817501-1
98751231-12
752336-12-7
236352-723
6511-723-30
515023-30173
So our multiplicative inverse is -30 mod 173 ≡ 143
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 ≡ 510 × 273-1 (mod 647) ≡ 510 × 237 (mod 647) ≡ 528 (mod 647)
x ≡ 199 × 965-1 (mod 167) ≡ 199 × 9 (mod 167) ≡ 121 (mod 167)
x ≡ 300 × 98-1 (mod 173) ≡ 300 × 143 (mod 173) ≡ 169 (mod 173)


Now the actual calculation

  1. Find the common modulus M
    M = m1 × m2 × ... × mk = 647 × 167 × 173 = 18692477
  2. We calculate the numbers M1 to M3
    M1=M/m1=18692477/647=28891,   M2=M/m2=18692477/167=111931,   M3=M/m3=18692477/173=108049
  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
    647288910647010
    2889164744423101
    647423122401-1
    42322411991-12
    224199125-12-3
    199257242-323
    252411-323-26
    24124023-26647
    So our multiplicative inverse is -26 mod 647 ≡ 621
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1671119310167010
    11193116767041101
    167414301-4
    4131321-453
    3211-453-57
    212053-57167
    So our multiplicative inverse is -57 mod 167 ≡ 110
    Source: ExtendedEuclideanAlgorithm.com

    nbqr t1t2t3
    1731080490173010
    10804917362497101
    1739717601-1
    97761211-12
    7621313-12-7
    2113182-79
    13815-79-16
    85139-1625
    5312-1625-41
    321125-4166
    2120-4166-173
    So our multiplicative inverse is 66 mod 173 ≡ 66
    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
    =  (528 × 28891 × 621 +
       121 × 111931 × 110 +
       169 × 108049 × 66)   mod 18692477
    = 17882314 (mod 18692477)


    So our answer is 17882314 (mod 18692477).


Verification

So we found that x ≡ 17882314
If this is correct, then the following statements (i.e. the original equations) are true:
273x (mod 647) ≡ 510 (mod 647)
965x (mod 167) ≡ 199 (mod 167)
98x (mod 173) ≡ 300 (mod 173)

Let's see whether that's indeed the case if we use x ≡ 17882314.