Chinese leftovers II
Let An be the smallest positive integer satisfying An mod pi = i for all 1 ≤ i ≤ n, where pi is the
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.