# Problem 552: Chinese leftovers II
Let An be the smallest positive integer satisfying An mod pi = i for all
1 ≤ i ≤ n, where pi is the i-th prime. For example A2 = 5, since this is
the smallest positive solution of the system of equations A2 mod 2 = 1
A2 mod 3 = 2 The system of equations for A3 adds another constraint.
That is, A3 is the smallest positive solution of A3 mod 2 = 1 A3 mod 3 =
2 A3 mod 5 = 3 and hence A3 = 23. Similarly, one gets A4 = 53 and A5 =
1523. Let S(n) be the sum of all primes up to n that divide at least one
element in the sequence A. For example, S(50) = 69 = 5 + 23 + 41, since
5 divides A2, 23 divides A3 and 41 divides A10 = 5765999453. No other
prime number up to 50 divides an element in A. Find S(300000).