`solve-chinese``as``ns`procedure- as
- (list-of integer)
- ns
- (list-of integer)

Given a length-k list of integers as and a length-k list of coprime moduli ns, (solve-chinese as ns) returns the least natural number x that is a solution to the equations

x = a₁ (mod n₁) ... x = aₖ (mod nₖ)

The solution

`x`is less than`(* n1 ... nk)`.The moduli

`ns`must all be positive.What is the least number

`x`that when divided by 3 leaves a remainder of 2, when divided by 5 leaves a remainder of 3, and when divided by 7 leaves a remainder of 2?> (solve-chinese '(2 3 2) '(3 5 7)) 23

Wikipedia: Chinese Remainder Theorem