In modular arithmetic, the method of successive substitution is a method of solving problems of simultaneous congruences by using the definition of the congruence equation.

For example, consider the simple set of simultaneous congruences

x ≡ 3 (mod 4)

x ≡ 5 (mod 6)

Now, for x ≡ 3 (mod 4) to be true, x=3+4j for some integer j. Substitute this in the second equation

3+4j ≡ 5 (mod 6)

since we are looking for a solution to both equations.

Subtract 3 from both sides (this is permitted in modular arithmetic)

4j ≡ 2 (mod 6)

We simplify be dividing by the greatest common divisor of 4,2 and 6. Division by 2 yields:

2j ≡ 1 (mod 3)

The Euclidean multiplicative inverse of 2 mod 3 is 2. After multiplying both sides with the inverse, we obtain:

j ≡ 2 × 1 (mod 3)

or

j ≡ 2 (mod 3)

For the above to be true: j=2+3k for some integer k. Now substitute back into 3+4j and we obtain

x=3+4(2+3k)

Expand out

x=11+12k

to obtain the solution

x ≡ 11 (mod 12)

In general:

• write the first equation in its equivalent form
• substitute it into the next
  • simplify, use the multiplicative inverse if necessary
• continue until the last equation
• back substitute, then simplify
• rewrite back in the congruence form

If the moduli are coprime, the chinese remainder theorem gives a straightforward formula to obtain the solution.

See also

• simultaneous equations