Chinese remainder theorem

func chinese_remainder(*n) {
    var N = n.prod
    func (*a) {
        n.range.sum { |i|
            var p = (N / n[i])
            a[i] * p.invmod(n[i]) * p
        } % N
    }
}
 
say chinese_remainder(3, 5, 7)(2, 3, 2)

Output:

23